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Algebra I
Volume 1
Wickenburg Unified School District
Algebra I
Volume 1
Authors John Douthat, Ben Newman, Sheri Canham
License CC BY-NC-SA
Preface
This textbook is intended to be a living document. The movement towards more
rigor and depth of knowledge in secondary mathematics education requires that a
textbook grow and adapt to these changes. This is our first version of an Algebra I
textbook. We expect to update the text on a yearly basis to better reflect the content
and instructional practices of the Common Core and ACT Quality Core® . We also hope
to produce accompanying resources that include formative and summative assessments,
interactive online delivery, unwrapped documents, task analysis, recommended activities
and other supplementary materials.
This book was written in a style that we believe is student friendly. The text is
intended to speak to the student as a teacher holding a conversation. The voice and tone
of the text are focused on leading the student through the concepts and processes with
examples and detailed explanations. Many textbooks have a focus on the content;
however, we believe the focus should be on the student’s thought processes and the
connections needed for success in Algebra.
We would like to thank the Wickenburg Unified School District for funding this
effort and for the district’s commitment to quality education. We would like to thank in
particular Dr. Howard Carlson, Ranae Macias, and Lynn Greene for their contributions.
Table of Contents
Unit 1
The Tools of Mathematics
Section
Section
Section
Section
Section
Section
Section
Section
Section
Section
Section
Unit 2
Finding and Using Measures of Central Tendency
Creating Bar and Line Graphs, Frequency Tables & Histograms
Creating and Interpreting Circle Graphs, Stem and Leaf Plots
Interpreting Box and Whisker Plots
Creating and Interpreting Scatter Plots
Choosing a Graphic Representation
Using the Cartesian Coordinate System
Applying Properties of Real Numbers and Like Terms
Translating Words to Expressions
Converting Visual Patterns to Expressions
Converting Numeric Patterns to Expressions
Solving Linear Equations and Inequalities
Section
Section
Section
Section
Section
Section
Section
Section
Section
Unit 3
1
2
3
4
5
6
7
8
9
10
11
1
2
3
4
5
6
7
8
9
Solving One-Step Equations
Using Properties of Equality to Solve Two-Step Equations
Solving Equations Using Like Terms
Solving Equations Using the Distributive Property
Solving Equations with Variable Terms on Both sides
Solving Equations Containing Absolute Value Expressions
Solving and Applying Literal Equations
Solving and Graphing Linear Inequalities
Using Conjunctions and Disjunctions of Inequalities
Graphing Functions
Section
Section
Section
Section
Section
1
2
3
4
5
Defining and Recognizing Relations and Functions
Domains and Ranges of Functions and Relations
Creating Graphs and Using the Vertical Line Test
Graphing by Substitution
Using Set Builder Notation
Unit 3
Graphing Functions (continued)
Section
Section
Section
Section
Section
Unit 4
Defining and Using Slopes and Y-Intercepts
Using the Slope-Intercept form of a Linear Equation
Isolating y to Graph Linear Equations
Interpreting Slopes as Rates of Change
Recognizing and Finding Equivalent Forms of Representation
Writing and Applying Linear Functions
Section
Section
Section
Section
Section
Section
Section
Section
Section
Unit 5
6
7
8
9
10
1
2
3
4
5
6
7
8
9
Deriving and Using the Slope Formula
Writing Linear Equations Given Slope and Other Information
Writing Linear Equations Given Two Points
Converting between Forms of Linear Equations
Finding and Applying X and Y Intercepts
Solving Contextual Problems with Linear Equations
Writing and Applying ‘nth’ Terms for Arithmetic Sequences
Finding and Applying Lines of Best Fit
Defining and Applying Direct Variation
Systems of Equations and Inequalities
Section
Section
Section
Section
Section
Section
1
2
3
4
5
6
Section 7
Section 8
Section 9
Solving Systems of Equations Graphically
Identifying Parallel and Perpendicular Systems
Solving Systems by Substitution with an Isolated Variable
Solving Systems by Isolating a Variable and Substituting
Solving Systems by Elimination
Solving Systems by Elimination Using the Multiplicative Property
of Equality
Graphing Inequalities in Two Variables
Graphing Systems of Inequalities
Solving Contextual Problems with Systems of Equations
Unit 1
The Tools of Mathematics
Introduction
Consider the four car buyers shown below.
I’ll buy it!
My friends
will love it!
I’ll buy it!
It has good
power and
feel
Buyer A
I’ll buy it!
the color and
styling are
great
I’ll buy it! It
has good gas
mileage, a low
interest rate, great
safety numbers, and
low insurance
costs
Buyer C
Buyer B
Buyer D
There are many reasons to buy a car. Four different buyers shown above each have a
different reason. Buyer D is making the decision to purchase the car based on numbers.
Not all decisions must be based on numbers, however when we use numbers we often
will make good and probably the best of decisions. Mathematical reasoning provides us
with the tools to think, analyze, model, and predict. Good decisions in business and life
depend on our ability to think logically and globally. Algebra will help train our thought
processes in addition to being a tool we use to analyze the world around us.
Unit 1
Vocabulary and Concepts
Coefficient A coefficient is the number in front of a variable that we multiply by.
Histogram A histogram is a bar graph that shows the distribution of a data set.
Line of Best Fit
A line drawn through the middle of the points on a scatter
plot that matches the general direction of the points and has
about as many points above as below the line.
Mean
A measure of central tendency calculated by dividing the sum of the
values by the count of the values.
Median
A measure of central tendency calculated by putting the values in order
and finding the middle value.
Mode
A measure of central tendency calculated by finding the value that
occurs most often.
Property
Something that is true about operations no matter what values are used.
Range
The range measures the spread of the data. The range is calculated
as the high value minus the low value.
Key Concepts for the Tools of Algebra
Measures of Central Tendency: These measures tell us what most of set is like.
We use measures of central tendency to compare two sets or to compare
an element of the set to the rest of the set.
Graphs:
Graphs are a way to display data so that it can be understood and
interpreted visually and quickly.
Justification: The process of justifying answer or the steps to finding an answer by
using properties and logical deduction.
Combining Terms: Terms are combined by adding the coefficients of like terms.
Patterning: Visual and numeric patterns can be represented by algebraic expressions.
Unit 1 Section 1
Objectives

The student will explain why measures of central tendency are important.

The student will list and explain how to find measures of central tendency.

The student will find measures of central tendency and use them to make
decisions.

The student will predict how changes to a data set will affect the measures
of central tendency.

The student will determine when measures of central tendency are used
deceptively.
Statistical Analysis is one of the most powerful tools in mathematics. Measures of
central tendency are the basis for all statistical analysis. The three measures of
central tendency are the mean, the mode and the median. The mean is also called
the average.
Measures of central tendency will tell us what most of a set is like.
When we know measures of central tendency we can use comparisons that will help
us make decisions. If a model of car has an average resale value after 2 years
of $14000 and another model has an average resale value of $18000 we might want
to purchase the car that will be worth more in the future.
Measures of central tendency also make it possible to compare an individual element
of the set to the average. If the average cost of a certain model of used car is
$12000 and the seller is asking $10500 then we would consider this to be a good
price. However, if the seller were asking $14500 then we would know the price is too
high. If we did not know the average price we could not determine what we should
pay for the car.
In short, measures of central tendency allow us to compare between types of things
and between the elements of a set and the set as a whole.
We need to be able to calculate and compare the three measures of central tendency.
The mean or average is calculated by adding up all the values in the set and
then dividing by the count of the values in the set.
The median is the value in the middle of the ordered set of values. To find this
value we must put the numbers in order and then count to find the middle number.
The mode is the value that occurs most often. There can be more than one value
if two or more numbers occur with the same frequency.
Another statistical measure we will often use is the range of a data set. We find the
range of a data set by subtracting the low value in the set from the high value. The
range tells us how spread out the set may be or its distribution.
The examples below show us how to find the three measures of central tendency.
Example A
Given the set of quiz scores 9, 11, 13, 17, 12, 10, 9, 14, 9, 11, 15, 14, 12
For the mean we find the sum 9+11+13+17+12+10+9+14+9+11+15+14+12.
The sum is 156, then we count the values, there are 13 and we divide to find
the mean or average which is 12.
To find the median we must put the values in order
9, 9, 9, 10, 11, 11, 12, 12, 13, 14, 14, 15, 17
Once the values are in order the middle value is 12. There are 6 values on each
side of the circled value in the middle. So the median is 12. If the values are in
order it is easy to see the mode is 9 because it occurs three times. The range
of values is 17 – 9 which is 8.
As we can see in this example the three measures can be different or they can
be the same.
If we had another student take the quiz and score 12 the mode would change
to 9 and 12 because both would now occur three times. However the average
would not change since the total would be 168 and the number of values would
be 14 and 168 divided by 14 is 12. The median also would not change because
adding another value that is median will not change the median as shown
below.
9, 9, 9, 10, 11, 11, 12,
12, 12, 13, 14, 14, 15, 17
median
The twelve we added is in bold. Because we now have 14 scores the middle value
must have seven numbers on each side. The number between the two twelve’s
must be twelve so the median stayed the same.
Example B
An accident/injury law firm wants to create an advertisement to run on TV. The
firm’s last ten clients have won the following awards of money.
0, 1500, 2000, 1000, 0, 1000, 1000, 3000, 2000, 250000
Find the mean, median and mode for this set. To find the mean or average we
begin by finding the total of the values. The total is 261500. When now divide
by 10 to find the mean which is $26150 dollars. The value 1000 occurs three
times and so is mode. To find the median we put the values in order as shown
below.
0, 0, 1000, 1000, 1000, 1500, 2000, 2000, 3000, 250000
Since there are ten values the middle must have 5 on each side. so we separate
the values to find the one in the middle.
0, 0, 1000, 1000, 1000,
1500, 2000, 2000, 3000, 250000
median
The value in between 1000 and 1500 can be found by adding the two numbers
together and dividing by 2. So the median is 1250. The range for this data
set is 250000 – 0 which is 250000.
When the advertisement is filmed the law firm tells the public that their average
award they win for their clients is $26150. While this is true it is deceptive. The
mean or average here is not the best measure of central tendency. There is
an outlier in the data set that skews the mean. An outlier is a value that is
far away from the rest of the set. Because 250000 is so much larger than the
other values in the set it makes the mean a poor measure of central tendency.
Most of the firm’s clients will not receive amounts close to $26150. Most clients
should expect to receive about $1000 or $1250 dollars. The mode and the
median are better measures of central tendency.
If the firms next client receives an award of $1000 this will affect the both the
mean and the median but not the mode. The mean will now be calculated with
a total of $262500 (we added the new value of $1000) and there will now be
11 numbers. By dividing we get $23863.64 for the mean or average. We can
now put the numbers order and find the median as shown below.
0, 0, 1000, 1000, 1000, 1000, 1500, 2000, 2000, 3000, 250000
Because there are now eleven values there must be five on each side of the
median, the circled value, 1000 is the median. The mode will not change since
it was 1000 because it occurred three times and it stays the mode because the
1000 now occurs four times.
Exercises Unit 1 Section 1
1. What are the names for the three measures of central tendency.
2. Why are measures of central tendency important? ( What does a measure of
central tendency tell us? )
3. How do we find the mean or average for a data set?
4. How do we find the median for a data set?
5. How do we find the mode for a data set?
6. What is an "outlier"?
7. What measure of central tendency is affected by an outlier and how does this
affect the accuracy of this measure?
8. The financial management firm of Acme Investments has produced the following
gains for its clients over a one month period.
$100, $10, $40, $50, $40, $80, $40, $200, $60, and $24000
a. What is the mean for this set of data?
b. What is the mode for this set of data?
c. What is the median for this set of data?
d. When they tape the TV commercial they tell people what the average income
is and do not tell people the median or the mode. What do you think of this?
9. In problem number #8 what is the outlier?
10. What is another word for mean?
11. How do we find the range of a data set?
12. Find the mean, median, mode and range for the set given below.
16, 10, 18, 24, 8, 28, 23, 16, 17, 30, 18, 19, 40, 23, 36, 14
13. If we start with the data set
8, 8, 10, 12, 14, 20
a. Find the mean, median and mode.
b. If we add the value 12 to the data set find the mean, median and mode.
c. Explain the changes we see in part ‘b’.
d. Why did the mean not change in part 'b'?
14. If we start with the data set
7, 7, 9, 11, 13, 19
a. Find the mean, median and mode.
b. If we add the value 22 find the mean, median, and mode.
c. Explain the changes we see in part ‘b’.
d. Why did the mode not change in part 'b'?
15. If we start with the data set
9, 9, 11, 13, 15, 21
a. Find the mean, median and mode.
b. If we add two values of 13 find the mean, median and mode.
c. Explain the changes we see in part ‘b’.
d. Why did the mean not change in part 'b'?
Answer the following multiple choice questions
16. If we have a set of data with an average of 21 and a median of 19 and we add the
values 14 and 28 to the data set then
a.
b.
c.
d.
the mean will change
the median will change
the median and the mean will not change
the mean and the mode will stay the same
17. On his first 5 quizzes Sam scored 10, 14, 10, 12, 9. If Sam scores a 5 on his next
quiz then
a.
b.
c.
d.
The average will decrease and the median will also decrease
Only the average will decrease
Only the median will decrease
The average will increase and the median decrease
18. A survey of students and their tardies for their first hour classes produced the
following results: 0, 1, 3, 1, 2, 4, 6, 1, 2, 2, 0. If the next student surveyed
had 2 tardies then
a.
b.
c.
d.
The average will increase and the median will also increase
The mode will change and the median will decrease
Only the mode will change
The average will decrease and the mode and the median will also increase
19. Tom's set of test scores for first semester so far has been 86, 81, 79, 86, 91.
His sixth test score changed the mode but not the median. It could be
a.
b.
c.
d.
81
91
86
79
Unit 1 Section 2
Objectives

The student will interpret and create bar and line graphs.

The student will create and frequency tables and histograms.
When we need to analyze data sets the measures of central tendency are one important tool.
Another tool of great importance can be graphic display. One of the easiest ways to people
to interpret and use data is visually from graphs.
In order to create a graph we must have a set of data. Data sets are a group or collection
of ordered pairs. The data set below is the yearly enrollment for Wickenburg High from
2004 to 2012. It is ordered because the year is always on the left and the number of
students is always on the right.
{(04, 550), (05, 560), (06, 580), (07, 600), (08, 635), (09, 650), (10, 695), (11, 680) (12, 680)}
This data set could also have represented in a table form as given below.
Year
04
05
06
07
08
09
10
11
12
Enrollment
550
560
580
600
635
650
695
680
680
To construct a graph we must choose a title for the graph, scales for the each axis,
and labels for each axis. The horizontal axis should always have the values from the
left hand side of the ordered pairs which would be the top or left side of a table.
The vertical axis is the right hand values in the ordered pairs which is the bottom or
right hand side of a table. Below is an example of a bar graph created from this data.
Enrollment at WHS
# of
students
700
650
600
550
500
450
400
’04
’05
’06
’07
’08
Year
’09
’10
’11
'12
Choosing a scale for an axis means deciding what we will count by in order to label
the axis. The horizontal axis is easy because the values for the years are just counting
by ones. Finding the scale or interval on the vertical depends in part on the range
of the data. The range of the data is from 550 to 680. We should not start at 550
so we picked 400 as a reasonable first value. We must go up to 700 because we want
a value bigger than the highest number in the data set. The easiest way to count from
400 to 700 is by 50’s. In general we want between 5 and 10 numbers in our scales so
counting by 50 gives us seven numbers in the vertical axis.
A bar graph is a good choice for this type of graph because there are no data points
that could exist in between the given values.
Another type of graph that we will often see is a line graph. Line graphs should be used
when there are data points that exist in between the points given in the roster or table.
The data set below tracks the gas mileage we could expect from the average luxury car
in the year 2012. As the car’s speed increases the gas mileage changes. The data set
from which this graph was created is given on the left of the graph.
Speed MPG
Fuel Consumption for Luxury Cars in 2012
Average MPG 30
40
24
45
28
50
27
20
56
23.5
15
68
19
72
16.5
25
10
5
40
45 50 55 60 65 70 75
Speed in mph
For this data set we should use a line graph since we know a car will have other data
points besides those listed in the table. For instance from the graph we can estimate
that at 47 mph a luxury car would get 26 mpg.
The are some data sets that are very large and we often create categories for these sets
and then find the frequency ( how often ) data points occur for each category. When
we condense the data in this way we use a frequency table with tally marks that we
later will turn into a graph so that it can be interpreted easily.
An example data set, frequency table, and graph follow in the next example.
A survey of the ages of people who own tablet computers produced the following list
of ages.
21, 14, 12, 17, 18, 24, 23, 19, 32, 41, 31, 30 28, 27, 23, 16, 51, 18, 20, 18,
17, 22, 24, 23, 27, 29, 36, 37, 15, 28, 25, 42, 47, 38, 21, 20, 19, 13, 28, 33
The range of ages is from 13 to 51. When we subtract we find the answer is 38. If we
set up categories from 10 to 60 and count by tens we will have a reasonable distribution.
So our frequency table will have the age groups in the left column. We will put one
tally mark in the second column for each data element that fits into the category. After
we have completed the second column we will count to find the frequency with which
a category shows up in the data set.
Age Group
Tally
Frequency
10 – 19
12
20 – 29
17
30 – 39
7
40 – 49
3
50 – 59
1
We can now create a histogram. A histogram is a bar graph that show the distribution
of a set because the categories are arranged in a sequential order.
Number of People Owning Tablet Computers by Age
20
16
12
8
4
0
10-19
20-29
30-39
40-49
50-59
Exercises Unit 1 Section 2
1. Why are graphs a useful or important way of representing data?
2. What is the general process we go through to create a graph?
3. What is a data point?
Use the graph below to answer the questions that follow.
Monthly Sales for The Flagstaff Ski Shop
70,000
Amount
of
Sales
60,000
50,000
40,000
30,000
20,000
10,000
Jan
Feb
Mar
Apr
May
Jun
Jul
Aug
Sep
Oct
Nov
Dec
Month
4. What month had sales of
5. What were the sales for
a) $34000 ?
b) $20000 ?
c) $50000?
a) February? b) June?
6. Which month had the highest sales? Explain why this month was the best month
for sales?
7. Which month had the lowest sales?
8. Explain why the graph has a kind of ‘V’ shape? ( Why is the graph low in the middle
and high on the outsides? )
9. For what months should the store hire part time sales clerks? ( Explain your answer! )
10. What is the average sales per month for the year?
11. When should most of the employees take a vacation?
12. Given the data set below represents the average high temperature in Phoenix
for 9 months last year, construct a bar graph for the data.
Month
Average Temperature
Jan Feb Mar Apr May Jun Jul Aug Sept
68
74 81 85
93 105 108 103 100
Use the graph below to answer questions 13 and 14.
Wickenburg Widget Manufacturing Company
Six Month Income and Expenses
$30000
$25000
$20000
Income
$15000
Expenses
$10000
$5000
0
Jan
Feb
Mar
Apr
May
June
13. Based on the graph above the average profit per month is
a. 22000
b. 15500
c.
9500
d.
6500
14. Based on the graph above, the percent that profit would be of the income
is about
a. 30%
b. 42%
c. 70%
d. 142%
There are 6 Algebra I classes at WHS. The enrollment in each class is shown below.
# of students
35
30
25
20
15
10
5
0
Enrollment in Algebra I classes
5
1st
2nd
3rd
4th
5th
6th
Hour/period
15. In order to balance the size of classes 5 students are moved from 6th hour to 5th hour
and another 5 students from 6th hour go to 4th hour, then for the class sizes
a.
b.
c.
d.
the average will change and the mode will change
the mode and the median will change
the average and the median will change
only the mode will change
16. Construct a frequency chart in the space provided below. The data set represents
the ages of the viewers of MTV for a sample taken last week.
12, 21, 23, 18, 14, 17, 18, 32, 15, 42, 21, 22, 25, 18, 19, 17, 23, 28, 35, 16, 14
11, 24, 18, 14, 17, 16, 23, 26, 28, 29, 33, 31, 38, 39, 18, 22
Age
Tally
Frequency
11-15
16-20
21-25
26-30
31-35
36-40
41-45
Answer questions 17 and 18 using the table from problem 16.
17. If the Micro-Gadget Corporation wants to sell their new X-Gamer system
to 10-15 year olds using the frequency table above do you think MTV is a
good place to advertise? Answer yes or no and explain your answer.
18. If the Hip Cosmetics Corporation wants to sell their new perfume to 16-30 year olds
using the frequency table above do you think MTV is a good place to advertise?
Answer yes or no and explain your answer.
19. The list of data below represents a survey of the sophomores for the number
of tardies they had during the 2012-2013 school year.
3, 0, 5, 8, 9, 11, 4, 1, 0, 5, 0, 14, 3, 2, 7, 1, 12, 6, 2, 5, 0, 1, 4, 13, 3, 9,
10, 7, 12, 11, 8, 5, 3, 4, 10, 11, 14, 0, 3, 2, 9
Construct a frequency table for this set of data.
20. A histogram is a special kind of bar graph, what makes it special?
21. What are the three measures of central tendency that we have discussed?
22. How do we find the median for a set of data?
23. What is the mode for a set of data?
24. Find the mean, median, mode and range for the set given below:
12, 22, 13, 15, 12, 24, 22, 12, 18, 20
25. What is an outlier?
Unit 1 Section 3
Objectives

The student will interpret circle graphs.

The student will create and interpret stem and leaf plots.
There are many types of graphs and each can be used to represent different aspects of
sets. Circle graphs or pie charts should be used when we know 100% of a set and the
data can be divided into discrete categories. Below are some examples of how circle
graphs can be interpreted.
Example A
The graph below shows what the members of the class of 2013 intended to do after
graduation.
Go into the work force
Technical/Trade School
33
21
Four year University
26
8
Military
44
10
Other/Undecided
Junior College Full/Part Time
This chart begins with the actual numbers of graduates in each category. We could be
asked to find several quantities from this graph. First we would need to find the total
number of graduates in the class of 2013. If we add up all the categories we get 142.
This number can then be used to find the percentage of any category. Percentages are
found by dividing the part by the whole. To find the percentage of students going into
the military we divide 8 by 142 which results is 0.0563. We multiply this by 100 and
round it to the nearest tenth to get a final answer of 5.6%.
Example B
The chart below show the break-down of employment in Arizona by categories.
Medical Industry
Service Industry
8% 14%
Information Technology
10%
15%
Manufacturing & Construction
State/Local Government
19%
8%
Agriculture
26%
Retail
If in this example there are 2800000 people employed in the state of Arizona we should
be able answer several types of questions. We can find the number of people in each
category. This is done by multiplying the percent times the total number of people
represented in the graph. The number of people employed in Manufacturing and
Construction can be found be taking 2800000 times the 19%. The percent must first be
turned into a decimal number which would be .19 and then we multiply with a result of
532000 people. We can also find differences in categories. If we wanted to know
how many more people were employed in Service Industry than in Information
Technology we would begin by subtracting the two percentages. So 14% - 10% leaves
us with 4%. We can then multiply 4% by 2800000. The result of the multiplication
is 112000.
We have seen two ways of using a circle graph or pie chart. When a circle graph starts
with numbers we will be asked to find percentages. Finding percentages means we will
divide the part by the whole and turn the quotient into a percent. When a circle graph
starts with percentages we will be asked to find the number of elements in each
category. This is done by multiplying the percent as a decimal number times the total
number of elements in the set.
Another type of graphic representation is called a stem and leaf plot. The stem and leaf
plot allows us to condense the data to make finding some measures of central tendency
easier without losing track of the individual elements of the set. This means
the way data is organized is easy to interpret, and all the numbers are still listed. An
example of this type of graph follows.
Example
The latest unit test in Pre-Calculus produced the set of test scores below.
82, 96, 77, 61, 79, 88, 86, 84, 92, 90, 91, 77, 84, 83, 78, 77, 72, 77
In this type of graph the leaf’s are the last digit from each element. The stems
are the number with the last digit removed. So the first number in our list,
the '82', has a leaf of '2' and a stem of '8'. The stem and leaf plot consists of
a vertical line with the stems on the left side and the leaf's on the other side as
shown below.
6
7
8
9
1
2 7 7 7 7 8 9
2 3 4 4 6 8
0 1 2 6
When data is organized into a stem and leaf plot it is easy to find the mode,
the median and the range. In our example the four values of 77 are quite
apparent so 77 is the mode. We can count the values quickly and in this set
there are 18 test scores. The median will be between the 9th and 10th
values. The median is between 82 and 83 so the median is 82.5. The
range is the highest value minus the lowest value so we calculate the range as
96 – 61; and the range is 35.
Exercises Unit 1 Section 3
1. What percent of a set does a circle graph show?
2. When a circle graph contains the percentages for each category how do we
find the number of elements in each category?
3. When a circle graph contains the number of elements in each category how do we
find the percentages for each category?
Given the circle graph below shows the ways students commute to school answer the
questions that follow:
Ways WHS Students Get to School
Parent Provided 15%
Walk 5%
Bus 40%
Ride with a friend 15%
Drive their own Car 25%
4. How do most students get to school?
5. If there are 650 students at WHS, how many of them walk to school?
6. How many student cars will there be in the parking lot on a normal school
day if there are 650 students?
7. At least what percentage of students who drive do NOT transport a friend to and
from school with them? How many students is this if there are 650 students at WHS?
The graph below represents the Budget for the Wickenburg Super Slide Park.
Wickenburg Super Slide Park Budget
2013
Other 7%
Employee Wages
35%
Utilities 15%
Maintenance 20%
Finance Charges 13%
Insurance 10%
If the WSSP has a budget of $8,000,000, answer the following questions.
8. What does the park pay in wages (salaries and compensation) to its employees?
9. What does the park pay for finance charges in one year? in one month?
10. How much does it cost to maintain the park for one year?
11. What category has the least money spent on it?
12. If the park is only open 8 months of the year, how much is spent per month
on its employee wages?
13. How much more money is spent on Utilities than on Insurance for one year?
a.
b.
c.
d.
$200,000
$30,000
$10,000
$100,000
Given the graph below shows us a Student's Monthly expenses that he must pay from a
job that he has, answer the questions that follow.
Mike’s Monthly Expenses
Savings
$20
Entertainment
$85
Lunches
$100
Gas
$65
Insurance
$115
Clothes
$35
14. What is the total amount of money Mike expects to make each month at his job?
15. What percent of his income does Mike expect to pay for car insurance?
16. What percent does he expect to save?
17. If there are 4 weeks per month and 5 days a week, how much money does he
expect to spend daily on lunch at school?
18. A good estimate for the fractional part of his money spent on car insurance
would be
a. 1/10
b. 1/2
c. 1/4
d. 2/3
19. If we have a set of data which we want to organize in a stem and leaf plot, how do we
find the leaf for each of the values?
20. The set of ages for Psychology 101 class at Glendale Community College is given
below. Construct a stem and leaf plot for this data.
19, 23, 18, 25, 31, 26, 21, 22, 21, 34, 41, 18, 32, 30, 31, 20, 20, 23, 29, 27
21. The top twenty batting averages to the national league west are given in the stem
and leaf plot below.
Top twenty Batting Averages for the National League West
29
30
31
32
33
34
35
36
a.
b.
c.
d.
e.
f.
g.
h.
8
1
1
4
2
3
9
3 3 3 6 99
5 7 8
6 7
8
2
What is the mode for this set ?
What is the median for this set?
What is the highest batting average?
What is the lowest batting average?
How many players have a batting average of 309?
How many players have a batting average over 320?
What is the range of the data?
List batting averages are under 300 in this data set.
22. Given that the table of data below represents the turnovers per game of the
best point guards in the PAC 12, which of the following is true?
3.2
3.3
3.4
3.5
3.6
3.7
a.
b.
c.
d.
6
8
9
8
1 3 3 4
9
The mode for this set is 3.38.
The mode for this set is 3.63.
The modes for this set are 3.38 and 3.63.
The mode for this set is 3.6.
Unit 1 Section 4
Objectives

The student will interpret box and whisker plots.
Another type of graph that we will see is called a box and whisker plot. Box and whisker plots
are used when we have all of a set that is not broken down into categories and
where the focus is on displaying the distribution. The distribution in a box and whisker
plot is shown by quartiles. A quartile is one-fourth of the set. The set is divided into
quartiles using medians. The original set is divided in half when we find the first
median. Then we treat each of the two halves as if they were complete sets and find
the median for each of the halves. This process is shown below.
Example A
Given the data set of quiz scores 10, 14, 12, 13, 9, 16, 16, 17, 12, 10, 13, 14
We put them in order to find the median.
9, 10, 10, 12, 12, 13, 13, 14, 14, 16, 16, 17
Since there are 12 scores there will be 6 scores on each side of the median,
and we split the original set as shown below.
median
9, 10, 10, 12, 12, 13
13
13, 14, 14, 16, 16, 17
We now split each half into two parts. Each of the sides has 6 elements or scores
so the new medians must have three values on each side.
median
9, 10, 10, 12, 12, 13
9, 10, 10 11 12 , 12, 13
median
13
13, 14, 14, 16, 16, 17
13, 14, 14,
15
16, 16, 17
median
The box and whisker plot we would create from this set of quiz scores is given
below.
0
2
4
6
8
9
10
12
14
16
18
In the example box and whisker plot we have four quartiles as labeled below.
0
2
4
6
8
9
quartile 1
10
12
quartile 2
14
16
quartile 3
18
quartile 4
median
The whiskers in the plot are quartiles 1 and 4. The box consists of quartiles
2 and 3. In each of the quartiles we will have one-fourth or 25% of the set.
The median for the set as a whole is the line that divides the box into two
separate pieces.
Example B
The box and whisker plot below represents the height in centimeters of the students
in a PE class.
140
145
150
155
160
165
170
175
180
185
190
From the graph we should be able to find some statistical data quickly. First the
median of the set is 160 cm. The range of values for the set as a whole is 30 cm.
The box in the plot contains 50% of the heights so half of the class has a height
between 156 and 166 cm. Why do you think the whisker that represents the
4th quartile is so long?
Exercises Unit 1 Section 4
1. A quartile ( as a fraction ) is ____________ of the number of values in a data set
or ( as a percent ) it is _____________.
2. In a box and whisker plot we will always find the range divided into _________
pieces
which we call ________________.
3. Where do we find the median in a box and whisker plot?
4. Given the box and whisker plot below represents scores on the test
for English 101 answer the questions that follow.
40
45
50
55
60
65
70
75
80
85
90
95
100
a. What is the highest score on the test?
b. What is the median for the data?
c. At what values does first quartile start and end?
d. Which quartile would we find a score of 82 in?
e. What is the lowest score a student could have and still be in the second quartile?
f. What is the range of scores for the whole data set?
5. Given the box and whisker plots below represents the scores for two bowling teams
over the course of a season, answer the questions that follow.
100
115
130
145
160
175
190
205
235
Team A
Team B
a.
b.
c.
d.
e.
f.
What is the range of each team's scores?
Which team has the higher median?
Which team had the lowest score? and what was it?
If a member of team B had a score of 172 which quartile would that be in?
What percent of Team B's scores were between 132 and 190?
Which Team do you think is better? Justify your answer with details.
6. If a data set has an average of 35 and we add the number 20 to the set what will
happen to the average? Justify your answer with details.
7. If a data set has an average of 18 and we add the numbers 15 and 21 to the
set what will happen to the average? Justify your answer with details.
Unit 1 Section 5
Objectives

The student will create and interpret scatter plots.
The problem solving process that is often used in the world around us consists of
collecting data as a table or roster, turning the data into a graph, then using the
graph to create an equation and make predictions. We are going to examine the first
two steps of this process in Section 5. Below is an example of how to create a scatter
plot.
Example
The Algebra I teachers recently conducted a study. They wanted to see
exactly how missing assignments affected student grades on a unit test. The
data set below was collected. The number on the left is how many assignments
were missed during the unit by a student. The number on the right is the test
score the student earned on the unit test.
(4, 68) (1, 88) (0, 93) (3, 72) (1, 91) (0, 89) (6, 41) (7, 34)
(2, 80) (3, 76) (4, 61) (1, 86) (0, 98) (5, 52) (4, 58) (1, 42)
When we set up our axes the number of assignments missed belongs on the
horizontal axis. The percentage on the test should be on the vertical axis.
A Survey of Missing Assignments and Grades
Test Percent
100
90
80
70
60
50
40
30
0
1
2
3
4
5
6
7
8
Number of Missing Assignments
After constructing the grid and the scales, we plotted all the points. This graph
is called a scatter plot. It is a primary tool for analyzing data. In this graph
we can see a definite pattern. As the number of missing assignments goes
up the percentage earned on the test goes down. In order to accurately
interpret this graph we can draw a line of best fit. A line of best fit should go
through the middle of the points in the same direction as the data set. Since
not all the points will be on the line we should make sure that about the same
number of points are above the line and below the line. Using our previous
example we can draw a line of best fit as shown below.
100
A Survey of Missing Assignments and Grades
Test Percent 90
80
70
60
50
40
30
0
1
2
3
4
5
6
7
8
Number of Missing Assignments
This data set has an outlier which should be apparent as well. The point (1, 42)
is the outlier and does not need to be taken into account when drawing a line of
best fit. We can use this line of best fit to make predictions. If we were to
extend the line we would see it pass close to the point (8, 27). So we can
predict that the grade a student with 8 missing assignments would earn is going
to be very close to 27.
Data sets can be described based on the direction of their lines of best fit. If we draw
lines of best fit in the two graphs below we would say they are going up. This means
that as x -values go up the y-values go up. We call this a positive correlation.
y
y
x
x
If as the x-values go up and the y-values go down on our line of best fit, then we call
this a negative correlation. Some examples of negative correlations are shown below.
y
y
x
x
There is also the possibility that there is no correlation. There are two ways in which
a graph would have no correlation: First, if the points formed no pattern and were
chaotically scattered, we could not draw a line of best fit; Second, if the pattern
had a line of best fit that was perfectly flat or horizontal. Each of the graphs below
show one of these cases.
y
y
x
x
Exercises Unit 1 Section 5
1. Create a scatter plot for a survey's data set listed below. In the data points, the left
hand coordinate is the age of the person and the right hand coordinate is the
number of hours spent on homework per week. Use the grid provided by your
teacher.
(8, 5) (10, 11) (11, 10) (8, 9) (9, 11) (10, 13) (14, 17) (13, 12) (9, 8) (8, 7)
(14, 16) (11, 14 ) (10, 12 ) (9, 9) (10, 9) (14, 12) (15, 14) (14, 3) (11, 12)
Answer the questions ‘a’ through ‘g’ using the scatter plot.
a. Draw a "best fit line" for this set of data.
b. Is there an outlier in this set of data? _________ ( yes/no ) If your answer is
"yes" then list your coordinates for the outlier. _____________
c. The scatter plot from the data set above has
a. a negative correlation between age and hours of studying
b. no correlation between age and hours of studying
c. a positive correlation between age and hours of studying
d. a complex correlation between age and hours of studying
d. The coordinates that would make the endpoints of the best fit line for this data
set are closest to:
a. (8, 4) and (13, 11)
b. (8, 7) and (14, 15)
c. (8, 10) and (14, 12)
d. (8,16) and (14, 3)
e. The range of the data set in the above scatter plot is
a. 14
b. 17
c. 16
d. 20
f. Describe the general tendency shown in the graph.
g. If we were to use this graph to predict the number of hours a 16 year old would
study, the best estimate from those given below would be
a. 19
b.
4
c. 12
d. 15
2. Examine each scatter plot below and decide if there is a positive correlation, a negative
correlation, or no correlation ( answer positive, negative or none )
a)
c)
b)
d)
3. Why are measures of central tendency important? ( What does a measure of
central tendency tell us? )
4. How do we find the median of a set?
5. Find the mean, median, mode, and range of the data set below.
24, 36, 28, 30, 32, 32, 29, 29, 25, 37, 27, 28, 32
The graph below represents the time and distance for a 100 meter race between a man
and a horse.
Man vs Horse
Distance in Meters
120
100
80
60
Series1
Man
Series2
Horse
40
20
0
1
2
3
4
5
6
7
8
9
10
Time in Seconds
6. The graph above shows that
a. a man can run faster at the start of a race but the horse will catch and pass him.
b. a horse can run faster at the start of a race.
c. a man can run faster than the horse.
d. a man and a horse run at exactly the same speed.
Unit 1 Section 6
Objectives

The student will distinguish between discrete and continuous sets.

The student will choose the best type of graph to represent a given
data set.
The sets we see in graphs can be either continuous or discrete. The discrete sets
have numbers that are isolated. For instance, a set of years will be discrete because
there is no year in between 2012 and 2013. In continuous sets if we pick any two
numbers, we can always find a number in between them. For instance, a set of
temperatures would be continuous. If we were to pick 30o and 31o we would find that
there could be many temperatures between the two original values. No matter what two
temperatures we pick there will always be another number in between them that we can
find.
When we are deciding what type of graph to use for a particular set, whether the values are
discrete or continuous is important. Below are some of the criteria we can use to select the
type of graph that best represents a data set.
Bar/Histograms work best with discrete sets of data that can be grouped
into intervals or categories.
Line Graphs work best with sets that are continuous and are often
open ended. (When we are given data points that could have had
points collected in between the given points, and perhaps before
or after the points that are listed in the sets. )
Circle Graphs work best when we have 100% of a set known, and
the data can be divided into discrete categories.
Box and Whisker Plots work best with when we have all of a set that is not
differentiated into categories and where the
distribution is what we are interested in.
Stem and Leaf Plots work best when we need to condense the data
into categories but we do not want to lose track of
individual elements. This makes it possible to find some
measures quickly – mode, median, and range.
Frequency Tables normally are turned into histograms but they help analyze
data to see the distribution.
Scatter Plots are used to analyze a data set to see correlations ( positive or
negative, ) and make predictions ( usually based on linear
regression which we will do in the future! )
The examples below show how data can be displayed with several types of graphs.
Mr. Smith’s geography class is studying a number of states in detail. The class has
has collected the data below and organized it into a table. We will construct graphs
to display and interpret the graph.
State
Time Zone
Area*
Population
Year of Admission
Georgia
Eastern
58,390
9,919,945
1788
Maryland
Eastern
10,455
5,884,563
1788
New York
Eastern
49,112
19,570,261
1788
Texas**
Central
266,874
26,059,203
1845
Florida**
Eastern
58,681
19,317,568
1845
Montana
Mountain
147,047
1,005,141
1889
California
Pacific
158,648
38,041,430
1850
New Mexico
Mountain
121, 599
2,085,538
1912
Washington
Pacific
68,126
6,897,012
1889
Arizona
Mountain
114,007
6,553,255
1912
* Area is measured in Square miles.
** Small portions of these states are in other time zones.
States per Time Zone to be Studied
5
4
3
2
1
Eastern
Central
Mountain
Pacific
Here we have a bar graph showing the number of states in each time zone we are
studying.
There are slightly over 135,000,000 people in the states that the class will be studying.
Below is a circle graph showing how the populations of each time zone in the study relate to
the total.
Eastern Time Zone
40.4%
Mountain Time Zone
7.1%
Central Time Zone
Pacific Time Zone
19.3%
33.2%
The box and whisker plot below shows the distribution of the states in the study by area.
Sizes of States in the study in Thousands of Square Miles
0K
25 K
50 K
75 K
100 K
125 K
150 K
175 K
200 K
250 K
275 K
Exercises Unit 1 Section 6
1. What does it mean for a set to be discrete?
2. Decide whether each set is continuous or discrete. ( Answer continuous or discrete )
a. The set of years from 2000 to 2012
b. The set of colors { red, blue, green, yellow, white }
c. The set of elapsed time as a race is being run
d. A set of unrounded percentages on a test
e. The set of months in a year
f.
A set of temperatures
g. A set of unrounded lengths measured in centimeters
h. A set of candidates for student council offices
3. Fill in the blanks
a. __________________ work best with sets that are continuous and are often
open ended.
b. _________________ work best when we need to condense the data into
categories but we do not want to lose track of individual elements.
c. ____________________ work best when we have 100% of a set known and
the data can be divided into discrete categories.
d. ____________________ are used to analyze a data set to see correlations
and make predictions.
e. ____________________ work best with discrete sets of data that
can be grouped into intervals or categories.
f. ____________________ work best with when we have all of a set that is not
differentiated into categories and where the distribution is what we are
interested in.
g. ____________________ normally are turned into histograms but they help
analyze data to see the distribution
4. If we have a set of bowling averages for a whole league and we are interested in the
distribution so that we can determine the scale for handicaps we should use a
a) circle graph
b) line graph
c) stem and leaf plot d) box and whisker plot
5. Mike wants to plan a budget based on his monthly salary. He needs to pay rent,
insurance, car maintenance, food, entertainment, and savings. What is the best
way to display the data?
a) circle graph
b) line graph
c) stem and leaf plot
d) frequency table
6. The computer science class recently took a test. The scores were 62, 88, 67, 91,
76, 83, 93, 77, 72, 83, 90, 80, 78, 84, and 83. The best way to organize this data
and still show the individual scores is with a
a) circle graph
b) line graph
c) stem and leaf plot
d) frequency table
7. WHS is collecting data to relate the scores on the ACT test to grade point average.
The best way to organize this data is with a
a) bar graph
b) circle graph
c) scatter plot
d) box and whisker plot
8. Sam is doing a science experiment trying to relate the temperature of hot water to the
time it takes it to cool off to room temperature. The best way to represent the data
is with a
a) line graph
b) circle graph c) stem and leaf plot
d) box and whisker plot
9. Sarah is measuring how fast her hair is growing with the following table.
week
length in mm
1
80
2
83.5
3
87
4
90.5
5
94
Which of the representations below is the best way to organize the data?
a) circle graph
b) frequency table
c) stem and leaf plot
d) line graph
10. Find the mean, median, mode and range for the data set below.
15, 8, 13, 9, 10, 9, 14, 12, 4, 8, 9, 11
11. Use the graph below to answer the following questions.
Membership in SADD
25
# of Members
20
15
10
5
0
1
2
3
4
5 6 7
Time in Weeks
8
9
10
a. What type of graph is shown above? And why should it NOT be
drawn as a bar graph?
b. What was the membership after 2 weeks?
c. When did the membership get above 15?
d. What do you think the membership will be in 11 weeks?
Explain how you decided on this number.
12. Given the box and whisker plot below represents speeds recorded on the Loop 101
in Phoenix answer the following.
30
a.
b.
c.
d.
e.
40
50
60
70
80
90
100
110
120
130
What is the highest speed on the Loop 101?
What is the median for the data?
What is the range of speeds in the first quartile?
What percentage of the cars are going over 70 mph?
What is the lowest speed a person could be driving and still be in the
second quartile?
f. What is the range of speeds for the whole data set?
140
Unit 1 Section 7
Objectives

The student will list, identify and label the parts of the x-y coordinate
plane.

The student will identify and plot points on the x-y coordinate system.
The graphs we studied in the previous sections represented data sets that used only
positive values. But there will often be times when we must work with both positive and
negative values. Temperatures can be positive or negative. A business that makes a
profit will have positive values in the data while a business that loses money may have
negative values. We will be making graphs for equations that use the variables ‘x’ and
‘y’ in much of Algebra I. We will need to be able to construct a general type of graph
called the x-y coordinate plane. It is also called the Cartesian Coordinate system named
after a French mathematician, Rene’ Descartes who invented the system.
The coordinate plane is formed by two real number lines. One that is horizontal and the
other that is vertical. The horizontal real number line is the x-axis. The vertical real
number line is called the y-axis. In the ordered pair (x, y) the value that is on the left is
called the x-coordinate or the abscissa, while the value on the right is called the ordinate
or y-coordinate. The diagram below shows the axes with scales.
y
4
3
2
1
-4 -3 -2
-1 -1
1
2
3
4
x
-2
-3
-4
The point at which the two real number lines cross is called the origin and has coordinates
of (0, 0). The numbers are the scales. The x-axis is the horizontal real number line and
has its negative values to the left and positive numbers to the right as with a normal
real number line. The y-axis is vertical and the scale shows us the top part of the axis is
positive and the bottom part is the negative. The axes above show us some of the
parts of our x-y coordinate plane. These are the x-axis, the y-axis and the scales.
Another important part of the plane is the origin which is (0, 0).
The x-y coordinate plane is divided into four parts called quadrants. The quadrants
are labeled with the Roman numerals I, II, III, and IV. Below is the coordinate plane
with the quadrants labeled.
y
4
3
II
2
-4 -3 -2
-1 -1
III
I
1
1
2
-2
3
x
4
IV
-3
-4
The coordinate plane will often appear with a grid as shown below.
y
x
When the grid lines are drawn on the plane a scale is optional. We need to be able to
identify locations or points in the plane. Below are some points that have been plotted,
identified by their coordinates.
Below is a list of the
y
coordinates
C
for each point.
A
G
D
B
F
E
x
A
B
C
D
E
F
G
(-4, 3)
(-5, -1)
(2, 4)
(0, 0)
(0, -5)
(1, -3)
(3, 0)
Some of the points are in quadrants and some are on the axes. Point A is in the 2nd
quadrant, B in the 3rd, C in the 1st, F in the 4th. E is on the y-axis and G on the x-axis.
When we are given a point we should be able to plot or locate the point. We always start
at the origin (0,0) when we are plotting a point. The first coordinate of an ordered pair
is the x-coordinate. The x-axis is horizontal which means we will use the x-coordinate
to tell us how far to move left or right from the origin. After we have graphed the
x-coordinate then the y-coordinate is used to go up or down. The left-right or up-down
movement depends on the sign of the coordinate. The examples below show points
being plotted.
Example A
Plot the point A(6, 5)
y
y
y
A
5
x
x
x
6
Example B
Plot the point B(-5, -4)
y
y
y
-5
x
x
x
-4
B
Exercises Unit 1 Section 7
1. What is the number line that is vertical called?
2. What is the number line that is horizontal called?
3. What are the coordinates of the origin?
4. The number on the left of the ordered pair is called the _______________ or
the __________________.
5. The number on the right in an ordered pair is called the ________________ or
the ___________________.
Given the Graph below answer the questions that follow. ( The scale on the
coordinate plane is one square equals one unit )
N
M
P
S
R
T
V
W
Z
6. List the coordinates with their names ( letters ) for each of the plotted above.
7. What point is on the x-axis?
8. What point is closest to the origin?
9. Which point is in the third quadrant?
10. Which point has the same y-coordinate as point 'S'?
11. What are the coordinates of the point that would form a rectangle with
points 'V', 'S' and 'R'?
12. What point lies on the negative y-axis?
13. If we plot a point at (-4, 0) we can create a square. What are the names of the
other three points that would form the square?
14. Draw the x-y coordinate plane and label the x-axis, the y-axis, put a scale on each
axis, and put the appropriate Roman numeral in each quadrant.
15. Draw the x-y coordinate plane and plot each listed point on the coordinate plane
given below. Be sure to label each point with its name ( letter ).
A(3, 5)
B(-2, 6)
C(0,0)
D(-5, -1)
E(1.5, -4)
16. Point 'T' has been plotted on the grid to the right .
What point in the 3rd quadrant has an x-coordinate
whose absolute value is the same as point T's x-coordinate.
a)
b)
c)
d)
(5,4)
(-5, -4)
(5, -4)
(4, -5)
F(0, 3)
8
6
4
2
-8 -6 -4 -2
-2
-4
-6
-8
2
4
6
8
T(5,-4)
17. Point 'P' has been plotted on the grid to the right .
What point has coordinates that are the opposites
of point P's coordinates.
8
W
X
a)
b)
c)
d)
W
X
Y
Z
-8 -6
Z
6
4
2
-4 -2-2
-4
-6
-8
Y
2
4
6
8
P(2,-6)
18. If a point has a negative x-coordinate and a positive y-coordinate then
it lies in which quadrant?
a) I
b) II
c) III
d) IV
19. Using the graph below as a guide and assuming there are 640 people at WHS answer
the following.
Student Enrollment by District
Wickenburg Unified SD
Aguila SD
45%
5%
Morristown SD
11%
32%
7%
Congress SD
Nadaburg SD
a. How many people at WHS are from the Congress School District?
b. Which district has the second highest number of students?
c. How many people at WHS are from the Wickenburg Unified SD ?
d. How many people at WHS are NOT from Aguila?
e. Explain how to find the answer in part 'd'?
Unit 1 Section 8
Objectives

The student will state and apply the properties of real numbers.

The student will combine like terms.
You have known many of the properties of real numbers since you first started
learning how to add. You may not have known the names for them but you could
always apply them. For instance, you have always known that 3 + 2 is the same
as 2 + 3. The values in an addition problem can have their order changed and we
will still get the same answer. The property is known as the “Commutative Property
of Addition”.
Definition
A property is something that is always true no matter what values are used.
.
Below is a list of the real number properties we will be using.
The COMMUTATIVE PROPERTY OF ADDITION
In Words: We can switch the values when adding. In Variables: a + b = b + a
The COMMUTATIVE PROPERTY OF MULTIPLICATION
In Words: We can switch the values when we multiply. In Variables: a  b = b  a
The ASSOCIATIVE PROPERTY OF ADDITION
In Words: We can switch the parentheses in adding. In Variables: (a + b) + c = a + (b + c)
The ASSOCITIVE PROPERTY OF MULTIPLICATION
In Words: We can switch the parentheses in multiplying. In Variables: a  (b  c)=(a  b) c
The DISTRIBUTIVE PROPERTY
In Words: We can multiply through parentheses. In Variables: a( b + c ) = a  b + a  c
The IDENTITY PROPERTY OF ADDITION
In Words: Any number plus zero equals that number. In Variables: a + 0 = a
The IDENTITY PROPERTY OF MULTIPLICATION
In Words: Any number times one equals that number. In Variables: 1  a = a
The INVERSE PROPERTY OF ADDITION
In Words: Any number plus its opposite equals zero.
In Variables: a + -a = 0
The INVERSE PROPERTY OF MULTIPLICATION
In Words: Any number times its reciprocal equals one.
In Variables: a 
1
=1
a
The ZERO PROPERTY of MULTIPLICATION
In Words: Any number times zero is 0.
In Variables: a  0 = 0
The MULTIPLICATION PROPERTY OF -1
In Words: Any number times negative one equals its opposite. In Variables: -1  a = -a
These properties can be used to make numeric problems easier or to give us alternate
ways to simplify or evaluate expressions. Below are some examples of using these
properties.
Example A
Example B
Example C
Example D
13 + 29 + 7
13 + 7 + 29
20 + 29
49
5 . 38 . 2
5 . 2 . 38
10 . 38
380
(19 + 24) + 6
19 + (24 + 6)
19 + 30
49
5(x + 4)
5.x+5.4
5x + 20
In this example we
use the Commutative
Property of Addition.
The 29 and 7
exchanged places.
In example B we
use the Commutative
Property of Multiplication.
The 38 and the 2
exchanged places.
In example C we
use the Associative
Property of Addition.
The parentheses
were moved.
In example D we
use the Distributive
Property. We multiplied
through the
parentheses.
We often use these properties to simplify variable expressions. We are now going to look
at one critical operation which we will use to simplify variable expressions. This technique
is called combining like terms. To use this tool of Algebra we first need to understand
what like terms are.
Definition
A term is part of an expression separated by plus or minus.
Definition
Like terms have the same variable with the same exponents.
Below are some examples of terms that are like and some that are not like.
Example A
3x + 5x
These two terms are “like” because they have the same
variable ‘x’ with the same exponent ‘1’.
Example B
4a2 – 7a2
These two terms are “like” because they have the same
variable ‘a’ with the same exponent ‘2’.
Example C
x + 12y
These two terms are NOT “like” because they do NOT have
the same variables.
Example D
8y2 - 11y
These two terms are NOT “like” because they do NOT have
the same variables with the same exponents. The
exponents
are different.
Example E
Example F
-3xy + 9xy These two terms are “like” because they do have the same
variables ‘x’ and ‘y’ with the same exponents ‘1’.
13 + 9
These two terms are “like” since they both are constants.
Neither has a variable or an exponent.
We need to be able to identify like terms in larger expressions as well. Below are some
examples of expressions with three or more terms.
Example A
3x + 7y – 6x The “3x” and the “-6x” are like terms. Notice also that the
negative sign stays with the six.
Example B 8y – 3 + y + 2 There are two pairs of like terms in this expression. The
“8y” and the “y” are like. The “-3” and the “2” are like
terms as well.
Example C 5a2 – 7a2 + 2a In this expression there is only one pair of like terms.
The “5a2” and the “-7a2” are like but the “2a” does not
have an exponent and so is NOT like the other two terms.
If we can recognize like terms we can combine them. When we combine like terms it
means we will add or subtract them. To understand how this works with the variables
we will first examine how we can add “like” constants.
The arithmetic problem below can be done in two ways.
2+2+2+2+2+2+2
The first way is to do all the addition 2 + 2 = 4, 4 + 2 = 6, 6 + 2 = 8, 8 + 2 = 10,
10 + 2 = 12, and finally 12 + 2 = 14. This process is simple but long. The other
method is to count the number of two’s and multiply. There are 7 two’s so the
arithmetic problem is 7(2) = 14. All of the problems below can be done the same way.
3+3+3+3+3
5(3)
7+7+7+7+7+7
6(7)
4+4+4+4+4+4+4+4
8(4)
These problems can be changed into multiplication because all the values we are
adding up are the same. Let’s look at the same kind of problem with variables.
x+x+x+x+x+x+x
7x
We can change the addition into multiplication problem because whatever number we
substitute for the first ‘x’ we must substitute for every other ‘x’ so all the numbers will
be the same. We can also reverse this process. If x + x + x gives us 3x, then if we
start with 3x we can write it x + x + x. Some examples of this are given below.
9z
z+z+z+z+z+z+z+z+z
5xy
xy + xy + xy + xy + xy
4a2
a2 + a2 + a2 + a2
This means that the coefficient (the number in front of the variable) tells us how many
variables are being added together.
We can now investigate how like terms are combined. Let’s examine the problem below.
3x + 2x
(x + x + x) + (x + x) We first make the multiplying into addition
(x + x + x + x + x) The Associative Property of Addition lets us change the ( )
5x
Adding all the x’s together
The properties we studied earlier help justify each step in the process above. We don’t have to
show all the steps in the process however. To combine like terms we should add or subtract
their coefficients. The examples below show us how to combine like terms.
Example A
Example B
Example C
9a + 4a + 2a
15a
3z – 8z – z
3z – 8z – 1z
-6z
4x – 2 + 7x – 10
11x – 12
We added the
coefficients,
9 + 4 + 2 = 15
We put in the
understood coefficient
of 1. Which is
using the Identity
Property of Multiplication
Example D
We can combine
the ‘4x’ and the ‘7x’.
The ‘-2’ and ‘-10’ are
also like. So we add
then the coefficients.
3x2 + x + x2 – x
3x2 + 1x + 1x2 – 1x
4x2
The ‘x’ terms and the
‘x2’ terms are not like
and must be added
separately. The ‘1’ and
‘-1’ add up to 0. So we
don’t have any plain x’s.
Exercises Unit 1 Section 8
List the property demonstrated in each problem below
1. -1
.
x = -x
3. 0 = a
5.(11
.
.
6)
2. a = a
0
.
2 = 11
.
1
4. (9)(3) = (3)(9)
.
(6
.
2)
6. a(b + c) = ab + ac
7. 9 = -1
9. 4
.
.
-9
2 = 2
8. a + 0 = a
.
4
10. 1
.
12 = 12
1
12. b . = 1
b
11. a + -a = 0
List the property that would justify changing the top expression into the bottom
expression.
13.
2a + 8 + 5a
2a + 5a + 8
14. (13 + 29) + 1
13 + (29 + 1)
15.
5(2x – 13)
10x – 65
16.
5x – x
5x – 1x
List the property that would justify each transformation.
17.
(a + b) – b
a + (b – b)
a+0
a
_________________
_________________
_________________
18.
xy + yx
xy + xy ______________
1xy + 1xy _______________
(1 + 1)xy _______________
2xy
19. Write each of the following as a sum of variables.
a. 6z
b. 3a + 2y
20. Write each of the following using a variable and a coefficient.
a. x + x + x + x + x + x + x + x
b. y + y + y + y + y + y
21. Write each expression using coefficients.
a.
a+a+a+x+x+x+x
b. y + z + z + z + z + y + y + z
22. Simplify.
a. 3x + 6x
b. 2y - 6y
c. 2a + 5a + 7a
d. 12x - 3x
e. 4t - 5t + t
f. 6y + 7y + 5z - 11z
g. 12s + 4s - 5s
h. 13z + 4z + 1
i. 4x + 8x + 3 + 11
j. 3a + 2b - 6b + 6a
k. -5y + 6 - 2y + 11
l. 4x + 2y + x + 4 + 5y +
9
m. 5a + 7 - x + 8a - 2 - 4
23. Define like terms.
24. How do we combine terms?
n. 11 + 3z - 2 + 12z + 5
25. What is the understood coefficient?
26. Given the expression 2w + 5x + 7y + z + 8
a. How many terms does the expression have?
b. What is the coefficient of y?
c. What is the third term?
d. What is the coefficient of z
27. What does it mean to substitute?
28. What does it mean to evaluate?
29. Given x = 5 and y = -3 evaluate ( Show the Work )
a. 6x + 2y
b. 6y2
d. 5(x + y)3
e.
30. What is the value of the expression
a.
b.
c.
d.
449
89
129
1649
y 1
x 1
4xy2 + (x - y)2
c. 3x2y - x
f. (x + y)(x - y)
for x = 5 and y = -2
Unit 1 Section 9
Objectives

The student will translate phrases and sentences written in words into
algebraic expressions.
Variables and expressions are the most important tools in Algebra. We must be able
to convert problem and situations into expressions and equations. One method for
doing the translation is to recognize key words. Below is a list of how words and
phrases can be translated into operations.
Addition is indicated by: plus, more, sum, total, increased by, added to
Subtraction is indicated by: minus, less, difference, decreased by, subtracted from
Multiplication is indicated by: times, product, twice ( this is special because it also
gives us a number ‘2’ )
Division is indicated by: quotient, divided by, half ( this is special because it also gives
us a number ‘2’ )
There are also special words that tell us other things.
Than - this word is used with "more" and "less" it tells us to switch the order of the
numbers or expressions involved.
A Number - is replaced by a variable.
Below are some examples of translations.
Example A
“twice a number plus fifteen” twice means to multiply by 2, plus
means we should add, a number can be any variable so we use ‘n’.
“twice a number plus fifteen”
2 .
n
+
15
So the translation is
Example B
The translation is shown below
each word.
2n + 15
“Six times a number is twenty.” times means to multiply, ‘is’
will become ‘=’, and a number can be any variable so we pick ‘n’
“Six times a number is twenty.” The translation is shown below
.
6
n
=
20
each word.
So the translation is
6n = 20
Example C
“two more than three times a number” times means to multiply,
more than will become ‘+’, ‘than’ will mean to switch terms, and a
number can be any variable so we pick ‘x’
“two more than three times a number” The translation and the
.
2
+
3
x
switch are shown below.
switch
So the translation is
Example D
3x + 2
“Five less than half a number is nine” less than will become ‘-’,
‘than’ will mean to switch terms, half is to divide by 2, a number can
be any variable so we pick ‘y’, and ‘is’ will become ‘=’
“Five less than half a number is nine” The translation and the
5
−
y ÷ 2
= 9
switch are shown below.
switch
So the translation is
y
− 5=9
2
We will also have to translate contextual problems. Examples of this type of problem
are given below.
Example A
If Mike has ten less than five times as many songs on his MP3 player
as Sam. Write an expression for the number of songs Mike has.
We know that Sam has a certain number of songs on his mp3 player.
We can let that number be a variable ‘n’. So that the expression would
be 5n – 10.
Example B
Maria sells ‘x’ number of raffle tickets and Sandy sells ‘y’ number
of raffle tickets. If the tickets price is ‘p’ dollars each write an expression
for the amount of money they made selling raffle tickets.
If we look at the problem conceptually we know how to find the total
number of tickets sold. For instance if Maria sold 4 and Sandy sold 5,
then together they sold 9 tickets. We added the numbers to get the
total, so if we use ‘x’ and ‘y’ and we add them, “x + y” is the
expression we want.
If we consider how we generally use the price and number of items
we are buying, we multiply. For instance, if we buy 4 candy bars for
$0.90, then we multiply to find the cost so 4($0.90) is $3.60.
So then we must multiply the price times the total number of tickets.
Our expression then is p(x + y)
Exercises Unit 1 Section 9
Translate each sentence or phrase into an EXPRESSION.
1. nine less than a number "a"
2. seven more than a number "b"
3. a number "w" decreased by twelve
4. a number "t" increased by sixteen
5. thirty less than a number 'x'
6. four less than twice a number "n"
7. half a number "x" decreased by a number “a”
8. five more than eight times a number "y"
9. six times the sum of a number "z" and three
10. the difference of twice a number 'n' and that same number
Translate each sentence into an EQUATION.
11. Half a number "x" increased by five is thirty-two.
12. Fourteen less than a number "a" is twice the number.
13. Four times a number "w" is three less than the square of the number.
14. Five times a number "x" decreased by four is six more than twice the number.
15. Six times the difference of a number "s" and nine is the number squared.
16. Half the sum of a number "x" and eleven is four less than twice the number.
Write variable expressions for the following.
17. The number of eggs in ‘z’ dozens.
18
If Mike can change the oil in 3 cars per hour, write an expression for the number
of cars he can service in 't' hours.
19. Tabatha is an inspector on an assembly line. If she can check 'a' items per hour
and works 'w' hours, write an expression to model this situation.
20. Juan and Derrick play basketball. Juan averages 'x' points per game and Derrick
averages 'y' points per game. Write an expression for the number of points they
will score together in "g" games.
21. Maria has y% of the votes in an election. If there are "v" number of votes cast,
write an expression for the number of people who voted for Maria.
22. Ruben bought 'a' tickets to a movie for his family. If the tickets cost 'n' dollars each
and he spent 'r' dollars on refreshments, write an expression for the amount of money
he spent.
23. If Max can hike at a rate of "x" miles per hour and he covered "d" distance, write an
expression for the number of hours he spent hiking.
24. At the football game the gate receipts were $2000. Adult tickets were sold for
'w' dollars each and the number of adults tickets sold was 't' tickets. The expression
for the amount of money collected for student tickets is best represented by
a. 2000 + w + t
c. 2000 – wt
b. 2000 + wt
d. 2000wt
25. At WHS x% of our students ride the bus. WHS has 's' number of students enrolled
in the school. The best expression for the number of students that ride the bus is
a. s - x
x
s
c.
100
b. sx
d. s +
x
100
26. At a recent concert floor seats were sold for 'a' dollars each. The seats in the stands
were sold for 'b' dollars each. There were 2000 seats sold for the concert. If there
were 'n' number of floor seats sold which expression below is the best representation
for the amount of money that was made on seats in the stands?
a. b(2000 - n)
c. 2000 – bn
27.
b. a(2000 - n)
d. (a + n)(2000 - b)
Sam and Dave are installing wood flooring in a large reception hall. Sam can install
'x' number of square feet in an hour. Dave can install 'y' number of feet in an hour.
Which of the expressions below best represents the number of square feet can they
install together in 't' hours?
a. x + y + t
c. txy
b. t(x + y)
d. t + xy
Unit 1 Section 10
Objectives

The student will convert visual patterns into numeric patterns
and algebraic expressions.
Frequently in the problem solving process we will start with data in one of several
representations. We will need to find patterns within our data and use these patterns
to write expressions or equations and then make predictions using the Algebraic
expressions. Below are examples of visual patterns converted to expressions and
used to make predictions.
Example A
Given the visual pattern below answer the questions that follow.
Element 1
Element 2
Element 3
Element 4
a) Draw the next element in the pattern.
This would be the answer to part ‘a’
b) Describe in words how the 20th element in the pattern would look.
The 20th element would have 20 columns of six squares.
( This can be determined from the fact that element 1 had 1 column,
element 2 had 2 columns, element 3 had 3 columns, etc. )
c) Describe in words how the number of squares is growing in the pattern.
The number of squares grows by 6 squares as we go from the previous
element to the next.
d) Represent the pattern numerically ( show how the number of squares grows ).
__6__, __12__, __18__, __24__, __30__, …
e) Describe how the numerical pattern in part ‘d’ is growing.
The numbers in the pattern go up by six for the next element.
f) How many squares would the 30th element have? Show your arithmetic.
6 . 30 = 180
g) How many squares would the nth element have? Show your arithmetic.
6 . n = 6n
Example B
Given the visual pattern below answer the questions that follow.
Element 1
Element 2
Element 3
Element 4
a) Draw the next element in the pattern.
This would be the answer to part ‘a’
b) Describe in words how the 30th element in the pattern would look.
The 30th element would have 30 rows of five squares and 2 extra on top.
`
c) Describe in words how the number of squares is growing in the pattern.
The number of squares grows by 5 squares as we go from the previous
element to the next.
d) Represent the pattern numerically ( show how the number of squares grows ).
__7__, __12__, __17__, __22__, __27__, …
e) Describe how the numerical pattern in part ‘d’ is growing.
The numbers in the pattern go up by five for the next element.
f) How many squares would the 100th element have? Show your arithmetic.
5 . 100 + 2 = 502
g) Write an expression for the number of squares the nth element would have.
Show your arithmetic.
5 . n + 2 = 5n + 2
h) Use the expression to find the number of squares in the 45th element.
5(45) + 2
225 + 2
227
Exercises Unit 1 Section 10
1. Examine the pattern below and answer the questions that follow.
Element 1
Element 2
Element 3
Element 4
a. Draw the next element in the pattern.
b. Describe in words how the 10th element in the pattern would look.
c. Describe in words how the pattern of circles is growing.
d. Represent the pattern with numbers as shown below.
Element
1
Element
2
Element
3
Element
4
Element
5
_____, ______, ______, ______, ______, ...
e. Describe in words how you would extend the number pattern.
f. How many circles would the 80th element have? Show your arithmetic.
g. Write an expression for the number of circles the ‘nth’ element would have.
h. Use the expression to find the number of circles the 19th element would have.
2. Examine the pattern below and answer the questions that follow.
Element 1
Element 2
Element 3
Element 4
a. Draw the next element in the pattern.
b. Describe in words how the 20th element in the pattern would look.
c. Describe in words how the pattern of circles is growing.
d. Represent the pattern with numbers as shown below.
Element
1
Element
2
Element
3
Element
4
Element
5
_____, ______, ______, ______, ______, ...
e. Describe in words how you would extend the number pattern.
f. How many circles would the 50th element have? Show your arithmetic.
g. Write an expression for the number of circles the ‘nth’ element would have.
h. Use the expression to find the number of circles the 37th element would have.
3. Examine the pattern below and answer the questions that follow.
Element 1
Element 2
Element 3
Element 4
a. Draw the next element in the pattern.
b. Describe in words how the 25th element in the pattern would look.
c. Describe in words how the pattern of circles is growing.
d. Represent the pattern with numbers as shown below.
Element
1
Element
2
Element
3
Element
4
Element
5
_____, ______, ______, ______, ______, ...
e. Describe in words how you would extend the number pattern.
f. How many circles would the 60th element in the pattern have? Show your arithmetic.
g. Write an expression for the number of circles the ‘nth’ element in the pattern
would have.
h. Use the expression above to find the number of circles the 43rd element would have.
4. Examine the pattern below and answer the questions that follow.
Element 1
Element 2
Element 3
Element 4
a. Draw the next element in the pattern.
b. Describe in words how the 40th element in the pattern would look.
c. Describe in words how the pattern of circles is growing.
d. Represent the pattern with numbers as shown below.
Element
1
Element
2
Element
3
Element
4
Element
5
_____, ______, ______, ______, ______, ...
e. Describe in words how you would extend the number pattern.
f. How many circles would the 60th element in the pattern have? Show your arithmetic.
g. Write an expression for the number of circles the ‘nth’ element in the pattern
would have.
h. Use the expression above to find the number of circles the 38th element would have.
5. Examine the pattern below and answer the questions that follow.
Element 1
Element 2
Element 3
Element 4
a. Draw the next element in the pattern.
b. Describe in words how the 40th element in the pattern would look.
c. Describe in words how the pattern of circles is growing.
d. Represent the pattern with numbers as shown below.
Element
1
Element
2
Element
3
Element
4
Element
5
_____, ______, ______, ______, ______, ...
e. Describe in words how you would extend the number pattern.
f. How many circles would the 60th element have? Show your arithmetic.
g. Write an expression for the number of circles the ‘nth’ element would have.
h. Use the expression to find the number of circles the 31st element would have.
6. Examine the pattern below and answer the questions that follow.
Element 1
Element 2
Element 3
Element 4
a. Draw the next element in the pattern.
b. Describe in words how the 35th element in the pattern would look.
c. Describe in words how the pattern of circles is growing.
d. Represent the pattern with numbers as shown below.
Element
1
Element
2
Element
3
Element
4
Element
5
_____, ______, ______, ______, ______, ...
e. Describe in words how you would extend the number pattern.
f. How many circles would the 70th element have? Show your arithmetic.
g. Write an expression for the number of circles the ‘nth’ element would have.
h. Use the expression to find the number of circles the 52nd element would have.
Unit 1 Section 11
Objectives

The student will convert numeric patterns into ‘nth’ terms and tables.
Often when data is collected we will look for numeric patterns. These patterns can be
based on various operations such as addition, multiplication or combinations of these
arithmetic operations. In numeric patterns the terms or elements of the patterns can be
numbered. There will be a first term, a second term, a third term and so on. The set of
counting numbers is called the Natural Numbers. The set of Natural Numbers is defined
by the roster { 1, 2, 3, 4, … }. Numeric patterns are also called sequences. The sequence
below shows how these patterns are named and represented.
an = 6, 12, 18, 24, 30, …
The “an” is the name for the sequence or pattern. The subscript “n” indicates that the
pattern can be indexed with natural numbers. This means the ‘6’ is the first term or “a1”,
the 12 is the second term or “a2”, and so on. Names for sequences or patterns can be
either capital letters or lower case letters with an optional subscript of ‘n’. Numeric
patterns can be extended and can be represented by expressions called ‘nth’ terms.
Below are some examples of numeric patterns.
Example A
2, 5, 8, 11, 14, … This pattern is based on adding three to go from
one term to the next.
Example B
bn = 5, 10, 20, 40, … The pattern is based on multiplying by two to go
from one term to the next. This pattern is
named “bn”.
Example C
An = 1, 4, 9, 16, 25, … The sequence is based on the squares of the
natural numbers.
These patterns can all be extended by using the operation that takes us from one term
to the next. In example A because we are adding three, the next two terms would be
17 and 20. Example B can be continued by multiplying by 2 so the next two terms are
80 and 160. Finally example C has the squares of the number 1 to 5 so to extend the
pattern we can square 6 and 7 which gives us 36 and 49.
Any given sequence or pattern can have an expression that will generate the terms
by substituting the natural numbers. This expression is called the ‘nth’ term.
The examples below show us starting with an ‘nth’ term and generating the first
five terms.
Example A
cn = 5n – 1
We calculate c1
c2
c3
c4
c5
Example B
=
=
=
=
=
5(1)
5(2)
5(3)
5(4)
5(5)
so cn = 4, 9, 14, 19, 24, …
–
–
–
–
–
1
1
1
1
1
=
=
=
=
=
4
9
14
19
24
tn = (2)n = 2, 4, 8, 16, 32, …
We calculate t1 = 21 = 2
t2 = 22 = 4
t3 = 23 = 8
t4 = 24 = 16
t5 = 25 = 32
Sequences or patterns can also be represented in tables. The examples below
show our two previous examples in table form.
Example A
n
Example B
cn
n
tn
1
4
1
2
2
9
2
4
3
14
3
8
4
19
4
16
5
24
5
32
These tables show the natural numbers in the left hand column. The right hand column
has the numbers from the pattern generated by ‘nth’ terms from above. We should be
able to extend the tables if we are asked to do so. In example A, the next two lines
would be 6 and 29, then 7 and 34. In example B, the next two lines would be 6 and 64,
then 7 and 128. The left hand column does not always have to be the natural numbers
in consecutive order. The tables in the next examples show tables with patterns in both
the left and right hand columns.
Example A
Example B
x
y
P
Q
1
10
2
16
4
15
5
14
7
20
8
12
10
25
11
10
____
____
____
____
____
____
____
____
____
____
____
____
In these two examples the blanks indicate that we need to extend the pattern. The
left hand column and the right hand column both need to be extended. In example
A, the extension would be 13 and 30, then 16 and 35. In example B, the extension
would be 14 and 8, then 17 and 6.
Exercises Unit 1 Section 11
1. Extend the numerical patterns below
an = 5, 7, 9, 11, _____, _____
bn = 3, 6, 9, 12, _____, _____
Cn = 100, 95, 90, 85, _____, _____
dn = -4, -6, -8, -10, _____, _____
Tn = 2, 4, 8, 16, 32, _____, _____
pn = 3, 6, 12, 24, 48, _____, _____
rn = 320, 160, 80, 40, 20, _____, _____
wn = 5, -5, 5, -5, _____, _____
Zn = 2, 7, 12, 17, 22, _____, _____
kn = 0, 11, 22, 33, _____, _____
vn = 1, 4, 9, 16, 25, _____, _____
2. Given the sequence Wn = 3, 6, 9, 12, 15, …
a. What is the
3rd term?
b. What will the 6th term be?
c. What will the 8th term be?
d. What will the 100th term be?
e. What will the general or ‘nth’ term be? _________
3. Extend the tables below:
a.
b.
c.
n
Bn
n
cn
n
an
1
6
1
12
1
2
2
12
2
9
2
11
3
18
3
6
3
20
4
24
4
3
4
29
5
____
5
____
5
____
6
____
6
____
6
____
7
7
7
d.
e.
f.
x
y
P
Q
t
w
2
10
3
6
1
0
4
20
7
14
2
1
6
30
11
22
3
2
8
40
15
30
4
3
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
4. Write an equation to show the relationship between 'x' and 'y' in table 3d.
5. Write an equation to show the relationship between 'P' and 'Q' in table 3e.
6. Write an equation to show the relationship between 't' and 'w' in table 3f.
7.
List the Natural Numbers ____, ____, ____, ____, ____, ____, ____, ____, ….
8.
The general term or ‘nth’ term of a of a sequence is an equation that
generates the sequence using _________________ of the Natural Numbers.
9. Find the first five terms of each sequence or pattern below using the
given general terms or ‘nth’ terms.
a.
an = 4n - 1
b.
bn = 2n + 3
c.
cn = -3n
d.
an = n + 4
e.
Tn = 3 + (n - 1)(4)
f.
y = n÷2
g.
v = u2
h.
P = -Q + 1
10. Create a table of values for the given general or ‘nth’ term.
a.
bn = 6n
b.
Q = 2P + 4
c. w = t ÷ 3
11. Find the relationship between the variables 'u' and 'v' in the tables below and express
it in words and as a general or ‘nth’ term.
a.
u
v
1
3
2
6
3
9
4
12
u
v
1
4
2
8
3
12
4
16
words: I can ________ the variable 'u' by _____ to get 'v'
‘nth’ term/equation
v = ______________
b.
words: I can __________ the variable 'u' by _____ to get 'v'
‘nth’ term/equation
v = ______________
c.
u
v
1
7
2
8
3
9
4
10
u
v
1
6
2
11
3
16
4
21
u
v
2
6
4
8
6
10
8
12
u
v
5
11
6
13
7
15
8
17
words: I can add _____ to the variable 'u' to get 'v'
‘nth’ term/equation
v = ______________
d.
words: _________________________________________
_________________________________________
‘nth’ term/equation
v = ______________
e.
words: ________________________________________
________________________________________
‘nth’ term/equation
v = ______________
f.
words: _________________________________________
_________________________________________
‘nth’ term/equation
v = ______________
Unit 2
Solving Linear Equations and Inequalities
Introduction
Formulas are one of the most useful tools we find in mathematics. We have all used
formulas from geometry to find perimeters, areas and volumes. There are two ways to use
formulas. The first way in which we use formulas involves substitution and simplification.
To illustrate this we will apply the formula for the volume of a sphere as given below.
V=
4
 r3
3
In this formula the ‘V’ stands for volume and the ‘r’ stands for radius. If we are given the
size of the radius we can use the formula to find the volume. An example is shown below.
If r = 10 in. then
So
V=
4
 (10) 3
3
( we substitute 10 for ‘r’ )
4
 (1000)
3
V = 4188.8 in3
V=
And using our calculator
Follow the above example and find the volume for spheres with a radius as given below:
Problem A: Find the volume of a
sphere with r = 15”
Problem B: Find the volume of a
sphere with r = 8 cm
V = ___________
V = ____________
The other way a formula is used requires solving an equation. In this situation we know how
big the sphere should be in terms of its volume and we need to find the radius. We can use
the formula with the method of guess and check to find the radius required to produce a
volume of 30 m3. We start by substituting the 30 for the ‘V’ in the volume formula.
30 =
4
 r3
3
Find the value of the radius to the nearest hundredth using the method of guess and check.
r = _____________
If we compare the two ways to use a formula which do you think is easiest?
Do you think we need a better method than “guess and check” to solve equations?
Unit 2
Vocabulary and Concepts
Equation
An equation is two equal expressions.
Ex: 3x + 2 = 5x – 24
Inequality
An inequality is a mathematical sentence that says two expressions
are not equal. Inequality symbols are >, <, >, <, ≠
Ex: 3x + 1 > 7
Isolate
To isolate a variable means to get the variable by itself on one side
of the equation.
Solution
A solution is a value that makes an equation true.
Solve
To solve an equation means to find the number that makes the equation
true.
Verify
To verify a solution is to check your answer by substituting the answer
into the original equation.
Key Concepts for Solving Equations
Goal:
The goal of solving an equation is to isolate the variable.
Method:
The method for solving an equation is to reverse the operations we see
in the original equation and the order of operations. (The keyword here
is reverse. If the equation has addition in it then we will need to subtract.)
Process:
The process for solving an equation is doing the same operation with the
same values on both sides of the equation. This means using a Property
of Equality.
Properties of Equality
Addition Property of Equality
This is used when we add the same value to both sides of an equation.
Subtraction Property of Equality
This is used when we subtract the same value to both sides of an equation.
Multiplication Property of Equality
This is used when we multiply the same value to both sides of an equation.
Division Property of Equality
This is used when we divide by the same value on both sides of an equation.
Unit 2 Section 1
Objectives

The student will explain the process of solving an equation.

The student will use the properties of equality to solve one step equations
The solving of an equation is the reverse of evaluation. To illustrate this, let’s examine the
problems below.
Example A
We start knowing the value of a variable and an expression, then we must find the overall
value of the expression by substituting and simplifying.
Evaluation:
given x = 4
with 3(x + 6)
then we substitute and simplify
3(4 + 6)
3(10)
30
Example B
When we solve an equation we start knowing the expression and the overall value of the
expression, then we find the value of the variable. An example is shown below.
Solving:
3(x + 6) = 30
…
x=4
(Notice the expression and overall value.)
(We will follow the solving process here.)
(We finish with a value for x.)
If we look at the two examples we can see that what we start and what we end with are in
reverse order!
Evaluation:
Solving:
Starts with “x =” and expression
Starts with overall value and expression
Finishes with the overall value
Finishes with “x =”
So solving means we will REVERSE the operations and the order of operations used in the
process of evaluation.
Definition:
An Equation is a mathematical sentence that says two expressions are equal.
Equations can be true or false. To illustrate these possibilities we can examine the
equations below.
Example A
42 – 1 = 3(6 – 1)
42 – 1 = 3(5)
16 – 1 = 3(5)
16 – 1 = 15
15 = 15
Example B
2(4 + 1)2 = 72
2(5)2 = 72
2(5)2 = 49
2(25) = 49
50 = 49
True!
False!
Definition:
A Solution to an equation is a value that makes the equation true.
The examples below show us how we can determine when a value is a solution to an
equation and when it is not.
Example: Determine which if either of the values below is a solution to the equation.
x2 + 1 = 3x + 5
with 3, 4
First we can try 3
Second we try 4
32 + 1 = 3(3) + 5
9 + 1 = 3(3) + 5
9+1=9+5
10 = 14
42 + 1 = 3(4) + 5
16 + 1 = 3(4) + 5
16 + 1 = 12 + 5
17 = 17
This is false, ‘3’ is not a solution!
This is true, ‘4’ is a solution!
There will be times when an equation can have two or more solutions but there will also
be times when an equation will have no solutions. (For our purposes at this time we will
see problems for which “none” of the listed values is a solution and problems where there
can be more than one solution in the list of possible values.)
Definition:
Solving an equation is the process of finding a value that makes the equation true.
Solving an equation involves a goal, a method and a process. These are:
Goal:
The goal of solving an equation is to isolate the variable.
Method:
The method for solving an equation is to reverse the operations we see in the
original equation and the order of operations.
Process:
The process for solving an equation is doing the same operation with the same
value on both sides of the equation.
First we need to understand the process using the Properties of Equality. In the
process, if we start with a true equation, then the equation must remain true at each
step from the start to the end, otherwise our result will be invalid. The examples below
illustrate the Properties of Equality.
In each example we will start with a true equation and then do the same operation on both
sides of the equation.
Example A
Example B
Example C
Example D
12 = 12
+3
+3
21 = 21
-4
-4
8 = 8
5(8)
5(8)
30 = 30
15 = 15
17 = 17
40 = 40
Addition Property
Of Equality
6 = 6
Subtraction Property
Of Equality
Multiplication Property
Of Equality
5
5
Division Property
Of Equality
If we were to add only to one side of the equation, or if we added different values to the left
and right sides of the equation, then we would end up with a false equation. So we would
not find the solution to the equation.
Solving one-step equations means we need to recognize the operation that is in the
original equation and then perform the reverse operation with the same value. The
operations of addition and subtraction are reverses and multiplication and division are
reverses. Below are examples with explanations of this process.
Original equation
Original equation
x + 12 = 21
-12 -12
(This equation has addition of 12 in it.)
(So we subtract 12 from both sides.)
x = 9
(12 and -12 add up to 0 and are gone)
3.5y = 23.45
3.5
3.5
(This equation has multiplying by 3.5 in it.)
(So we divide by 3.5 on both sides.)
y = 6.7
Original equation
a
= 18
3
(This equation has division by -3 in it.)
a .
-3 = 18 . -3
3
(So we multiply by -3 on both sides.)
a = -54
Exercises Unit 2 Section 1
Define or complete the following
1. Define Equation
__________________________________________________
2. If we want to solve an equation we want to _______________ the operations
we see in the original equation.
3. The goal of solving an equation is to _______________________________.
4. If an equation is true then to keep it true we should _____________________.
5. Decide whether each equation is true or false.
a. _____ -2 + 7 = 32 - 4
b. _____ 6 - 16 = -5(3) + 5
c. _____ 1 + 42 = 1 + (3)(5)
d. _____
-6 + (3)(3) = 22
6. What does it mean to SOLVE an equation? ___________________________.
7. Decide which of the listed numbers is a solution to the equation.
a. _______
2x + 4 = 14
c. _______
3a - 1 = 3a + 5
5 or 6
b. _______
1 or 3
x2 + 5 = 9
d. _______ x + 2 = 2x – 1
1, 2, or -2
2, 3, or 4
Solve the following equations and list the property that is used. Show your WORK.
8. y + 6 = 23
9.
z - 11 = 33
10. 34 = z + 15
11. -17 = a - 9
12. w - 15 = 22
13. a + 6 = -20
14. a + 24 = 24
15. 5 = z – 5
16. 28 = 4w
17.
-4x = 36
18.
20.
-18 =
23.
1
2
y
2 .5
= a -
1
4
26. y - 1.09 = -1.09
a
2
= 18
19. 11 =
21.
3t = -20
24.
-10 = -2x
27.
-24 = 5a
z
9
22. y - 4.56 = -6.2
25.
2
v
=
5
3
Solve the following equations and CHECK your solutions. Show your WORK.
28.
12a = 108
29. -23 = x – 4
30.
55 = -10y
31.
a
4
= -11
Translate and solve:
32.
6 less than a number is 18. Find the number.
33.
If a student finds of
34.
The quotient of a number and 5 is -6. What is the number?
35.
Four times a number is 50. Find the number.
1
a number, the result is 7.
3
Unit 2 Section 2
Objectives

The student will explain the process of solving a two-step equation.

The student will use the Properties of Equality to solve two-step
equations.
The solving of an equation is the reverse of evaluation. In this section we will examine
how the order of operations must also be reversed to solve two step equations.
In the order of operations used to evaluate expressions we always do any multiplication
or division before we perform the addition or subtraction. Below is a list of steps we will
follow to solve multi-step equations. This list will be expanded as we continue through
Unit 2. This is our first list of steps there will be others that are more complete.
The Algorithm for Solving Multi-Step Equations
1. Reverse any addition or subtraction we see in the original equation.
2. Reverse any multiplication or division we have left in the equation.
When we solve equations we will often be asked to list the properties that we use
to transform the equation step by step. The examples below show the method and the
listed properties in the solution process.
Example A
2x + 11 = 37
- 11
-11
Subtraction Property of Equality
2x = 26
2
2
Division Property of Equality
x = 13
Checking the answer
2(13) + 11 = 37
26 + 11 = 37
37 = 37
In the above example the addition of 11 in the original equation was reversed by
subtracting 11 from both sides. This means we used the Subtraction Property of Equality.
And the step is labeled appropriately. The second step in solving the equation is to
reverse the multiplication by 2. We did this by dividing both sides of the equation by 2.
This step is also labeled appropriately as the Division Property of Equality. Note also that
the equal signs are lined up to help keep our work straight.
Example B
w
- 3.6 = -4.5
 2 .2
+ 3.6
+ 3.6
Addition Property of Equality
w
= -0.9
 2 .2
-2.2 .
w
= -0.9
 2 .2
.
-2.2
Multiplication Property of Equality
w = 1.98
Checking the answer
1.98
- 3.6 = -4.5
 2 .2
 0.9 - 3.6 = -4.5
-4.5 = -4.5
Example C
7
x +2 = 6
3
-2
-2
3.
7
x
3
= 4
7
x
3
=4.3
7x
= 12
7
7
x =
Subtraction Property of Equality
Multiplication Property of Equality
Division Property of Equality
12
7
Checking the answer
7 . 12
+2 = 6
3
7
4 + 2 = 6
6 = 6
Note that there are other ways to solve Example C but clearing the denominator
by using the Multiplication Property of Equality is one of the most straight forward.
Exercises Unit 2 Section 2
Solve the following equations. Show all the WORK. List the Property of Equality for
each step and check your answers for the equations in this section.
1. 3x + 2 = 23
4. 5x - 7 = 3
7. 8 = -2y - 16
2.
14 = 2x - 12
3.
4=
z
-2
7
z
+ 4 = -9
2
5. -33 = 6x + 3
6.
8. 4x - 14 = 0
9. 3z + 31 = 39
Solve the following equations. Show all the WORK.
10. 13 =
13.
a
+3
9
w
+ 3.2 = 2.6
 1.3
16. a - 71 = 5
11. 34 = -x - 2
14. -
4
2
16
x=
5
3
3
17. -3.8 =
12.
48 = 11y + 48
15.
-9x = 18
a
 6 .5
18. Define EQUATION.
19. What is the expression on the right side of the equation?
5x2 – x = -11x + 3
20. Decide whether each equation is true or false.
a. _____ (2)(4) + 1 = 32
b. _____ 40 - 6 = 5(7)
21. What does it mean to SOLVE an equation?
22. What is the SOLUTION of an equation?
Write mathematical expressions for each problem.
23.
seven less than a number 'x'
24.
five more than twice a number 'y'
25. half the sum of eight and a number 'z'
Translate and solve the following problems:
26. The product of 3 and a number decreased by 7 is 11. Find the number.
27. When a student divided a number by 2 and increased that by 5, the result is -8.
What is the number?
28. Peter subtracted 3 from the product of 4 and a number, the result was 17.
What was the number?
Unit 2 Section 3
Objectives

The student will list the algorithm for solving a multi-step equation
and explain them.

The student will combine like terms on each side of an equation before
solving a two-step equation.
Definition:
Terms are considered to be "like" when they have the same variables with
the same exponents.
The examples below contain terms that are considered to be like and some that are not
like. We will apply the definition above in finding the sets of like terms in each
expression.
Example A
3x + 2y - 5x + y
In the above expression the "3x" and the "-5x" are like since they have the
same variable. Also the "2y" and the "-y" are like for the same reason.
Example B
9 - 2.5a + 6 - a - 4
In the above expression the "-2.5a" and the "-a" are like since they have
the same variable. Also the "9", "6" and the "-4" are like because they are
constants or plain numbers. ( They do not have a variable. )
Example C
4z2 + 12 - 3z + 7z2 + 8z
In the above expression the "4z2" and the "7z2" are like since they have
the same variable with the same exponent. The "-3z" and the "-5z" are
like because they also have the same variable with the same exponent (the
understood exponent of one ). The constant 12 does not have a matching
like term. The terms with "z2" and "z" are not like since they have
different exponents.
Combining terms is done by adding the coefficients of like terms. In Example C the
simplifying would be:
4z2 + 12 - 3z + 7z2 + 8z
11z2 + 5z + 12
The 4z2 plus the 7z2 is 11z2 and the -3z + 8x is 5z with the 12 remaining unchanged.
We will frequently need to simplify equations as we solve them by combining like
terms. This will add a step to our algorithm for solving a multi-step equation.
The Algorithm for Solving Multi-Step Equations
1. Simplify the equation by combining like terms as needed.
2. Reverse any addition or subtraction we see in the original equation.
3. Reverse any multiplication or division we have left in the equation.
The examples below show equations solved including the combining of like terms.
Example A
3x + 11 - x = -17
2x + 11 = -17
- 11 - 11
( We could insert the understood coefficient: 3x + 11 - 1x = -17 )
( 3 - 1 is 2 so we have 2x )
Subtraction Property of Equality
2x = -28
2 = 2
Division Property of Equality
x = -14
Example B
x + x + 15 +2x +4 = 3x + 14 - x - 2x
4x + 19 = 14
- 19 - 19
( We combined all the like terms on each side. )
Subtraction Property of Equality
4x = -5
4 = 4
x =-
Division Property of Equality
5
4
In example B it is critical to note that we cannot combine terms across the equal sign.
Each side of the equation MUST be simplified separately!
Exercises Unit 2 Section 3
Solve the following equations. Show all your WORK CORRECTLY
1. 4x - 7 = 21
2. 24 = -3y - 9
3. 4 =
z
- 7
5
4. –x + 4 = 15
5. -8z + 4 + x = -24
6. 10 =
1
x+4
3
7. 8 = y + 9 + y
8. 1.7x - 15 + .8x = 10
9. 16 = a - 2 - 8a
10. 21 = -x + 5 + 3x
11. z + z + z + z - 3 = -1
12.
13. 20 = 4y + 12 + 8 + y
14. 6x + 15 = x + 27 - x
15. 2a + 7 - 3a = 11
16. 2x + 2 - x - x = 15
17. x2 + 2x - x2 - x2 + 8 = 5
18. 18.7 = 3.4a + 7.5 - 2.1a + 6
1
1
1
y+ y-4 + y=5
2
2
2
20. 4a + 17 + 2a + 1 - 11a = 28
19.
2
1
= 6z 5
5
List the properties of equality you would use to make the top equation into the bottom
equation in each problem. Write the name of the property in FULL.
a
21. 12x + 5 = 41
22.
= -5
4
12x = 36
a = -20
________________________________
23. 5x - 13 = 70
________________________________
24.
5x = 83
________________________________
-24 = -3x
8=x
________________________________
25. Describe how we can check our answer after we solve an equation.
26. What does it mean to SOLVE an equation?
27. What is the goal of solving an equation?
Unit 2 Section 4
Objectives

The student will state and apply the Distributive Property

The student will use Distributive Property in solving multi-step equations.
The Distributive Property is one of the most useful concepts we find in Algebra. When
we evaluate expressions we must do operations that are in parentheses first. However when
one of the values in the parentheses is a variable it is not possible to perform the operation
in the parentheses first. If we are multiplying the parentheses by a value we can still simplify
the expression using this Distributive Property. The property is shown below.
The Distributive Property
a(b +c) = ab + ac
The Distributive Property allows us to multiply through parentheses.
Below are examples of how the Distributive Property can be used.
Example A
3(x + 5)
The arcs show our multiplication through the parentheses.
3x + 15
We first multiply 3 times 'x' and get '3x', then 3 times 5 for 15.
Example B
8(x - 11)
8x - 88
8 time ‘x’ is ‘8x’
8 times -11 produced the -88.
Example C
-2(3a - 7)
-6a + 14
-2 times ‘3a’ is ‘6a’
The process is the same but we must be careful of the signs.
Example D
z(z - 2y)
z2 - 2yz
z time ‘z’ is z2
We can multiply by variables as well as constants.
Example E
-(4w - 9)
-1(4w - 9)
-4w + 9
In this example we used the understood coefficient of one
in order to make our multiplying easier. The overall effect
of the negative is to change the signs of all the terms in the
parenthetical expression.
There will be times when our equations will contain parentheses. This will mean once
again that we need to add to our list of steps in the algorithm for solving equations.
The Algorithm for Solving Multi-Step Equations
1.
2.
3.
4.
Use the Distributive Property to remove parenthetical expressions.
Simplify the equation by combining like terms as needed.
Reverse any addition or subtraction we see in the original equation.
Reverse any multiplication or division we have left in the equation.
The examples below show how we use the Distributive Property in solving equations.
Example A
4(x + 12) = 36
4x + 48 = 36
- 48 -48
4x = -12
4
4
x = -3
This is where we used the Distributive Property
Subtraction Property of Equality
Division Property of Equality
Example B
-2(x - 5) + 3x = 17
-2x +10 + 3x = 17
x + 10 = 17
- 10 -10
x= 7
This step is where we used the Distributive Property.
Here we simplified the left hand side by combining terms.
Subtraction Property of Equality
Example C
3(2x + 7) + x - 1 = -x -(4 - x)
6x + 21 + x - 1 = -x - 4 + x
7x + 20 = -4
- 20
-20
7x = -24
7
7
24
x=7
We used the Distributive Property on both sides!
We simplified both sides.
we used the Subtraction Property of Equality.
We used the Division Property of Equality.
Exercises Unit 2 Section 4
Solve the following equations. Show all your work correctly. Show also the checks
for your answers in this section.
1. -4(m + 12) = 36
2. 3(y + 8) - 7 = 33
4.
-35 = 5(3k + 2)
5.
7.
6x -2(3x + 1) = x + 4
8. 12 = 2(z - 5)
1
(4x - 8) = 17
2
3. -(a + 4) + 6 = 10
6. -24 = 2a - 6(a - 12)
9. x + 9 - x = 3(2x -1)
Solve the following equations. Show all your work correctly.
10. -2(x - 3) + 2x - 6 = x - 1
11. 5 = 3(4a + 2)
12. -8 = 2(y - 5) + 10
13. -(2x - 21) + 2x = 4x - 5
14. 2a + 5(a + 3) = 32
15. 2.5(x + 7) = 28.75
16.
15 =
3
1
(z - 1) + z
2
2
Translate and solve:
17. The sum of a number and 2 multiplied by 3 is 18. Find the number.
18. Tina solves the equation 8(x - 2) + 2x = 74. What is the value of x?
a) 10
b) 5.8
c) 9
d) 7.6
Unit 2 Section 5
Objective

The student will solve equations with variables terms on both sides.
There will be equations in which there are variable terms on both sides of the equal sign.
When variable terms occur on the same side of the equal sign we can combine them
because the terms will be either positive or negative and their coefficients can be added.
However, when the terms are on opposite sides of the equal sign we have a different
situation, as illustrated below.
Example A
Example B
x+x=8
2x = 8
x=x
In example A the plus sign between the variables tells us exactly what to do, so we can
add the x's and solve the equation. In example B the equal sign creates an equation that
can be true or false but there is NO arithmetic to do! So we can say the equation is true
but we can't add or subtract the x's.
The goal of solving an equation is to isolate the variable on one side of the equation.
When an equation has variable terms on both sides we will need to find a way to eliminate
one of these terms. We can use the properties of equality to make one of the two terms
go away. The example below shows us how to use the Subtraction Property of Equality
to turn a more complicated equation into a simple two-step equation.
5x + 7 = 3x + 25
- 3x
- 3x
2x + 7 = 25
-7
-7
2x = 18
2
2
x=9
Subtraction Property of Equality
Subtraction Property of Equality
Division Property of Equality
In the example above the positive ‘3x’ and the ‘-3x’ add up to zero and are gone. The ‘5x’
minus the ‘3x’ is ‘2x’ and this creates a two-step equation which we can routinely solve.
To solve equations with variable terms on both sides of the equal sign we should always
use the Addition or Subtraction Properties of Equality to eliminate the smaller of the
two variable terms.
One of the keys to solving equations with variables on both sides is recognizing which term
is smaller. To illustrate this idea we will examine the pairs of terms below, some are simple
but others we need to examine closely.
Example A
3x
5x
Example B
Example C
Example D
-8x 2x
-4a -11a
-6y 6y
In example A, both of the coefficients are positive and the smaller term is '3x'. In example B,
we must keep in mind that any negative number is less than a positive value so '-8x' is
smaller. With example C we have two negative coefficients but -11 is smaller than -4 so
the '-11a' is the smaller of the two terms. The final example D shows numbers that initially
look similar but again the negative is always less than the positive so '-6y' is the smaller term.
The example below shows another equation with variable terms on both sides being solved.
2y - 31 = -9y + 24
+ 9y
+9y
11y - 31 = 24
+31 +31
11y = 55
11
11
y=5
Addition Property of Equality
In the above example the '-9y' was smaller than the '2y' so to cancel the -9y out we had to
use the Addition Property of Equality and add a positive 9y to both sides. By doing this we
eliminate the ‘y’ term on the right hand side of the equation.
The situation with variable terms on both sides of the equal sign means we need to add one
more step to our algorithm for solving multi-step equations. The new list is given below.
The Algorithm for Solving Multi-Step Equations
1.
2.
3.
4.
5.
Use the Distributive Property to remove parenthetical expressions.
Simplify the equation by combining like terms as needed.
Use Properties of Equality to eliminate extra variable terms.
Reverse any addition or subtraction we see in the original equation.
Reverse any multiplication or division we have left in the equation.
The example below shows us how to apply the algorithm for solving multi-step equations.
4(2a + 1) - 6a + 3 = -(a + 19)
8a + 4 - 6a + 3 = -a - 19
2a + 7 = -a - 19
+a
+a
3a + 7 = -19
- 7
- 7
3a = -26
3
3
26
a=3
First we apply the Distributive Property
Second we combine like terms
Third use the Addition Property of Equality
Fourth use the Subtraction Property of Equality
Fifth use the Division Property of Equality
Exercises Unit 2 Section 5
Solve the following equations. Show all the work correctly. If there is no solution, write
NO SOLUTION.
1. 4x + 5 = 7x - 28
2.
5a + 7 = a - 41
3. 2 = -5(a + 2)
4.
5(a + 3) = 27 - a
5. 2(x + 7) = 2x + 14
6.
5y - 2 = 2(3y + 9) - 1
7. z + 7 = z - 12
8.
-2w + 33 = 4(3w - 3) + w
9. -y + 16 = 5(y - 4)
10. 4x + 25 = 7(x - 2) + x - 3
11. 3(a + 15) = 2(a + 1)
12. x - 21 = x + x + x + 49
13. 3a + 85 + a = 85
14.
-11 = y + 20
a
16. 11 =
+7
3
18. -(x + 4) + x = x + 15
15. -3a + 51 = 0
17.
14 - 3x = 23 - 7x
19. Which of the values below is the solution to the equation 3x + 4 = 25
a) 6
b)
7
c) -7
d) no solution
20. Which of the values below is a solution to the equation
a) 5
b) -5
c) 15
d)
-2x + 8 = x - 7
15
2
21. Which of the values below is a solution to the equation 4x - (x + 2) = 3(5 - x) - 12
a)
29
6
b) -
5
6
c)
5
6
d)
7
6
Unit 2 Section 6
Objectives

The student will define and find the absolute value of real numbers.

The student will solve equations with absolute values.
There are problems that we will need to solve that must have only positive values as
answers. In Algebra we have an operation called the absolute value function that
produces positive answers.
Definition:
The absolute value of a number is the positive distance between the number
and zero.
We use two vertical bars (
|| )to indicate when we want to find the absolute value of a
number. Below are examples of the notation used for the absolute value.
Example A
Example B
Example C
|10|
|-2.4|
|x |
Example D
|z - 3 |
The examples above show the notation used to tell us to find the absolute value of '10',
'-2.4', 'x' and the expression 'z - 3'. One way to find the absolute value of a number is
to count the number of units between the number and zero. Below are examples of
using the real number line to find the absolute value.
Example A Finding |4|
-7
-6
-5
-4
-3
-2
-1
as we count
0
1
1
2
2
3
3
4
4
5
6
7
So the |4| = 4
Example B
Finding the |-5|
-7
-6
-5
-4
as we count
Example C
-3
-2
-1
0
1
2
3
4
5
-1
0
1
2
3
1
2
3
4
5
6
So the |-5| = 5
Finding the |2.5|
-7
-6
-5
-4
-3
-2
as we count
1
2
.5
4
5
6
7
So the |2.5| = 2.5
From these examples we can conclude two things. First the absolute value of any
number is always POSITIVE. Second it does not matter if the value is an integer,
fraction, decimal, or irrational number, the answer maintains all the digits. This means
finding the absolute value of number is relatively straight forward.
To find the absolute value of a number we must make the value in the vertical
bars positive. Whether the number starts as positive or negative, the answer is
positive.
The examples below illustrate this principle.
|11| = 11
|-43| = 43
|-8.25| = 8.25
|-| = 
3
3
We will find at times that there are absolute values in the equations we need to solve.
An absolute value in an equation can cause an equation to have more than one solution
or it can also mean there will be no solution. The examples below illustrates these
possibilities.
Example A
Given the equation |x| = 9, both x = 9 and x = -9 are solutions
since both values make the equation true. We can demonstrate
this by substitution as shown below.
|-9| = 9
Example B
|3 1 | = 3 1
and
|9 | = 9
Given the equation |x| = -3, we CANNOT find a solution. Since
the answer to an absolute value problem is always positive there
is no way to get a negative answer.
7
The fact that we can have two solutions means we must actually treat an equation that
contains an absolute value function as two separate equations. We create these two
separate equations because the number in the absolute value could be positive or
negative and make the equation true. The examples below show how to create the two
equations.
|2x + 3|
Example A
= 13
2x + 3 = 13
-3
-3
2x + 3 = -13
-3
-3
2x = 10
2x = -16
2
2
2
x = 5
2
x = -8
The solutions are x = 5 and x = -8. Frequently we will show the answer in
what is called a solution set {-8, 5}.
These answers should also be checked as with all solutions to equations.
|2x + 3|
|2(5) + 3|
|10 + 3|
|13|
|2x + 3| = 13
|2(-8) + 3| = 13
|-16 + 3| = 13
|-13| = 13
= 13
= 13
= 13
= 13
13 = 13
13 = 13
When an equation has terms or factors outside the absolute value sign, these must
be reversed before we create our two separate equations. (These outside operations
would be done last when evaluating and so must be reversed first when solving.)
Example B
3|z| + 5 = 26
-5
-5
Subtraction Property of Equality
3|z| = 21
z=7
3
3
|z| =
7
Division Property of Equality
The absolute value is isolated so
we create the two separate equations.
x = -7
So the solution set is { -7, 7 }
( We should check the solutions. )
Example C
In this example we must first isolate the absolute value before we can
create our two equations!
|x - 5| + 3x =
-3x
0
-3x
|x - 5| = -3x
x - 5 = 3x
-x
-x
x - 5 = -3x
-x
-x
- 5 = 2x
- 5 = -4x
2
2
x = -
-4
5
2
so the solution set could be
-4
5
= x
4
{- 5 , 5 }
2
4
BUT we still need to check our answers.
|- 5
2
|x - 5| - 3x = 0
. 5
- 5| - 3 - = 0
|x - 5| - 3x
| 5 - 5| - 3 . 5
2
|- 15 | + 15 = 0
2
2
15
15
+
=0
2
2
30 = 0
2
4
=0
=0
4
|  15 | - 15 = 0
4
4
15 15
=0
4
4
0 = 0
15 = 0
Only one of the solutions is valid! The - 52- did not give us a true
equation when we checked the answer! It is called an "extraneous"
solution and cannot be part of our solution set.
So the solution set is
{5 }
4
Exercises Unit 2 Section 6
Solve the following equations. Show all the work correctly.
List your answers as solution sets. If there is no solution, answer NO SOLUTION.
1.
|x| = 3
2.
|x| = -2
3.
4.
|y| - 2 = 8
5.
|z| + 10 = 28
6. -7 + |s| = 0
7.
6 - |t| = 14
8.
|a| - (-2) = 4
9. | 2y + 1 | = 7
10. | -3x + 2 | = 5
11.
13. –(|x| + 2) = -6
14.
16. -6|4x - 1| + 7 = -179
17. -3 – (15 - |a|) = 12
18. 2|x| + 5 = 19
19.
20. |w| + 3 = 17
2
21. 10 – 4|x| = -2
3
23.
24. |3/4 – x| = 6
22.
5 = -3|a| + 17
7 – 3|x| = 1
2
2 = 6 + |t|
|t| = 13
12. -|x| = -14
4 – (2 - |n|) = 2
15. 3|x + 1| = 36
4|t| = 16
7
25. (9|x| + 3) – 5|x| - 3 = 12
26. |x + 2| = 3x
27. |3x - 1| + x = 2x + 13
28.
|4x - 1| = |5x + 2|
29. The solution set for the equation 2|x + 5| = 16 is
a) {3,9}
b) {3, -13}
c) {13}
d) {9, -13}
30. The solution set for the equation 5 - |2x + 7| = x is
a)
{2,

2
}
3
b)
{-12}
c)
{-12,

3
}
2
d)
{-12,

2
}
3
d)
{-4,
22
}
7
31. The solution set for the equation |6x + 1| = |x + 21|
a)
{4,

22
}
7
b)
{4 }
c)
{  22 }
7
Unit 2 Section 7
Objectives

The student will explain why solving equations for a given variable is
helpful.

The student will isolate a variable in formulas using the properties of
equality.
The formula for the perimeter of a rectangle is given as P  2l  2w . There is a
formula from basic geometry which we have used in the past. If we know that the
length is 4 ft and the width is 3 ft then we can substitute these values into the formula
easily and find the perimeter as shown below.
P  2l  2w
P  2(4)  2(3)
P  86
P  14 ft
P
l

 1.
2w
w
In this form finding the perimeter is much more difficult.
However this formula could also be written as
P
l

1
2w
w
P
4
 1
2(3)
3
P
4

1
6
3
P
7

6
3
P  14 ft
When we need to use a formula it is very helpful if one of the important variables
is isolated. In this section we will be using the Properties of Equality to isolate a
variable in a formula or equation. Once we have isolated the desired variable we
can then use the new equation to find quantities by substitution.
The process of isolating a given variable in a formula or equation is called "solving a
literal equation". The examples below show us how to use the Properties of Equality.
Example A
Given the equation
Isolate h
V  lwh
( Volume = length . width . height )
V  lwh
V
lwh

lw
lw
V
 h
lw
Using the Division Property of Equality
Example B
C  p  tp
Given the equation
Isolate t
C  p  tp
p
( Cost = price + tax rate . price )
Using the Subtraction Property of Equality
p
C  p  tp
p
Using the Division Property of Equality
p
Cp
t
p
Example C
Given the equation V 
Isolate h
V 
3 V 
 r 2h
3
(Volume = (  . radius2 . height) / 3 )
( This is the Volume of a cone. )
 r 2h
3
 r 2h
3
3
Using the Multiplication Property of Equality
3V   r 2 h
3V  r 2 h

 r2  r2
3V
h
 r2
Using the Division Property of Equality
The equations we will be given do not have to be from geometry or finance.
The example below is of a more general nature.
Example C
Given the equation
y=
2
x+c
3
Isolate x
2
x+c
3
-c
-c
2
y-c= x
3
2
3 . (y - c) = x . 3
3
3y - 3c = 2x
y=
Using the Subtraction Property of Equality
Using the Multiplication Property of Equality
Using the Distributive Property
3y - 3c = 2x
2
2
Using the Division Property of Equality
3 y  3c
=x
2
If after we isolate x were we told that y = 4 and c = 8 we could use
substitution and find the value of x as shown below.
3 y  3c
=x
2
3( 4)  3(8)
=x
2
12  24
=x
2
 12
=x
2
-6 = x
Since we had x isolated the arithmetic to find its value was simple substitution.
Exercises Unit 2 Section 7
Solve for 'x' in each equation. Show your work correctly.
1.
x–k=g
2. x + 3c = b
4.
x
z
a
5.
r = sx + t
8.
nx – m = -p
9. 5x – 8a = 2x + 7a
11.
b b x
  0
4 5 9
12. 4x + b = 0
7.
10.
x
3

a
4

b
6
–3 =
1
xy
3. –bx = d
6.
ax + b = c
Solve the following problems, show all work and answer all questions in full sentences
where appropriate.
13. Given the formula for the area of a triangle is A = 1/2 bh with 'A' the area, 'b'
the base and 'h' the height, answer the questions that follow.
a. Solve the formula for 'b' and explain how you found the answer.
b. If the area (A) of a triangle is 50 cm2 and the height is 4 cm use the equation
from part 'a' above and find the base (b). Show your work algebraically and
explain how you found your answer.
c. Check your work for part 'b' algebraically by using the area and the height in
the original equation and find the base. Show your work.
14. Given the formula for the converting from Fahrenheit to Celsius is C = 5/9 (F - 32)
with C the temperature measured in Celsius and F the temperature measured in
Fahrenheit, answer the questions that follow.
a. Solve the formula for 'F' and explain how you found the answer.
b. If the temperature is 40o Celsius us the equation from part 'a' to convert it into
a temperature in Fahrenheit. Show your work algebraically and explain how
you found the answer.
c. Check your work algebraically by using the 40o and the original equation and
solving for 'F'. Show your work.
15. Given the formula for the Circumference of a circle is C = 2r with 'C' the
Circumference and 'r' the radius, answer the questions that follow.
a. Solve the formula for 'r' and explain how you found the answer.
b. If the Circumference of a circle is 320 cm use the equation from part 'a'
above and find the radius. Show your work algebraically and explain how
you found your answer.
c. Check your work algebraically by using the circumference in the original
equation and find the radius. Show your work.
16. The surface area (A), of a right circular cylinder with radius r and height h is
given by
A = 2r 2 + 2rh.
a. Solve the formula for h. Show your work algebraically, and explain how
you found your answer.
b. Use your equation from part 'a' to find the height of a right circular cylinder
with surface area of 30 cm2 and radius 3 cm. Show your work algebraically,
and explain how you found your answer.
c. Check your answer to part 'b' by using the original equation to find the height
of the cylinder. Show your work algebraically.
Solve the following equations. Show all the work correctly.
17. y + 12 = -3(y - 4)
18. 2x + 15 = 9(x - 2) - x - 3
19. 2(a - 18) = 4(a + 3)
20. x + x - 1 = x + 2x + x + 8
21. 3a + 5 + 2a = 4a + 5
22.
23. -2a + 41 = 39
2
1
= y 3
5
a
24. 17 =
+ 21
3
25. |z| + 5 = 19
26. |2x + 4| = 14
27.
28.
-3|x + 2| + 1 = -29
|x + 3| = |3x - 7|
Unit 2 Section 8
Objective
 The student will solve and graph linear inequalities in one variable.
There are times when the problems we solve do not require equations but inequalities
instead. These kinds of problems would be similar to:
If Sam is paid $200 dollars a week and $15 for every car stereo he installs, find the
number of stereos he must install to make more than $650 a week.
The phrase “more than” means that there will be a range of values beginning with 31 and
going up from there for which Sam can make more than $650. We could say that Sam
must install more than 30 stereos. This can be expressed symbolically with x > 30.
There are four inequality symbols that we will use in Algebra I, these are:
Symbol
>
<
>
<
Meaning
greater than
less than
greater than or equal to
less than or equal to
Inequalities can be true or false just like equations can be true or false. The examples
below show these possibilities.
Example A
9>5
We translate this into words as “nine is greater than five”.
This inequality is true.
Example B
-6 > 5
Example C
5>5
We translate this into words as “five is greater than five”.
This inequality is false.
Example D
2<5
We translate this into words as “two is less than or equal to five”.
This inequality is true.
Example E
6<6
We translate this into words as “six is less than or equal to six”.
This inequality is true.
We translate this into words as “negative six is greater than five”.
This inequality is false.
We found out in Unit 1 that graphs help us to interpret many ideas and relationships in
mathematics. When we work with inequalities graphs can be an important way to
visualize our answers. Inequalities are graphed on the real number line. We will use
a method called “split point and test point” to graph our inequalities. Below are
examples of this method.
Example A
Given the inequality x > 3 the ‘3’ is the split point. The ‘3’ will
divide the real number line into two regions, a region to the left
of ‘3’ and a region to the right of ‘3’. One of these regions
contains numbers that will make the inequality true and the other
region has values that will make the inequality false.
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
Our first step in the process is to find ‘3’ on the real number line and place
an open circle on it. This circle is open in order to indicate that the ‘3’ is
the split point but it will not make the inequality true. ( 3 > 3 is false. )
Next we pick a test point. This can be any point other than ‘3’, but we
should choose a number that is easy to work with like ‘1’. Then we
substitute this value into the inequality.
1>3
This inequality is false, so the left region does not make the inequality
true. We can now pick a point from the right side region like ‘5’ and
substitute the ‘5’ into the inequality.
5>3
The inequality is now true, so the right region contains all the values that
make the inequality true. So we shade the right side of the real number
line as shown below.
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
The shading is a dark line with an arrow indicating that all the numbers to
right from 3 to infinity will make the inequality true.
Example B
Given the inequality y < -2 the ‘-2’ is the split point. The ‘-2’ will
divide the real number line into two regions, a region to the left
of ‘-2’ and a region to the right of ‘-2’. One of these regions
contains numbers that will make the inequality true and the other
region has values that will make the inequality false.
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
Our first step in the process is to find ‘-2’ on the real number line and place
a closed circle on it. This circle is closed in order to indicate that the ‘-2’
is the split point and it will make the inequality true. ( -2 < -2 is true. )
Next we pick a test point. This can be any point other than ‘-2’, but we
should choose a number that is easy to work with like ‘1’. Then we
substitute this value into the inequality.
1 < -2
This inequality is false, so the right region does not make the inequality
true. We can now pick a point from the left side region like ‘-4’ and
substitute the ‘-4’ into the inequality.
-4 < -2
The inequality is now true, so the left region contains all the values that
make the inequality true. So we shade the left side of the real number
line as shown below.
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
The shading is a dark line with an arrow indicating that all the numbers to
the left from -2 to negative infinity will make the inequality true.
Example C
Given the inequality a < 1.5 the graph for this inequality is shown below.
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
We estimate the location of the circle for the split point ‘1.5’ as accurately
as possible.
Example D Given the inequality w > -4
-7
-6
-5
-4
-3
-2
2
the graph for this inequality is shown below.
3
-1
0
1
We estimate the location for the split point ‘-4
2
3
4
5
6
7
2
’ as accurately as possible.
3
We have looked at graphing the simplest of inequalities. Inequalities however can
have expressions as well as isolated variables. Below is an example of an
inequality that has more complicated expressions.
3(2y -1) + 5 > 7y + 6
In order to graph this inequality we must first isolate the variable. When the
variable is by itself the inequality will look exactly like the problems in the
previous examples A through D. In the previous examples the split point was
easy to locate. To change the complicated expression into a simple inequality
we use the Properties of Inequality. The Properties of Inequality operate in
a similar way to the Properties of Equality. This means that whatever we do to
one side of the inequality we must do the other side as well. There is one
difference in the properties of Inequality that we must investigate.
The inequality -2 < 3 is true. We can demonstrate this by plotting the points
on the real number line as shown below.
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
If we take the original inequality and we multiply -1 times both values, let’s see
where they end up on the real number line.
(-1)(3)=-3
(-1)(-2)=2
-7
-7
-6
-6
-5
-5
-4
-4
-3
-3
-2
-2
-1
-1
0
0
1
1
2
3
4
5
6
7
2
3
4
5
6
7
Our two points ended up switching places on the real number line which means
when we multiply or divide by any negative number the values switch places
and so we must also switch the inequality symbol as shown below.
-2 < 3
(-1)(-2) < (3)(-1)
2 > -3
The Properties of Inequality tell us that we must always perform the same operation
on both sides of the inequality and IF we multiply or divide by a negative value then
we must switch the inequality symbol.
Below are some examples of solving inequalities using the Properties of Inequality.
Note how similar this process is to solving an equation.
Example A
3x + 11 > -4
- 11
-11
3x > -15
3
3
x > -5
Subtraction Property of Inequality
Division Property of Inequality
When we are graphing a more complicated inequality it is always a good
practice to substitute our test points into the original inequality. We will
put an open circle on '-5' and then test the value '1' as below.
3(1) + 11 > -4
14 > -4
is true so we shade the right side of the real number line. and our graph is:
-7
-6
Example B
-5
-4
-3
-2
-1
0
-4(x + 1) -2 > -14
-4x - 4 - 2 > -14
-4x - 6 > -14
+6
+6
-4x > -8
-4
-4
x<2
-7
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
7
Distributive Property
Combining like terms
Addition Property of Inequality
Division Property of Inequality
We divided by a negative value so
we had to switch the sign.
0
1
2
3
4
5
6
7
Example C
5x - 6 > 7x - 4
-5x
-5x
-6 > 2x - 4
+4
+4
-2 > 2x
-2
-2
-1 > x
-7
-6
-5
-4
-3
-2
Subtraction Property of Inequality
Addition Property of Inequality
Division Property of Inequality
-1
0
1
2
3
4
5
6
7
Exercises Unit 2 Section 8
Decide which of the listed values from the given set makes each inequality true and list
that as your answer.
1.
{-9, 4, 8}
3. {-2, 0, 1, 5}
x>6
2.
{-7, -1, 0, 3}
a<0
4. {-2, -1, 0, 1}
-1 > y
-1.2 > z
Graph the inequalities below.
1
< x
2
5. x > 4
6. y < -2
7. -1.25 > a
8. 3
9. z > -1
10. z < -5
11. 0.9 < a
12. -
1
> w
4
Solve and graph the following inequalities. Show your work correctly. If there is a false
inequality, show your work to prove it is false.
13. 4x > 12
14. 16 > y + 10
15.
2x - 7 < 4
16. -2x < 8
17.
2y - 5 + y > 10
18.
-8 < 4(w + 4)
19. x + 1 > x
20.
y-1>y+1
21.
5x + 2 < -2(x + 6)
22. 3x + 1 < x
23.
1 < 5y + 1
24.
3x + 4 > -(x + 1) + 2x
25. -4x < 0
26.
41 < 3y + 2
27.
7(x - 4) > 3(x + 2) + 2x
28. Which of the following is the graph of the solution set for
-6x + 14 > 2
a.
0
1
2
3
4
b.
0
1
2
3
4
c.
0
1
2
3
4
d.
-4
-3
-2
-1
0
Unit 2 Section 9
Objective

The student will solve and graph inequalities with conjunctions and
disjunctions.
There are two key words in logic and mathematics. These are words we frequently use in
every day conversation. Their meanings and uses in mathematics are very special however.
The two words are "or" and "and". These two words can be used to link two inequalities
together. The mathematical sentence created by using the word "or" is called a
disjunction. The mathematical sentence created by using the word "and" is called a
conjunction. We will begin by investigating the disjunction.
A disjunction is formed by linking two inequalities with the word "or". A disjunction
will be true when a value makes either of the two inequalities true.
To illustrate this principle we need to examine the statement below.
The assembly will be on Thursday or the assembly will be on Friday.
Here we have two statements that can be true or false joined by the word "or". This is a
disjunction. As long as one of the statements is true, the overall sentence is true. If both of
the statements are false, then the overall sentence is false. A similar sentence in Algebra
would be:
x > 7 or x < 2
This sentence will be true or false based on the value of 'x'. Let’s look at how different
values for 'x' effect the outcome of the sentence.
if x = 0 then we have the sentence
less than or equal to two.
0 > 7 or 0 < 2 this is true because zero is
if x = 5 then we have the sentence
inequalities are false.
5 > 7 or 5 < 2 this is false because both
if x = 9 then we have the sentence
greater than seven.
9 > 7 or 9 < 2 this is true because nine is
In mathematics we use symbols for words. For instance, ">" means "greater than".
We have symbols for all the inequalities. In Algebra we use the symbol "  " for the
word "or".
The disjunction from our example above can be written in two ways as shown below.
x > 7 or x < 2
x>7  x< 2
Like a simple inequality a disjunction can be graphed on the real number line. Below is
an example of solving and graphing a disjunction.
Example
Given 2x + 1 < 3  22 < 5x + 2 we can solve each inequality separately.
- 1 -1
-2
2x < 2
2
-2
20 < 5x
2
5
x<1 
5
4< x
These two inequalities give us two split points which we must plot on the real
number line.
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
These two split points create three regions on the real number line. We will
need to pick a test point from each of the regions and substitute the point into
the disjunction to see which region or regions to shade.
Left Region
(test point is 0)
Middle Region
(test point is 2)
Right Region
(test point is 5)
0<1 
2<1 
5<1 
0< x
This is true since zero
is less than one.
4<2
This is false since
both are false.
4<5
This is true since four is
less than or equal to five
Because the left region and the right region are true we must shade these two
sections of the real number line as shown below.
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
The word "and" forms a conjunction of inequalities and must be evaluated differently.
A conjunction is formed by linking two inequalities with the word "and". A
conjunction will be true only when a value makes both of the two inequalities true.
To illustrate this principle we need to examine the statement below.
The assembly will be on Thursday and we will need to change the bell schedule.
Here we have two statements that can be true or false joined by the word "and". This
is a conjunction. If both of the statements are true then the overall sentence will be true.
If either of the statements are false then the overall sentence is false. A similar sentence
in Algebra would be:
-1 < x and x < 6
This sentence will be true or false based on the value of 'x'. Let’s look at how different
values for 'x' effect the outcome of the sentence.
if x = -3 then we have the sentence -1 < -3 and -3 < 6 this is false because negative
three is NOT greater than negative one.
if x = 4 then we have the sentence -1 < 4 and 4 < 6 this is true because both
inequalities are true.
if x = 8 then we have the sentence -1 < 8 and 8 < 6
NOT less than six.
this is false because eight is
The symbol we use to represent the word "and" is "  ". The above conjunction can
be represented in the two ways shown below.
-1 < x and x < 6
-1 < x  x < 6
There is also a third way a conjunction can be shown. A conjunction is usually a set of
values that is between two split points. So it could be shortened to:
-1 < x < 6
If we were to translate the sentence into words it would be "x is greater than negative
one and x is less than six". Which would mean x is between the two numbers.
Graphing this conjunction would result in:
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
We would apply the same method as in the example for disjunction by picking points
from all three regions and testing them to see which region to shade.
Exercises Unit 2 Section 9
Decide which of the listed values from the given set makes each conjunction or
disjunction true and list that as your answer.
1.
{-3, 5, 10}
x > 6 or x < -2
3. {-4, -2, -1, 5}
5. {-4, -1,0, 3
a < -2  a > 4
1
}
8
-1 < w < 3
1
4
2.
{1, -1, 0, 4}
5 > y and y > 1
4. {-4, -3, -2, 2.4} z > -6  -2.4 > z
6. {-5, -4, -2, 3} z > 0 or -4
1
>z
2
Graph the inequalities below.
7. x > 3 or x < -1
8.
4 > y and y > -2
10. a < 2.5  a > -1.7
11. -5 < z < -1
9. a < -2  a > 2
Solve and graph the following inequalities. Show your work correctly.
12. 3x > 12 or x + 2 < 3
13. 3 > 2y + 5  -4y < -16
14. 5x - 7 < 23 and x + 9 < 2x + 10
15. -2(x + 1) < 6 
16.
1
x<3
2
9 > 4x - 1 > -9
17. Which of the following is the graph of the inequality 3x - 8 > 4 and
a.
c.
2
2
3
3
4
4
5
5
6
6
b.
2
d.
2
3
3
4
4
5
5
6
6
x
+ 1.5 < 4
2
Unit 3
Graphing Functions
Introduction
In Unit 1 we learned that graphs are an important tool in mathematics. Graphs
display data sets visually and make it easier to interpret the world around us. In Unit 3
we will learn to create graphs from equations.
Also in this unit we will be interested in a type of equation that we call a function.
Some equations will produce two solutions and some will result in only one solution or
answer. Functions are the type of equation that will produce only one answer. This can
be very critical when we are dealing with applications in the real world.
Before a motor cycle jumper performs a jump he must calculate the exact speed
at which the jumper is to leave the ramp. If the calculations were performed with
equations that were not functions we might get two or more answers. These answers
might be 75 mph and 85 mph. If we got these two answers how does the jumper know
what speed to use? If he goes too slowly he will crash before he gets to the ramp. If
he goes too fast he will overshoot the ramp and land badly. Having two possible
solutions creates real problems for the jumper.
Working with functions ensures that we will get one answer and this is best when
we are working with all applications. Having two or more solutions means there can
still be problems. We will learn to recognize and use functions in this Unit 3.
Unit 3
Vocabulary and Concepts
Domain
The domain of a function or relation is the set of the x-values we can use.
Function
A function is a relation in which any 'x' value is paired with at most
one 'y' value.
Range
The range of a function or relation is the set of the y-values we can
generate as answers.
Relation
A relation is a set of ordered pairs.
Slope
The slope of a line is the rate at which the line is going up or down.
It is represented by the letter m and calculated by:
m
rise
run
m
vertical change
horizontal change
m
change in y
change in x
m
y
x
Y-Intercept The y-intercept is the point where a graph crosses the y-axis.
Key Concepts for Graphing Functions
Representations of Relations and Functions
Relations and functions can be represented in three forms: rosters,
rules (equations) and graphs. Each has their uses and each form
contributes to problem solving. The same set of data can be shown
and used in all three forms.
Methods of Graphing
There are three methods we can use to graph equations, these are:
substitution, critical features and technology (graphing calculator).
Methods for Recognizing Functions
We can use the definition of a function when we examine rosters,
substitution when need to test rules and the vertical line test when
we examine a graph.
Unit 3 Section 1
Objectives

The student will define relation and recognize valid relations.

The student will list and apply the three forms of representations
for sets.

The student will define and recognize functions in roster and rule form.
As we learned in Unit 1 there are three ways to represent a data set. These are:
Roster - a list of all the elements of a set.
Rule - a description of the elements that belong in a set, usually an
equation.
Graph - a visual display of the set.
Rosters can be written as horizontal lists such as { (2, 4), (3, 6), (4, 8) }
Rosters can also be shown as a table. The example below uses the same data from the
horizontal list in table form.
x y
2 4
3 6
4 8
The Rule or equation for this set of data would y = 2x. We can confirm this by
substitution. If we replace the 'x' in the equation with the values from the x-column
in the table we will find the y-values in the right hand column by multiplying by 2.
A graph of this same data is shown below.
(4,8)
(3,6)
(2,4)
When we collect data we will use the roster form of representation. For instance we can
measure a person's height and the length of their foot for a set of people to see there is a
relationship between the two measurements. The data set we could get as a roster would
look like the example below:
{ (152, 23), (157, 24), (168, 26), (180, 30), (197, 33) }
This data set shows the relationship of a person's height in centimeters and the length of
their shoe in centimeters. The roster here is a list of ordered pairs of numbers. The
word "ordered" here means that we always put the height on the left side of the pair and
shoe length on the right. Keeping the measurements in this order makes it easier for us
to see the trend or pattern. The relationship we can see in this set of data can be
described as: the taller the person, the larger the shoe. When we have ordered pairs
that are related we actually call this a relation.
Definition:
A Relation is a set of ordered pairs.
In the example above the values on the left side of the relation were always the height of a
person. These values would form the set of heights for the people in our study. The name
for the set of values on the left side of a relation is the domain. In the above example
the domain is { 152, 157, 168, 180, 197 }. We also have a name for the values on the
right side of the ordered pairs. This set is called the range. The range for this example
is { 23, 24, 26, 30, 33 }.
The values that appear on the left side of the ordered pairs will be called the x-values. The
values on the left side will be used as x-coordinates when we graph our set or ordered
pairs. The values on the right side will be called y-values since they will be used as
y-coordinates when we graph the data set.
Definition:
The domain of a relation is the set of all the x-values we can use.
Definition:
The range of a function or relation is the set of the y-values we can
generate as answers.
This is a special class of relations that we call functions. We need to understand what it
means to be a special class. If we have a set of all motor vehicles, we can divide it up into
smaller classifications. The Venn diagram below illustrates some of these classifications.
All Motor Vehicles
pick-up
trucks
mini-vans
compact
cars
In the diagram above the rectangle represents all motor vehicles as the big set. Within
the set of motor vehicles there are special classes of other types of vehicles such as
pick-up trucks, mini-vans, and compact cars. There could be other classifications as well.
The diagram below shows us relations and functions.
Relations
Functions
Definition:
A Function is a special relation in which in which any 'x' value is paired
with at most one 'y' value.
.
To understand the difference between function and non-function relations we will need
to compare the examples below.
Example A
{ (4, 2), (5, -2), (6, 1.2) (7, ) }
Example B
{ (3, 5), (5, 7), (5, -4), (-8, 10) }
Example A is a function. In this example none of the x-values are repeated. So every
x-value is paired with only one y-value. Example B is not a function because the x-value '5'
is paired with '7' and with '-4' and so 5 gives us two answers.
From the two examples on the previous page we can conclude that the best way
to check for non-functions is to look for duplicate x-values with different y-values.
Some other examples with explanations are given below.
{(2,
{(6,
{(1,
{(2,
3),
3),
3),
3),
(-1, 5.6), (0, 2), (3, -7) } no duplicate x's - it's a function
(6, 5), (0, -2), (-3, -7)} '6' is paired with 3 and 5 - not a function
(-1, 3), (0, 3), (5, 3) }
no duplicate x's - it's a function
(-5, .5), (2, 3), (8, -7)}
no duplicate x's with different y's - it's a
function
Rosters can also be listed as tables. We should also be able to recognize functions and
non-functions in table form as well. Some examples are shown below.
x
3
4
5
6
y
9
16
25
36
function!
no duplicate
x values
x
3
-2
3
0
y
4
.15
5
6
x
-3
7
3.5
0
y
4
4
4
4
function!
no duplicate
x values
not a function!
duplicate 3's
& different 'y'
values
Another form of representation is the rule or equation form. To determine when an
equation is a function or not a function we should use the method of substitution. The
method of substitution requires that we pick a number for 'x' and put it into the equation
to find out how many y-values we get as answers. The examples below show us how to
use this method.
Example A
y
y
y
y
=
=
=
=
x2 + 1 we pick x = 4
(4)2 + 1
16 + 1
17
We got one answer for y so
we got the ordered pair (4, 17)
This is a function.
Example B
|y| = x + 1
|y| = 5 + 1
|y| = 6
y=6
we pick x = 5
y = -6
We got two answers for y. So we
have two ordered pairs (5, 6) (5, -6).
This is not a function!
Exercises Unit 3 Section 1
1. What are the three ways to represent a set of points?
Fill in the blanks; list your answer on your own paper.
2. A ______________ is a list of all the elements in a set.
3. A ______________ is an equation or description that tells us how 'x' and 'y' relate
4. A _______________ is a visual display of the points in a set.
5. List the definition we have for a relation.
6. What is the domain of a function or relation?
7. What is the range of a function or relation?
8. List the definition we have for a function.
9. What is the special or important feature of a function?
10. Convert the list into a table. {(3, 7), (4, 9), (5, 11)}
11. Convert the table into a list
x
2
-3
1.2
0
y
4
2.6

¼
12. Given the relation:
{ (3, triangle ), ( 4, quadrilateral ), ( 5, pentagon ), ( 6, hexagon ) }
a. List the domain
b. List the range
c. Each of the range elements is a name for a _______________ figure.
d. Each of the domain elements is a _____________ of sides.
e. Is this relation a function? _____________ ( yes/no)
Explain your answer.
13. Given the relation:
{ (9, freshmen), (10, sophomore), (11, junior), (12, senior) }
a. List the domain
b. List the range
c. Each of the range elements is a name for a _______________ .
d. Each of the domain elements is a _____________ level.
e. Is this a function? ______________ ( yes/n)
Explain your answer.
14. Given the two sets below
A = { May, July, December, February }
B = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 }
create a relation using elements from set A as the domain and elements from
set B as the range. ( Your answer must have at least 3 ordered pairs. )
15. Given the two sets below
C = { Fall, Spring }
D = { September, December, March, April, October, July }
create a relation using elements from set C as the domain and elements from
set D as the range. ( Your answer must have at least 3 ordered pairs. )
16. Given the relation:
{ (90, A ), ( 80, B ), ( 70, C ), ( 60, D ) }
a. List the domain
b. List the range
c. Is this relation a function? ___________ ( yes/no)
Explain your answer.
17. Given C = { John, Paul, Juan, Tim } and D = { Maria, Alexa, Joan, Sara }
Determine which sets below are valid relations - answer YES or NO
a = { (John, Maria), (Paul, Alexa), (Juan, Joan), (Tim, Sara) }
b = { (Sara, Maria), (Paul, Alexa), (Juan, Joan) }
c = { (John, Alexa), (Paul, Alexa), (Juan, Alexa), (Tim, Alexa) }
d = { (John, Maria), (John, Alexa), (John, Joan), (John, Sara) }
e = { (John, Maria), (Paul, Alexa), (Joan, Juan), (Sara, Tim) }
f = { (John, Maria), (Paul, Alexa), (Tim, Sara) }
g = { (John, Sara) }
h = { (John, Paul), (Paul, Juan), (Sara, Joan), (Tim, Sara) }
z = { (Maria, Tim), (Alexa, John), (Joan, Paul) }
18. Decide which of the following rosters in list form represent a function and which
are non-functions.
( Write yes for a function and no for a non-function )
a.
b.
c.
d.
e.
f.
{(2, 8), (9,-10), (4.5, ¾) (  , 0)}
{(5, -3), (2.1, 0), (5, 1) (4,  )}
{(6, 7), (-4, 7), (1.5, 7) (0, 7)}
{(4, 8), (-5,-1), (4, 8), (-7, 5)}
{(3, 5), (3,-11), (3, ½) (3, 0)}
{(3, 5), (5, 3), (-8.9, ½) (½, -8.9)}
19. Decide which of the following rosters in table form represent a function and which are
non-functions. ( Write yes for a function and no for a non-function. )
a)
x
y
9
4
-1
5
-2
1.3
0

b)
x
y
3
5
-1
3
6
1
8
¼
c)
x
y
11
-7
-2.8
0
4
4
4
4
d)
x
y
6
2
2
6
-12 0
0 -12
20. Use the method of substitution to determine which equations represent functions and
which do not. Show your work. ( Write yes for a function and no for a non-function. )
a. y = | x + 5|
b. y2 = x (substitute 4 for x)
c. y = .5x + 10
Unit 3 Section 2
Objectives

The student will use function notation.

The student will create tables for given domains and functions.
When we need to solve problems in real world applications we may find that several
equations are used to find the ultimate solution. Since we may need to use multiple
relations and functions they will need to be named and we will often have to use more
variables than just 'x' and 'y'. The most common way to name a relation or function is
with a lower case letter. The example below shows a function in roster form with its
name.
w = { (2, 3), (4, 5), (5, 6), (7, 8) }
The name for this particular relation is 'w'. This same function in rule or equation
form would be given as:
w(x) = x + 1
with the domain D = { 2, 4, 5, 7 }
The points in the roster are generated by substituting each domain element into the
equation and finding the y values, as shown below.
so
w(x) = x + 1
w(2) = 2 + 1
w(2) = 3
y=3
w(x) = x + 1
w(4) = 4 + 1
w(4) = 5
y=5
w(x) = x + 1
w(5) = 5 + 1
w(5) = 6
y=6
w(x) = x + 1
w(7) = 7 + 1
w(7) = 8
y=8
and the points are: (2,3), (4, 5), (5, 6), (7, 8)
In the above examples it should be noted that both the 'x' in the parentheses and the
'x' in the expression on the right were replaced by the value we were substituting.
There will frequently be times when we need to verbalize the notation. This means
we need to say the notation properly. Some examples of how we should put the
notation into words are given below.
for
for
for
for
g(x)
f(x)
v(t)
w(2)
we could say
we could say
we could say
we could say
"g of x"
"f of x"
"v of t"
"w of 2"
or
or
or
or
"g at x"
"f at x"
"v at t"
"w at 2"
Generally the words "at" and "of" are interchangeable; however the word "at" is most
often used when a number appears in the parentheses, and the word "of" is most often
used when a variable is in the parentheses that follows the name.
Below are more examples of using this notation.
Example A
Given g(x) = 3x2 + x then to find g(-1) we substitute and get g(-1) = 3(-1)2 + (-1)
then simplifying g(-1) = 3(1) + (-1)
g(-1) = 3 + (-1)
g(-1) = 2
Example B
Given f(x) = |1 - 2x| then to find f(4) we substitute and get f(4) =
then simplifying f(4) =
f(4) =
f(4) =
Example C
|1 - 2(4)|
|1 - 8|
| -7 |
7
Given p(x) = 5x + 7 then to find p(1.3) we substitute and get p(1.3) = 5(1.3) + 7
then simplifying p(1.3) = 6.5 + 7
p(1.3) = 13.5
In this unit we are going to graph various types of functions. The method we will
use to graph functions initially is called substitution. When we use this method we
can be given the domain and an equation to use that will help create a table of points.
Examples of how we create our tables are given below. The "D" stands for the domain
we must use. The examples show the calculations and how to fill in the table.
f(x) = 2x + 3
D = { -1, 2.1, 4 }
x
f(x)
g(x) =x2
D= {2, -4, 1.5}
x
g(x)
f(-1) = 2(-1)+3 = 1
g(2) = (2)2 = 4
f(2.1) = 2(2.1)+3 = 7.2
g(-4) = (-4)2 = 16
f(4) = 2(4)+3 = 11
g(4) = (1.5)2 = 2.25
x
f(x)
-1
1
2.1 7.2
4
11
x
g(x)
2
4
-4
16
1.5 2.25
Exercises Unit 3 Section 2
1. What is a roster?
2. What are the two ways a roster can be listed?
3. What is a rule?
4. Define relation.
5. What is the set of y-values that we are generating as answers called?
6. What is the set of x-values that we are allowed to use called?
7. List the definition we have for a function.
8. What is the special or important feature of a function?
9. Given the relation below:
{ (8, cup ), ( 16, pint ), ( 32, quart ), ( 128, gallon ) }
a. List the domain
b. List the range
c. Each of the range elements is a name for a ___________measurement.
e. Each of the domain elements is a number of ___________.
f. Is this relation a function? _____________ ( yes/no)
Explain your answer.
10. Given the relation:
{ (Jones, Rm100), (Johnson, Rm101), (Sanchez, Rm102), (Wilson, Rm103) }
a. List the domain
b. List the range
c. Each of the domain elements is a name for a _______________ .
e. Each of the range elements is a number for a _____________.
f. Is this a function? ______________ ( yes/n)
Explain your answer.
11. Given the two sets below
A = {horse,ram,monkey,rooster,dog,boar,mouse,ox,tiger,rabbit,dragon,snake}
B = {1990, 1991, 1992, 1993, 1994, 1995, 1996, 1997, 1998, 1999, 2000, 2001}
create a relation using elements from set B as the domain and elements from
set A as the range. ( Your answer must have at least 3 ordered pairs. )
12. Given C = { M, N, O, P } and D = { 5, 6, 7, 8 } determine which relations
below which are valid and which are not. (Answer yes for valid and no for invalid.)
a = { (M, 5), (N, 8), (P, 6), (O, 7) }
b = { (M, 5), (N, 5), (P, 5), (O, 5) }
c = { ((5, M), (6, P), (7, O), (8, N) }
d = { (M, 5), (7, P), (N, 6), (7, O) }
e = { (M, 5), (7, 8), (P, N), (6, 5) }
13. Decide which of the following rosters in list form represent a function and which are
non-functions. ( Write yes for a function and no for a non-function. )
a = {(5, 7), (-1, 10), (3.5, -2.6) (0, 0)}
b = {(-3, -3), (2, 2), (5, 5) (  ,  )}
c = {(6, 7), (-4, 2), (1.5, -½) (6, 8)}
d = {(4, 2), (-5,-2), (-9, 2 ), (  , 5)}
e = {(-2, 5), (-2,-1), (-2, ½) (-2, 0)}
f = {(2, 6), (6, 2), (9, ½) (½, 9)}
19. Decide which of the following rosters in table form represent a function and which are
non-functions. ( Write yes for a function and no for a non-function. )
a)
b)
x
y
9
8
9
7
-2
2.5
0

x
y
1
2
3
4
5
3
4
1¾
c)
x
y
11
4
5
12
4
11
12
5
d)
x
y
5
-3
4
0
2
2
2
2
20. Function notation is a way of naming an ______________________.
21. In the equation "f(x) = x - 4" the "f(x)" could be replaced by ________.
22. How should we verbalize (say) the equation "f(x) =" _________________________
23. How should we verbalize (say) the equation "g(2) =" _________________________
24. Given f(x) = x2 find: f(3), f(4) and f(-1)
25. Given g(x) = 2x + 1 find: g(5), g(4.5) and g(-½)
26. Given f(x) = 1.4x - 3.5 find: f(6), f(-2) and f(2.2)
27. Given h(x) =
3x  2
find: h(5), f(1) and f(-1)
x2
10
2
and the Domain D = {2, 3, -2.5, }
x
5
find g(x) for all values of the domain. (Use the proper notation for your answers)
28. Given g(x) =
29. Given k(x) = | x - 4 | and Domain D = {7, -1, 3.5, ¾}
find k(x) for all values of the domain. (Use the proper notation for your answers)
30. Given the domain 'D' and the equation fill in the table for each problem.
Draw the table on your own assignment.
a. f(x) = .5x + 3
D = { 4, 6, 1.4, ¼ }
x
f(x)
b. g(x) =
x2
x 1
D= {2, 5, 0, 1.5}
x
g(x)
c.
p(x) = (x + 1)2
D = {-1, -2, .5, -1,25}
Unit 3 Section 3
Objectives

The student will graph functions with a given domain.

The student will graph functions with the domain of all real numbers.

The student will use the vertical line test to determine when relations
represented as graphs are functions.
Functions operate as input-output machines. They are called input-output machines
because we can substitute one number in the function for ‘x’ and it will produce one
y-value out as the answer. The diagram below illustrates this principle.
the “machine”
f(x) = 2x - 5
input to the “machine”
x=4
output from the ”machine”
f(4) = 2(4) - 5
f(4) = 8 – 5
f(4) = 3
y=3
Graphing functions or equations with a given domain using substitution can be
described as step by step process. The steps are:
1. Create a T-chart with the x-values from the domain.
2. Substitute each x-value into the function or equation and find the y-values
3. Plot the points on the x-y coordinate plane.
Below are some examples of this process.
Example A
Given the equation with the domain
g(x) = 2x – 1
D = { -2, -1, 0, 1, 2 }
Create the Table
x
Find the y-values
y
Plot the points
x
y
-2
g(-2) = 2(-2) – 1 = -5
-2
-5
-1
0
1
g(-1) = 2(-1) – 1 = -3
g(0) = 2(0) – 1 = -1
g(1) = 2(1) – 1 = 1
-1
0
1
-3
-1
1
2
3
2
g(2) = 2(2) – 1 = 3
Example B
Given the equation with the domain
f(x) = |-2 – 2x|
D = { -2, -1, 0, 1, 2 }
Create the Table
Find the y-values
Plot the points
x
y
f(-2) = |-2 – 2(-2)| = 2
-2
2
-1
0
1
f(-1) = |-2 – 2(-1)| = 0
f(0) = |-2 – 2(0)| = 2
f(1) = |-2 – 2(1)| = 4
-1
0
1
0
2
4
2
f(2) = |-2 – 2(2)| = 6
2
6
x
-2
y
When we are given a functions and a domain that is a limited set of points, then our
graph will consist of a set of isolated points. Many of our functions will have a domain
of all real numbers. All real numbers will mean that every fraction, every decimal,
every irrational number, whether positive or negative, is included as an element of
the domain. In order to create a graph with a domain of all real numbers we must
choose enough points to see a pattern. These patterns can be straight lines,
V-shapes or U-shapes. Once we see the pattern we should connect our points
appropriately. The steps for graphing a functions with all real numbers as our domain
are given below:
Graphing functions or equations with a domain of all real numbers using
substitution can be described as a step by step process. The steps are:
1.
2.
3.
4.
5.
Choose points to substitute into the given function.
Create a T-chart with the x-values from the domain.
Substitute each x-value into the function or equation and find the y-values
Plot the points on the x-y coordinate plane.
Connect the points with an appropriate shape.
Below are some examples of this process.
Example A
Given the equation with the domain of all real numbers
g(x) = .5x + 1
We will pick the numbers -4, -2, 0 2, and 4 because they will be easier to use.
Create the Table
x
Plot the points
then connect them
Find the y-values
y
x
y
-3
-4
g(-4) = .5(-4) – 1 = -3
-4
-2
0
2
g(-2) = .5(-2) – 1 = -2
g(0) = .5(0) – 1 = -1
g(1) = .5(2) – 1 = 0
-2 -2
0 -1
2 0
4
g(2) = .5(4) – 1 = 1
4
1
In this example the pattern formed by the points was a straight line. This
pattern will appear in many of our equations and their graphs. We draw the line
because the points in between the numbers we picked will produce values that
lie on the line. For instance g(3) = .5 which is consistent with the graph above.
Likewise g(5.5) = 1.75 which is also on the line as we graphed it.
Example B
Given the equation with the domain of all real numbers
f(x) = 3 – x2
We will pick the numbers -2, -1, 0 1, and 2 because they will be easy to use.
Create the Table
x
-2
Plot the points
then connect them
Find the y-values
y
f(-2) = 3 – (-2)2 = -1
(-1)2 =
x
y
-2
-1
-1
0
1
f(-1) = 3 –
2
f(0) = 3 – (0)2 = 3
f(1) = 3 – (1)2 = 2
-1
0
1
2
3
2
2
f(2) = 3 – (2)2 = -1
2
-1
In this example the pattern formed by the points was a U-shape. This
pattern will appear in some of our equations and their graphs. This U-shape
is will appear whenever the functions have an ‘x2’ term in them.
Example C
Given the equation with the domain of all real numbers
y=5
( functions can be written as y = )
We will pick the numbers -2, -1, 0 1, and 2 because they will be easy to use.
Create the Table
x
Plot the points
then connect them
Find the y-values
y
x
y
-2
y=5
-2
5
-1
0
1
y=5
y=5
y=5
-1
0
1
5
5
5
2
y=5
2
5
In this example the pattern formed by the points was a straight line. The equation above
tells us that ‘y’ is always equal to 5. The table then is very easy to fill in. All the yvalues will be 5.
Example D
Given the equation with the domain of all real numbers
y = |2 – x|
We will pick the numbers -2, 0, 2, 4, and 6 because absolute values
will sometimes require a wider spread of values to see a pattern.
Create the Table
x
Plot the points
then connect them
Find the y-values
y
x
y
-2
y = |2 – (-2)| = 4
-2
4
0
2
4
y = |2 – (0)| = 2
y = |2 – 2| = 0
y = |2 – 4| = 2
0
2
4
2
0
2
6
y = |2 – 6| = 4
6
4
In this example the pattern formed by the points was a V-shape. The equation above
contains an absolute value. Any time we have an absolute value in the equation we
will see a V-shape in the graph.
If we are given a graph we need to be able to determine when the graph represents
a function. The method for deciding whether or not a graph represents a function is
called the vertical line test.
The Vertical Line Test tells us that if there is any place we can draw a vertical line
and pass through two or more points on the graph, then the graph does not represent
a function.
The examples below show us how the vertical line test works.
Example A
In example A if we draw a vertical
line through any of the points we
only pass through that one point.
Example A is a function.
Example C
In example C any vertical line
we can draw will only intersect the
graph in one point. The example
passes the Vertical Line Test and
does represent a function.
Example B
In example B if we draw the
vertical line shown it passes through
(4, 5) and (4, 2) so the graph can
not represent a function.
Example D
In example D any vertical line
that we draw from x = -3 to x = 3
will intersect the graph in two places
or points. This graph does not
represent a function.
Exercises Unit 3 Section 3
1. List our definition for a function.
2. What are the three ways a set of points can be represented?
3. Define relation.
4. How does an "input-output machine" work?
5. Describe the process of graphing a function with a given domain
using the method of substitution.
6. Describe the process of graphing a function with a domain of all real numbers
using the method of substitution.
7. Graph the following equations using the method of substitution with the
given domain. ( You create a table, fill in the table, and then graph the points. )
a.
y = 4x – 6
D = { 0, 1, 2, 3, 4 }
b.
y = x2 - 2
D = {-2, -1, 0, 1, 2}
c.
y = .5x - 1.5
D = { -6, -2, 0, 2, 6 }
d.
f(x) = | x + 2 | - 3
1
e. g(x) = - x + 5
4
f. f(x) = .5x2 - x
D = {-4, -3, -2, -1, 0}
D = {-4 , -2, 0, 2, 4}
D = {-2, -1, 0, 1, 3}
8. What is the difference we see in a graph between using a small finite domain and
a domain of all real numbers?
9. Graph the following equation or functions using all real numbers as the domain.
You must pick at least five values and get five points before you graph the functions.
( You should pick positive and negative numbers. )
a. y = 3x - 4
b.
y = -x2 +1
c. y = -2x + 5
d.
y = |x - 1| - 5
e. y = -.75x + 4
f.
g(x) = -|x| + 6
g. f(x) = 2x2 - 6
h.
g(x) = (x - 1)2 – 2
i. y = 3
j.
x = -2
10. Explain how to use the vertical line test to determine if a graph represents
a function or not.
11. Determine whether or not each graph represents a function. ( answer yes or no ).
a.
d.
b.
c.
e.
12. Use the vertical line test to determine which graphs are functions. ( answer yes or no )
a.
b.
c.
d.
e.
f.
13. Determine which equations represent functions, and which do not, by using substitution.
( Answer yes or no and show your substitution. )
a. y = 2x + 7
b. y = x2 + 2x
c. |y| = x + 1
Unit 3 Section 4
Objectives
 The student will isolate ‘y’ before graphing functions by substitution.
 The student will graph y = c and x = c where c is an element of
the real numbers
 The student will recognize and draw the general shapes for the
families of linear, quadratic and absolute value functions.
In the previous section we learned to graph functions or equations using substitution.
Based on our experience in graphing these functions or equations we should be able to
predict the pattern or shape we will see in a function's graph. When we see an equation
that has an 'x2' term we should know that the graph will have a U-shape. This U-shape
is called a parabola and we will study it in greater detail in later chapters. Equations
that have an 'x2' term are called quadratic equations. If an absolute value occurs in
the equation then the pattern or shape of the graph will be a V-shape. If the equation
has only an 'x' ( 'x' to the first power ) then the graph's pattern will be a straight line.
This family or category of equations is called linear.
Examples of each type are given below.
Example A
Example B
y = 2x - 1
y = -x2
The equation in Example A
has an 'x' that is to the first
power only and so is of the
family or category we call
linear and it graphs as a
straight line.
The equation in Example B
has an 'x2' in it. The equation
is a quadratic and has the
U-shape or parabola shape.
Example C
y = |x + 1|
The equation in Example C
has the absolute value of
an expression in it. In the
graph we see the V-shape
that is characteristic of
absolute values.
There are two types of linear equations that do not necessarily require substitution to
graph. These are equations of the form "y = c" or "x = c" where the 'c' is a constant.
The constant can be any real number so examples of these types of equations would be:
y=4
x = -2
y = 36.2
x=
1
2
y=0
x=
8
y= 
In the example below we see an equation of the form "y = c" graphed by substitution.
Example A
Given the equation with the domain of all real numbers
y=4
We will pick the numbers -2, -1, 0 1, and 2
Create the Table
x
Plot the points
then connect them
Find the y-values
y
x
y
-2
y=4
-2
4
-1
0
1
y=4
y=4
y=4
-1
0
1
4
4
4
2
y=4
2
4
In this example we see a straight line and since all the y-values have to be '4' the line is
horizontal. All lines of the form "y = c" will graph as horizontal lines.
If an equation is of the form "x = c" then all the x-values or coordinates must be 'c'.
As a result any line of this form will be vertical.
When we are graphing equations by substitution some of our equations will not be in
the best form for finding the y-values. If we compare the two equations below we
will see that finding the y-value when x = 5 is much easier in Example A than it is
in Example B. The key to A being easier is that 'y' is isolated.
Example A
y = 2x - 3
y = 2(5) - 3
y = 10 - 3
y=7
Example B
y3
2
y3
2
y  3. 2
2
=x
=5
= 5 .2
y + 3 = 10
-3
-3
y = 7
When 'y' is isolated the substitution and simplifying is much easier. When we want
to graph equations the best thing we can do is make sure we start with 'y' isolated or
we solve the equation for y before we create our table. The complete algorithm or
set of steps for graphing by substitution is given below.
The steps are:
1.
2.
3.
4.
5.
6.
Isolate the 'y' variable if needed.
Choose points to substitute into the given function.
Create a T-chart with the x-values from the domain.
Substitute each x-value into the function or equation and find the y-values
Plot the points on the x-y coordinate plane.
Connect the points with an appropriate shape.
The example below shows the complete process.
Example
Given the equation 2y - 4x = -10 construct the graph of the equation.
Step 1
2y - 4x = -6
+ 4x
+ 4x
2y = 4x - 6
2
2
2
y = 2x - 3
We will pick the numbers -2, -1, 0 1, and 2 because they will be easy to use.
Create the Table
x
Plot the points
then connect them
Find the y-values
y
x
y
-7
-2
f(-2) = 2(-2) - 3 = -7
-2
-1
0
1
f(-1) = 2(-1) - 3 = -5
f(0) = 2(0) - 3 = -3
f(1) = 2(1) - 3 = -1
-1 -5
0 -3
1 -1
f(2) = 2(2) - 3 = 1
2
2
1
Exercises Unit 3 Section 4
1. Graph the following equations using the method of substitution with the given domain.
( Show the tables as well as the graphs on your paper. )
a. y = -2x + 6
D = { 0, 1, 2, 3, 4 }
b.
D = {-2, -1, 0, 1, 2}
g(x) = x2 - 4
1
c. y = x
3
d. f(x) = -|x + 1| + 3
D = { -6, -3, 0, 3, 6 }
D = {-2, -1, 0, 1, 2}
2. Graph the following equation or functions using all real numbers as the domain.
You must pick at least five values and get five points before you graph the functions.
( You must pick positive and negative numbers. Show your table and the graph. )
a. y = 2x - 4
1
c. y = x + 1
2
e. y = -x2
b.
y = -3x + 7
d.
y = -x + 3
f.
g(x) = x2 - 5
g. f(x) = .5x2 + 1
h.
g(x) = (x + 1)2
i.
f(x) = |x|
j.
y = -2|x|
k. f(x) = |x| - 3
l.
y = |x + 1|
3. Isolate the variable 'y' in each equation then graph it by using substitution.
a. y + 4 = 2x + 1
b.
y - x = -2
c. -2y = 4x
d.
4y + x= 8
4. Graph the following equation or functions using all real numbers as the domain.
a. f(x) = 2
1
c. x = 2
b.
x=3
d.
y = -3.5
Unit 3 Section 5
Objectives
 The student will use set builder notation.
 The student will find domains and ranges for rosters and graphs.
In Algebra we use symbols in order to write sentences that would be long and difficult to
work with, and to make it easier to interpret and manipulate given information. For
instance an equation written in words is more difficult to solve than an equation written
symbolically. Consider the two ways of representing the same equation below.
2x = .5(x + 12)
Twice a number is equal to half the sum of
the number and twelve.
There will be times when the domain and range of a function or equation will be
critical in the problem solving process. Listing the domain and range symbolically
can make our work much simpler. We use a type of representation for the sets
of values in the domain and range called set builder notation.
An example of using set builder notation to describe a set of numbers is given below.
Example
A={x|x>3}
A = { x | x > 3 } translated into words is: "A = the set of elements x such that x > 3"
the vertical bar
is "such that"
A={x|x>3}
when we see
the { then we
say "the set of"
Whatever follows the
bar is the condition for
membership in the set
the first variable
is referred to as
the "elements"
For our purposes we will always be working within the real numbers so
these expressions can be translated into line graphs of inequalities.
so A = { x | x > 3 }
We should be able to translate between the set builder notation and line graphs.
Below are some examples of these conversions.
Example A
A={a|a<2}
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
The graph is the set of points defined by the inequality in the notation.
If we were to translate the set builder notation into words we would say:
" A is equal to the set of elements 'a' such that a is less than or equal to two".
Example B
B = { y | -2 < y < 5 }
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
The graph is the set of points defined by the inequality in the notation.
If were to translate the set builder notation into words we would say:
" B equals the set of elements 'y' such that 'y' is greater than negative two
and 'y' is less than five".
Finding domains and ranges for a function can be done in any of the three forms of
representation. We will begin by finding the domains and range in the roster form.
Example
Given the set of ordered pairs in roster form {(-1, 3), (0, 4), (2.5, 6.5) , (3, 7)}
The domain is defined as the set of x-values that we can use. The x-coordinates
are always found on the left side of the ordered pair so the domain is:
D = { -1, 0, 2.5, 3 }
The range is defined as the set of y-values that we can generate as answers. The
y-coordinates are always found on the right side of the ordered pairs so the
range is:
R = { 3, 4, 6.5, 7 }
Finding the domain when a function or relation is represented in a graph is done by
remembering that the x-axis is horizontal. To determine the domain we must
determine what part of the x-axis has points above or below it. The examples below
show us how to find the domain of a function from its graph.
Example A
-5
Example B
1 2.5 4 6
-6 -4 -2
In example A the domain
is { -5, 1, 2.5, 4, 6 }. These
are the x-coordinates of the
points that have been plotted.
The values at the top show
the x-coordinates found by
counting along the x-axis.
0
5
In example B the domain
is { -6, -4, -2, 0, 5 }. These
are the x-coordinates of the
points that have been plotted.
The values at the top show
the x-coordinates found by
counting along the x-axis.
We can also find the domain when the points are connected. The examples below
show us how to find and list the domains.
Example A
In Example A every point
on the x-axis will have a
point above it or below it
on the graph of the line.
As a result the domain is
D = all real numbers.
Example B
In Example B every point
on the x-axis that is greater
than or equal to -2 has a
point above it or below it.
As a result the domain is
D = { x | x > -2 }
Example C
In Example C every point
on the x-axis between -1
and 4 has a point above it
or below it. As a result
domain is
D = { x | -1 < x < 4 }
Finding the range when a function or relation is represented in a graph is done by
remembering that the y-axis is vertical. To determine the range we must find values by
counting up or down on the y-axis. The examples below show us how
to find the range of a function from its graph.
Example A
Example B
4
6
4
1
2
-2
-1
-5
-4
In example A the range
is { -5, -2, 1, 4 }. These
are the y-coordinates of the
points that have been plotted.
The values at the left show
the y-coordinates found by
counting up/down on the y-axis.
In example B the range
is { -4, -1, 2, 4, 6 }. These
are the y-coordinates of the
points that have been plotted.
The values at the left show
the y-coordinates found by
counting up/down on the y-axis.
We can also find the range when the points are connected. The examples below
show us how to find and list the ranges.
Example A
In Example A every point
on the y-axis will have a
point to the left or the
right of it that is on the
graph of the line. As a result
the range is
R = all real numbers.
Example B
In Example B every point
on the y-axis that is less
than or equal to 2 has a
point to the left or right
of it. As a result the
range is
R={y|y< 2}
Example C
In Example C every point
on the y-axis between -6
and -1 has a point to the
right of it. As a result the
range is
R = { y | -6 < y < -1 }
Exercises Unit 3 Section 5
1. Convert the set listed in set builder notation into a graph on the real number line.
a. { x | x < -2 }
b. { y | y > 3 }
c.
d.
{z|z< 1}
{ x | x > 1.5 }
2. Convert each graph into a rule in set builder notation.
a.
-3
-2
-1
0
1
2
3
-3
-2
-1
0
1
2
3
-3
-2
-1
0
1
2
3
-3
-2
-1
0
1
2
3
b.
c.
d.
3. Given the roster list the domain and range and determine whether the relation is a
function or not. ( List the domain and range in set brackets and answer yes/no for
function or non-function. )
2
)}
3
1
b. { (5, 1), (-6, 1), ( 2 , 1), ( , 1) }
5
c. { (3, -7), (-2, 0), (3, 4), (2.8, 3.9) }
d. { (2, -8), (2,  ), (2, 9), (2, 6.7) }
a. { (2, 4), (-1, -3), (2.5, 0), (  ,
e. { (5, -9), (-9, 5), (11, 4), (4, 11) }
4. Find the domain and range for each of the graphs below and determine
whether it is a function or not. ( List the domain and range in set brackets and
answer yes/no for function or non-function. )
a.
b.
c.
5. Find the domain and range for each of the graphs below. List your answers for the
domain and range in set builder notation. Then determine whether or not the
graph represents a function. ( answer yes for a function and no for a non-function. )
a.
b.
c.
d.
e.
f.
g.
h.
f.
6. Graph the following equation or functions using all real numbers as the domain.
You must pick at least five values and get five points before you graph the functions.
Remember to isolate 'y' where needed. (You should pick positive and negative values.)
a. y = 3x - 1
b. g(x) = -x - 1
d. f(x) = -x2 + 3
e. y - 2x = - 4
3
x+2
2
e. 2y - 4 = x
c. y =
Unit 3 Section 6
Objectives

The student will define slope and y-intercept.

The student will find slopes and y-intercepts from graphs.

The student will draw lines using slope and intercept or slope
and point.
The slope of a line tells us how fast a line in a graph is going up or going down. The
graphs below give us important information about businesses.
Business A Profit
Business B Profit
Business C Profit
$10000
$10000
$10000
$8000
$8000
$8000
$6000
$6000
$6000
$4000
$4000
$4000
$2000
$2000
$2000
0
1
2
3
4
5
0
1
Months
2
3
4
5
0
1
2
3
4
5
Months
Months
Business A increased their profits from $8000 to $10000 in 5 months which means the
line is going up at a rate of $400 per month. Business B started at $2000 a month and
in 5 months went up to $8000. The rate at which the line is going up or the slope of the
line is $1200 per month. Business C started with a profit $10000 and in 5 months
finished with a profit of $6000. The rate of change or slope of the line is −$800 per
month. We can use the slope of each line to accurately determine what each business
will have as a profit in the future. The place where each line begins is also important.
The vertical axis is called the y-axis and so the starting points for each line are called the
y-intercept. We should notice that lines that go up will have positive slopes while lines
that go down will have negative slopes. Below are the formal definitions for the slope
and the y-intercept. These two quantities are important because they allow us to make
accurate predictions about trends or patterns in the world around us.
Definition:
The slope of a line tells us how fast a line is going up or down. We use
the letter m to represent the slope and its symbolic representations
are given below.
m
rise
run
m
vertical change
horizontal change
m
change in y
change in x
m
y
x
Definition:
The y-intercept is the point where a graph crosses the y-axis.
We can think of the y-intercept as the starting point for lines which we will graph. The
letter b is used to represent the y-intercept.
When we have a graph of a line we can find the slope and y-intercept simply by counting.
The examples below show us how to find the slope by counting.
Example A
4
1
Example B
2
3
1
2
2
1
1
We should always count from the left point
to the right point when finding slope. So we
count up from 1 to 4. Then we start again
and count across and we get 2. The rise
is the '4' since we counted up four. The
run is the '2' since we went across two.
The y-intercept will be 3 because that is the
point at which the line crosses the y-axis.
The slope is m 
which is
m
rise
run
4
=2
2
The y-intercept is b = 3
2
3
Again we start counting from the left point.
But now we go down two. Then we count
across three. So in this case the 'rise' will
be a '-2' because we counted in a negative
direction, down. The run will still be
positive because we are counting across in
a positive direction. The y-intercept will
be -1.
The slope is m 
which is
m
rise
run
2
2
=−
3
3
The y-intercept is b = -1
We can find the slope and y-intercept if we are given a graph. We can also reverse this
process and draw a line if we are given a slope and an intercept. To draw the line
we must first plot the y-intercept and then count using the slope.
The example below shows us how this done.
Example
Given the slope m =
1. Plot the intercept
4
and the y-intercept b = -2 draw the line they determine.
3
2. Count to get the slope
3. Draw the line
3
4
To use the slope we counted up
4 and then across 3. We did it
again and got our third point.
this point is a check to make sure
we used the slope correctly.
There will be times when we know the slope or rate of change in a problem and we have
a data point. This is only a little different from our last type of problem. The process can
be described as: First plot the data point we are given, second use the slope to find
more points and third draw the line. The example below shows us how this is done.
Example
2
Given the slope m = − and the point (-2, 4) draw the line they determine.
3
1. Plot the point
2. Count to get the slope
-2
3
To use the slope we counted down
2 and then across 3. We did it
again and got our third point.
this point is a check to make sure
we used the slope correctly.
3. Draw the line
Exercises Unit 3 Section 6
1. Complete the definitions of slope
a.
m
run
b.
m


c.
m
vertical change
d.
m
change in
change in
2. What does the slope of a line tell us?
3. Explain how to find the slope of a line by counting from point to point.
4. If a line has a slope of
0
what does this mean? and what kind of line is this?
3
5. If a line has a slope of
5
what does this mean? and what kind of line is this?
0
6. What is a y-intercept?
7. Find the slope of each line by counting from point to point and list the y-intercept.
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
l.
m.
n.
o.
8. Draw a line with the listed y-intercept and slope.
a. m = 2 b = -4
a. m =
2
b = -3
3
b. m = -
1
b=3
2
b. m = -3
b=4
c. m = 0 b = -2
c. m = 1 b = 0
9. Given the listed point and the slope draw the line they determine.
a. (3, 1)
d. (-4, 5)
m=2
m = -3
b. (-2, 4) m = 
e. (-6, -3) m =
1
2
2
5
c. (-4, -2) m =
2
3
f. (1, -2) m = no slope
10. Graph the following equations or functions using all real numbers as the domain.
You must pick at least five values and get five points before you graph the functions.
Then find the slopes and y-intercepts
a. y = 2x + 1
b.
c. y = -3x + 5
d.
1
x-1
2
y=x-2
y=-
11. If we had the following two sets.
A = { Student Names from WHS }
B = { Student ID numbers}
and we created the relation with the domain as A and the range as B
would this be a function? _________ ( yes/no )
Explain your answer.
12. Find the domain and range for each of the graphs below. List your answers for the
domain and range in set builder notation. Then determine whether or not the graph
represents a function.
a.
domain _____________
range ______________
function _______ (y/n)
b.
c.
domain _____________
range ______________
function _______ (y/n)
domain _____________
range ______________
function _______ (y/n)
Unit 3 Section 7
Objectives


The student will list and explain how to use the slope-intercept form
of a linear equation.
The student will graph linear function using slope and y-intercept.
We can graph any function or equation using the method of substitution. However there
is a more efficient method for graphing linear functions or equations. This more efficient
method can be called "critical features". The two critical features for a linear equation
are the slope and the y-intercept. We found out in the previous section that graphing a
line could be done by plotting the y-intercept, counting to find more points on the line,
and then drawing the line through the points. We can use this method if we can find
the slope and intercept from the equation. When a linear equation has the 'y' isolated
it fits a pattern that makes finding these two numbers very easy. This pattern or form
of a linear equation is called the slope-intercept form, it is given below.
The slope-intercept form:
y = mx + b
From the previous section we should recall the letter 'm' represent the slope or rate of
change. And the letter 'b' represents the y-intercept or starting point. So the slope
will be the coefficient of 'x' and the y-intercept will be the constant or plain number in
the equation. We need to stress that the 'y' must be isolated in order to use method for
graphing a linear function. This process can be described in four steps.
1.
2.
3.
4.
Find the slope (m) and the y-intercept (b) from the equation.
Plot the y-intercept (b).
Count using the slope to find more points.
Draw the line through the points.
This is essentially the same process we followed in the previous unit to draw graphs
of lines in some of our problems. The first step of finding the slope and y-intercept has
been added. Below are some examples of finding the slope and intercept from a given
equation.
Example A
Find the slope and intercept from the equation y = 5x + 3
The slope is m = 5 and the y-intercept b = 3.
The 5 is the coefficient of 'x' and the 3 is the constant.
Example B
Find the slope and intercept from the equation y = x - 6
The slope is m = 1 and the y-intercept b = -6.
The understood coefficient of 'x' is 1 and so it is our slope. We must
also be sure to keep the negative sign with the constant so that the
y-intercept is negative.
Example C
Find the slope and intercept from the equation y = -2x
The slope is m = -2 and the y-intercept b = 0.
The y-intercept, b, is zero because our equation could have been
written as y = -2x + 0.
Example D
Find the slope and intercept from the equation y = 4
The slope is m = 0 and the y-intercept b = 4.
The slope, m, is zero because our equation could have written
as y = 0x + 4.
Example E
Find the slope and intercept from the equation
The slope is m = -
Example F
y=-
1
x+7
2
1
and the y-intercept b = 7.
2
Find the slope and intercept from the equation
y=
x
5
3
6
1
5
and the y-intercept b = .
3
6
1
x
1
The slope is
because the term
could have been written as
x.
3
3
3
The slope is m =
Example G
Find the slope and intercept from the equation y = -5.6x + 
The slope is m = -5.6 and the y-intercept b = .
If we know how to identify the slope and intercept in a linear equation then creating the
graph is a straight forward 4-step process.
Below are some examples of graphing linear equations using slope and intercept.
Example A
Graph the equation y = -2x + 6
1. m = -2 or m =
2. Plot the intercept
Example B
2
(to make counting easier) and b = 6
1
3. Count the slope
4. Draw the line
Graph the equation y =
2
and b = -1
3
2. Plot the intercept
2
x-1
3
1. m =
3. Count the slope
4. Draw the line
Example C
Graph the equation y = 2 - x
1. m = -1 and b = 2 The coefficient of 'x' is -1 since we must use the
understood coefficient and keep the sign of the term.
The constant here is 2. The equation could have been
written as y = -1x + 2
2. Plot the intercept
Example D
3. Count the slope
4. Draw the line
Graph the equation y = -3
1. m = 0 and b = -3 The equation could have been written as y = 0x - 3.
A slope of 0 can be thought of as
2. Plot the intercept
3. Count the slope
0
for counting.
1
4. Draw the line
Any equation of the form y = c where c is a constant such as y = -3 will graph as
a horizontal line. Equations of this form could be written as y = 0x + c or
as in our example y = -3. The slope is 0 or zero divided by 1 which equals 0.
There is one type linear equation that does not fit the pattern y = mx + b.
The equation is of the form x = c, where c is any real number. Examples of this type
of equation are: x = 5, x = -3.5 or x = 0. The equation does not have a 'y' in it!
If we consider the equation x = 2 the points that could make this equation true must
always have an x-coordinate of '2'. So some of the points on this line would be:
(2, 3), (2, -4), (2, 1.5), (2, -7) etc. This will create a vertical line with no slope.
There is no slope because when we try to count the slope for a vertical line we have a
zero value for the run in the definition of slope. ( Counting across produces a zero! )
If we place the zero into the definition of rise over run we get a fraction similar to
3
and any number divided by zero is undefined - it has no value so no
0
slope!
Example E
Graph the equation x = 2
Plot the value on the x-axis
Draw the vertical line
Any equation of the form x = c, where c is a constant such as x = 2 will graph
as a vertical line. When we try to count the slope of a vertical line we will get
a constant divided by zero. When we divide by zero the answer is undefined and
we have no slope.
Exercises Unit 3 Section 7
1. In the general equation y = mx + b
a. the “m” represents the ___________ of the line or the rate of ______________
b. the “b” represents the ________________ of the line.
2. The equation y = mx + b is called the ______________________________form of a
linear equation.
3. To graph a linear equation we should first ______________ the y variable.
4. If an equation is in the form y = mx + b explain ( list the steps ) for graphing it.
5. List the slope and intercept for each of the equations below.
a. y = 4x + 5
d. y = x +
1
2
g. y = 6.2x - 5.1
b. y = -3x + 11
c. y = -x
e. y = -8
f. y =
h. x = 3
i. y = 8x -
1
x-9
3
5
4
Using the slope/intercept form of a linear equation ( y = mx + b ) graph the following
equations. List the slope and intercept before making your graph.
6.
y = 3x - 5
7. y = -2x + 6
9.
y = x +2
10. y = -
1
x+5
2
11.
y=
12.
y=x
13. y = -
2
x+4
3
14.
y = -x + 4
15.
y = -2
16. y = -x
15.
x=4
18. y = 5x - 7
19. y = -2x
20.
y=
21.
22. y = -
23.
y = 2.5
y = .4x - 3
3
x+5
2
8.
y = 4x - 3
4
x-5
3
5
x-1
2
Unit 3 Section 8
Objectives

The student will isolate y in linear equations and then graph the
resulting equation using the slope and y-intercept.
In the previous section we learned how to graph equations of the form y = mx + b.
Many linear equations do not have the y variable isolated and as so we must first use
our skills for solving equations to get the y by itself before we graph the line. Some
examples are given below.
Example A
Graph y - 2x = -3
starting with y - 2x = -3
+ 2x + 2x
y is isolated
y = 2x - 3
so m = 2 and b = -3
Example B
Graph 2y + x = 8
starting with 2y + x = 8
-x
-x
y is isolated
so m = -
2y = -x + 8
2
2
y=-
1
x+4
2
1
and b = 4
2
2
Exercises Unit 3 Section 8
1. In the general equation y = mx + b
a. the “m” represents the ___________ of the line or the rate of ______________
b. the “b” represents the ________________ of the line.
2. To graph a linear equation we should first ______________ the y variable.
3. Isolate y in the equations below:
a. 2y = 6x - 8
d. -y =
1
x+3
2
b.
y + 7 = 3x - 2
c. 2y - 3x = 12
e.
-2y + 7 = 3x + 3
c.
1
y - 5x = 1
2
4. Using the slope/intercept form of a linear equation ( y = mx + b )
graph the following equations.
a.
y=x-5
b. y = -3x + 7
c.
d.
y=3
e. x = -4
f.
1
x-3
2
y = 5x - 6
y=
3
5. Isolate y and then use m & b to graph the equations.
a.
d.
2y = 4x - 10
1
y=x-3
2
b. y + 3 = -x + 5
c.
4y - 2x = 8
e. -y = 4x - 6
f.
y + 2 = 3(x + 1)
6. Complete the definitions of slope
a. m 
rise
b. m 


c. m 
horizontal change
d. m 
change in
change in
7. Draw a line with the listed y-intercept and slope.
a. m = -2 b = 7
b. m =
3
b = -2
2
8. List the definition we have for a relation.
9. What is the domain of a function or relation?
10. List the definition we have for a function.
c. m = 1 b = 0
Unit 3 Section 9
Objectives
 The student will interpret slopes as rates of change.
Previously in section 6 of this unit we defined the slope of a line. The slope of a line
tells us how fast a line is going up or down. When we deal with application problems
each axis will represent some quantity such as time, distance, money, people, etc.
In applications how fast the line goes up will tell us how fast a population is increasing,
or how the distance to a destination is decreasing, or a similar rate of change. The units
or name for the rate of change will depend on the quantity each axis represents. We
will investigate this with the examples below.
Example A
Paul works as a studio musician for a recording company. When he comes into work he
is paid $80.00. When Paul plays for a session then he receives an extra $40.00 for each
hour of the session. The graph below tracks how his pay grows
as he plays for each hour of a session.
B
$320
$240
A
$160
$80
$0
0
1
2
3
4
5
6 hours
Our task is to find the slope and y-intercept for the graph above. When each of
the axes are labeled then we must pay close attention to the quantity each axis
represents when we find m and b. The scale (number we count by) on the x-axis
is 1 hour. The scale that we count by on the y-axis is $40. We find that we go
up $240. If we count across we get 6 hours. The calculation for the slope
using rise over run is shown below. Notice that the units are important!
m=
240 dollars
which reduces to
6 hr
40
dollars
hr
or
40 dollars per hr
The units in the answer above are important to help us interpret the rate of pay!
The y-intercept is $80. This is the starting point for his pay calculation!
Example B
A large water tank is being drained. The graph below shows the amount of
water remaining in the tank after each hour. Find the slope or rate of change
at which the water is being drained from the tank and use the graph to
determine how much water was in the tank at the start of the operation.
A
Gallons
8000
7000
B
6000
5000
4000
3000
2000
0
1
2
3
4
5 hours
In this example we chose to count from point A to point B because these two
points are at integer intersections of the grid lines. This will make the counting
much easier. In this example we again must pay careful attention to the units
on the scales. The vertical axis in gallons and the horizontal axis is in hours.
If we count from the point on the left, point ‘A’, then we must go down 3
squares on the y-axis but this is actually 3000 gallons, and because we went
down it is-3000 gallons. In counting across we get 4 squares which is
4 hours on the x-axis. The calculation of the slope or rate of change is given
below.
m=
 3000 gallons
which reduces to
4 hr
 750
gallons
hr
or -750 gallons per hr
The units on the slope or rate of change are important in helping us interpret
how fast the tank is being drained.
The y-intercept is the starting point for our graph, so it is also the starting point
for the amount of water in the tank. To find the amount of water in the tank we
can work backwards from point ‘A’. If the rate of change is -750 gallons/hr then
the tank lost 750 gallons in the first hour so the tank had to start at 7750 gallons.
When we are using units for a rate of change we will usually use the word “per”
between the two quantities. The word “per” indicates division which is the operation
we use to calculate a slope or rate of change. Common rates of change are
‘miles per gallon, miles per hour, or dollars per hour, etc.
Exercises Unit 3 Section 9
1. Complete the definitions of slope
m
a.
b.
run
m


c. m 
vertical change
d.
m
change in
change in
2. The slope of a line can also be called a rate of _________________.
3. The small word that tells us when we are dealing with a rate of change is _______
4. What does the abbreviation "mpg" stand for? ______________________________
5. If a business wants to know how fast it is serving customers the rate of change
they would be interested in is ______________ per hour.
6. Given the graphs count to find the slope (use the scales). List your answer
with units. List the starting point or y-intercept with units as well.
a)
b)
Miles
Dollars Earned
d = distance
300
60
250
50
200
40
150
30
100
20
50
10
m = money
t = time
0
c)
feet
1
2
3
4
t = time
5 hours
0
1
2
3
4
5 hours
d)
Altitude of a plane on takeoff
s = s tudents in W HS
a = altitude
600
600
500
590
580
400
570
300
560
200
550
100
540
t = time
530
520
510
500
t = time
1 2
0
1
2
3
4
5 minutes
3
4
5 month of school
( January = 1)
e)
f)
Income
in Dollars
Wickenburg Jeep Tours
People in Wickenburg Amusement Park
# of People
$1000
1500
$800
1200
$600
900
$400
600
$200
300
$0
0
0
5
10
15
Customers
20
25
0
g)
2
4
6
8
Hours after Opening
h)
Commission
in Dollars
Sarah's earnings based
on Sales
Jim's Mileage on a trip
# of miles
$140
250
$120
200
$100
150
$80
100
$40
50
$0
0
0
5
10
15
# of Sales
20
25
0
4
8
12
14
gallons of gas
7. Find the slope and y-intercept of each line.
a.
10
b.
c.
16
8. Using slope/intercept form of a linear equation ( y = mx + b ) graph the following
equations.
4
a. y = -x + 4
b. y =
x-6
c. x = 3
3
9. Isolate y and then use m & b to graph the equations.
a.
-3y = 6x - 12
b. y - 2 = 3x - 5
c.
3y + 2x = 9
10. Graph the functions below using the method of substitution. You must create
a table, pick five points and the draw the graph.
a. y = (x + 1)2 - 2
b.
f(x) = -2|x -1| + 4
11. Decide which of the following rosters in list form represent a function and are
non-functions. ( Write yes for a function and no for a non-function )
a.
b.
c.
d.
e.
f.
{(4, 7), (-1, 11), (4, -2) (0, 3)}
{(6, -½), (-4, -½), (1.5, -½) (6, -½)}
{(4, 2), (-5,-2), (-9, 2), (4, 5)}
{(-4, -4), (7, 7), (2, 2) ( 5 , 5 )}
{(1, 6), (6, 1), (9,  ) (  , 9)}
{(-4, 5), (-4,-1), (-4, ½) (-4, 0)}
12. Find the domain and range for each of the graphs below. List your answers for the
domain and range in set builder notation. Then determine whether or not the
graph represents a function.
a.
b.
domain __________
range ___________
function ____ (yes/no)
domain _________
range ___________
function ___ (yes/no)
c.
domain __________
range ____________
function ____ (yes/no)
13. Find the domain and range for each of the graphs below and decide
if it is a function or not.
a.
b.
domain __________
range ___________
function ____ (yes/no)
c.
domain _________
range ___________
function ___ (yes/no)
domain __________
range ____________
function ____ (yes/no)
14. Convert the set listed in set builder notation into a graph on the real number line.
a. { x | x < -1 }
b. { y | y > 0 }
15. Convert each set represented as a graph into a rule in set builder notation.
a.
-3
-2
-1
0
1
2
3
-3
-2
-1
0
1
2
3
b.
Unit 3 Section 10
Objectives

The student will recognize equivalent representations of a set in
a roster, table, graph or equation form.
The key to recognizing equivalent forms is being able to determine when a point lies
on a line. A point is on a line when it makes the line’s equation true. Below are two
examples that show us how to substitute a point into and equation to see if it will make
it true.
Example A
Example B
Is the point (3, -5) on the
line whose equation is
y = -2x + 1
Is the point (-2, 4) on the
line whose equation is
y = -2x + 1
-5 = -2(3) + 1
-5 = -6 + 1
-5 = -5
4 = -2(-2) + 1
4=4+1
4=5
In example A the point
makes the equation true
so (3, -5) is on the line.
In example B the point
makes the equation false
so (-2, 4) is NOT on the line.
When comparing a set of points in a roster or table we must check each point to see
if it makes the equation true. The example below will check to see if the given
table or roster is equivalent to the given equation.
Example
Given the equation y = 3x – 7
decide if the given table and roster are equivalent.
{(5, 8), (-1, -10), (0, -7)}
8 = 3(5) – 7
-10 = 3(-1) – 7
-7 = 3(0) – 7
true
true
true
x
y
1
-2
6
-4
13
12
-4 = 3(1) – 7 true
13 = 3(-2) – 7 false
12 = 3(6) – 7 false
All of the points in the roster made the equation true and so the roster and equation
are an equivalent forms. The table, however, had only one point that made the
equation true so they are NOT equivalent forms. If even one point fails to make the
equation true then the forms are NOT equivalent.
If we are given a graph we can determine when an equation is an equivalent form
by comparing the slope and intercept from the equation to the slope and intercept
of the graph. The multiple choice problem below is an example of making this type
of comparison.
Which of the equations below best represents the graph?
a) y = 
4
3
x+2
2
2
2
x-2
3
3
c) y =
x+2
2
2
d) y =  x + 2
3
b) y =
-4
-2
2
4
-2
-4
The graph to the right crosses the y-axis at positive 2, so b = 2. To find the slope we
2
will count down 2 and across 3, which gives us a slope of m =  . The equation in
3
choice ‘d’ has the same slope and intercept so it is the equivalent form.
Exercises Unit 3 Section 10
1. Determine whether the listed point makes the equation true. Answer yes or no.
a. (3, 5)
y = 2x - 1
b. (-2, 4)
y = .5x + 5
c. (-3, -1) f(x) = 2x - 5
d. (1.5, 3.25) y = x2 + 1
e. (-5, 2) y = |x + 1| - 2
f. (6, 7.5)
2y + 1 = 3x - 4
1
h. ( , 4.5)
y = 5x + 2
2
g. (9, 5)
f(x) = -x + 4
2. How can we tell when a point lies on a given line?
3. Given the table to the right, which line will contain all the points in the table?
a) y = 2x + 2
b) f(x) = 4x - 1
c) 3y = x - 1
x
y
1
2
4
7
d) y = 3x + 1
3
10
4
13
4. Given the table to the right, which line will contain all the points in the table?
a) y = x - 1
b) f(x) = 2x - 5
c) 2y = x + 2
x
y
4
6
3
4
d) y = .25x + 2
0
1
5
3.5
5. Given the table to the right, which line will contain all the points in the table?
a) y = 2x + 2
b) f(x) = 4x + 6
c) y - 3x = 4
d) y + x = -4
x
y
-2
-1
-2
1
1
7
2
10
6. Which of the rosters below satisfies the equation y = 2x ( has points that make
the equation true ).
a)
b)
c)
d)
{ (3, 6), (4, 2), (1.5, .75) }
{ (1.5, 3), (4, 8), (0, 0) }
{ (0, 2), (-3, -6), (1.4, 2.8) }
{ (-2, -4), (1.4, .7), (-5, 10) }
7. Which of the rosters below satisfies the equation y + 2x = -3 ( has points that
make the equation true ).
a)
b)
c)
d)
{ (3, -3), (1, -5), (1.5, 0) }
{ (-1.5, 3), (-2, 0), (0, -3) }
{ (6, -15), (2.5, 8), (.5, -2) }
{ (0, -3), (.5, -4), (1.3, -5.6) }
8. Which of the rosters below satisfies the equation 3y - 2x = 1 ( has points that
make the equation true ).
a) { (-2, -1), (0,
1
), (-.5, 0) }
3
c) { (7, 5), (1.5, 1.5), (1, 1) }
1
1
), (2, 1 ) }
3
3
2
d) { (1, 1), (.5, 2.5), (0, ) }
3
b) { (1, 1), (-1,
9. Which of the equations below best represents the graph?
a)
b)
c)
d)
y = -x + 1
y=x-1
y-x=1
-y = -(x + 1)
4
2
-4
-2
2
4
-2
-4
10. Which of the equations below best represents the graph?
a)
b)
c)
d)
y = -(x + 2)
y - 3x = -2
3y = x + 6
y = -4(x + .5)
4
2
-4
-2
2
4
2
4
-2
-4
11. Which of the equations below best represents the graph?
a)
b)
c)
d)
y = -(x - 2)
y = -2x + 2
2y = -x + 4
y = .5(x + 4)
4
2
-4
-2
-2
-4
12. A relation is a _________ of _____________ pairs
13. What is the set of y-values that we generate as answers called?
14. What is the set of x-values that we are allowed to use called?
15. List the definition we have for a function.
16. What is the special or important feature of a function?
17. Graph the following equations using the method of substitution with the
given domain. ( You must pick points create and fill in a table then plot
the points. )
a. y = |x + 2|
D = { -2, -1, 0, 1, 2 }
b.
y = 2x2 - 5
D = {-2, -1, 0, 1, 2}
18. Use the vertical line test to determine which graphs below represent functions
a.
b.
c.
19. Determine which equations represent functions and which do not, using
substitution. Show the substitution and answer yes or no.
b. y2 = x2
a. y = -.5x + 3
c. y = |x + 3|
20. Complete the definitions of slope
m
a.
run
b.
m


c. m 
vertical change
d.
m
change in
change in
21. The slope of a line can also be called a rate of _________________.
22. What is the small word that tells us when we are dealing with a rate of change?
23. Using y = mx + b graph the following equations.
a.
y=x
b. y = -2x + 7
c.
d.
y = -1
e. x = 1.5
f.
1
x+4
2
y= 2x-4
y=-
3
24. Isolate y in the equations and then graph the equations using y = mx + b.
a.
2y = 3x - 8
b. y + 3 = -2x + 9
c.
-y + x = 2
Unit 4
Writing and Applying Linear Functions
Introduction
In mathematics we often learn to perform a process or operation, and once we
have mastered the operation we are asked to do the reverse. For instance after you
learned how to add you learned how to reverse addition, or subtract. Once you learned
how to multiply you then had to learn how to divide.
In Unit 3 we took equations and graphed them using the method of substitution
or the slope intercept form of a linear equation. The process we mastered was to start
with an equation and find its graph or points. Frequently in solving problems we will be
asked to start with a set of points or a table and asked to find the equation so that we
can make a prediction or find a solution.
For instance if we were to collect data on students to correlate the height of
the student to the length of the foot we might get a table similar to:
Height
Foot Length
Height
Foot Length
Height
Foot Length
132
21
164
28
181
30
153
23
165
29
182
31
154
23
166
24
183
31
159
24
171
30
184
32
160
25
172
28
186
30
161
27
173
28
187
32
163
28
177
30
190
34
This table can be turned into a scatter plot, and using the techniques we will learn in this
unit, the scatter plot will have a line of best fit; and the line of best fit will have an equation
that we can find. For the set of data above the line of best fit has an equation of
y = .25x – 50.4. The equation can be used to answer questions like:
What is the foot size for the tallest man in the world?
If we find a foot print that is 38 cm how tall is this person?
Forensic investigators would be very likely to use the second question and our equation
to find information that could help identify the owner of the foot print.
We will need to start with data sets or rates of change and find equations to make
predictions.
Unit 4
Vocabulary and Concepts
Constant of Proportionality
The constant of proportionality is the real number we
multiply an x-value by to produce the y-value.
Direct Variation Direct Variation occurs when a change in ‘x’ produces a
proportional change in ‘y’.
model
To model a situation means to create an equation that generates
the data points for the data set the situation describes or a study
collected.
x-intercept
The x-intercept is the location where a graph intersects the x-axis.
y-intercept
The y-intercept is the location where a graph intersects the y-axis.
Key Concepts for Finding and Applying Linear Equations
The Slope Formula: The slope of a line passing through two points (x1, y1)
y  y1
and (x2, y2) is given as: m  2
x2  x1
Forms of Linear Equations: the three forms are:
Slope-Intercept Form y = m x + b
m is the slope, b is the y-intercept
Point-Slope Form
y - y1 = m (x - x1) (x1, y1) are the coordinates
Standard Form
Ax + By = C
A,B, and C are integers and A is positive.
Finding Intercepts:
To find the x-intercept we set ‘y’ equal to zero and
simplify or solve.
To find the y-intercept we set ‘x’ equal to zero and
simplify or solve.
The General Equation of Direct Variation:
The equation is y = kx where ‘k’ is the constant of
proportionality
Finding the Constant of Proportionality:
To find the constant of proportionality we divide the
y-value or coordinate by the x-value or coordinate.
Unit 4 Section 1
Objectives

The student will use a formula to find slopes.
In Unit 3 we learned that we can find the slope of a line by counting. The two examples
below show us finding the slopes by counting.
Example A
4
1
Example B
2
3
2
1
1
2
1
The slope in example A is m =
2
3
4
2
or 2 and example B has a slope of m = - .
2
3
Counting is a method that we will not always be able to use. We will frequently be
given the coordinates of the two points and asked to find slope. The coordinates for
the points in example A are (-2, -1) and (0, 3). The y-coordinates give us the vertical
locations for the points and tell us how far we must go up. If we start at -1 and go up
to 3 the change in the y-coordinate is 4. This can be found by subtraction, 3 – (-1) = 4.
The x-coordinates give us the horizontal locations of the points and tell us how far we
must go across. If we start at -2 and go across to 0 the change in the x-coordinates
is 2. This can be found again by subtracting 0 – (-2) = 2. One of the definitions of
slope is change in y divided by change in x. When we do the division (4/2) our
slope m = 2.
In example B the coordinates are (-6, 3) and (-3, 1). If we subtract the y-coordinates
we get 1 – 3 = -2. When we subtract the x coordinates we get -3 – (-6) = 3.
2
Then using the definition again we find the slope is -2/3 so m = - .
3
So the process of finding the slope of a line passing through two points is to subtract
the x and y coordinates and divide the difference in the y’s by the difference in the x’s.
This process is best described by a formula. For the formula we must have two points
as our starting information. We will call these two points (x1, y1) and (x2, y2). The
subscripts tell us which point the coordinates come from. Point #1 has subscripts of 1
and point #2 has subscripts of 2. The formula then must subtract the y-coordinates
and divide this number by the difference of the x-coordinates. The formula is below.
y 2  y1
x2  x1
m
Below are some examples of finding slopes using the formula.
Example A
Find the slope of the line passing through the points (1, 3) and (4, 11).
First we label the points
x1 y1
x2 y2
(2, 3)
(4, 11)
Then we substitute into the formula
Last we do the arithmetic
Example B
m
m
8
2
and so m = 4.
Find the slope of the line passing through the points (-1, 9) and (4, 6).
First we label the points
x1 y1
x2 y2
(-1, 9)
(4, 6)
y 2  y1
69
and get m 
4  ( 1)
x2  x1
3
and so m   .
5
Then we substitute into the formula
Last we do the arithmetic
Example C
m
m
3
5
Find the slope of the line passing through the points (-1.5, 7) and (1, 8).
First we label the points
x1
y1
x2 y2
(-1.5, 7)
(1, 8)
y  y1
87
and get m 
m 2
1  ( 1.5)
x2  x1
Then we substitute into the formula
m
Last we do the arithmetic
Example D
y 2  y1
11  3
and get m 
42
x2  x1
1
2 .5
and so m  .4 .
Find the slope of the line passing through the points (5, -2) and (1, -2).
First we label the points
x1
y1
x2 y2
(5, -2)
(1, -2)
Then we substitute into the formula
Last we do the arithmetic
m
0
4
m
y 2  y1
 2  (2)
and get m 
1 5
x2  x1
and so m  0 .
Exercises Unit 4 Section 1
1. What is the formula for finding the slope?
2. Write the general equation for a line in slope-intercept form: y = _________________
3. In the y-intercept form, which variable represents the slope?
4. Find the rate of change or slope of the line between the listed points.
a. (6, 1) (8, 7)
b. (-2, 3) (1, -1)
c. (5, -1) (-2, -5)
d. (4, -2) (3, -1)
e. (5, 5) (8, 8)
f. (-1, -6) (2, 14)
g. (4.5, 6) (3, 0)
h. (2, 5) (8, 5)
i. (
j. (3.2, 1.4) (-1.3, 1)
k. (4, -5) (4, 2)
l. (0, 0) (-3, -1.5)
1
1
, 1) (- , -2)
2
2
5. Put the following expressions into slope-intercept form:
a. y + 6x -2 = 0
b. y – 4x + 5 = 0
c. 2y – 6x +4 = 0
d. 3y = 9x – 6
e. 2y = 3x + 8
f. 2y = -5x + 7
6. Complete the definitions of slope
a.
m
run
b.
m


c.
m
vertical change
d.
m
change in
change in
7. What does the slope of a line tell us?
8. Explain how to find the slope of a line by counting from point to point.
9. If a line has a slope of
0
what does this mean? and what kind of line is this?
3
10. If a line has a slope of
5
what does this mean? and what kind of line is this?
0
11. What is a y-intercept?
Unit 4 Section 2
Objectives

The student will write equations for linear functions in slope-intercept
form given slope and intercept.

The student will write equations for linear functions in slope-intercept
form given slope and point.

The student will write equations for linear functions in point slope
form given slope and point.

The student will write equations for linear functions in slope-intercept
form given the graph of the line.
Information or data will be given to us in many forms. We will be given contextual
problems, graphs, data sets, and individual data points. In most cases we will need to
develop an equation to make predictions and find solutions. Frequently writing an
equation requires that we know the slope and y-intercept of the line. If we know these
two quantities we can write the equation using the formula called the "Slope-Intercept
form" of a linear equation. This is one of the most important tools of Algebra.
The Slope-Intercept form of a Linear Equation is
y= mx+b
where ‘m’ is the slope of the line and 'b ’ is the y-intercept.
From our last unit we know we can find the slope and intercept from a graph. Below is
an example of finding the equation of a line from its graph.
In this example we can count to find the slope is 2 and the y-intercept is -1
so the equation is: y = 2x - 1
There will be times when we are given both the slope and the intercept and we need only
to substitute properly to write the equation of the line. Below are some examples of this
process.
Example A
Given m = -3 and b = 5 write the equation of the line. y = -3x + 5
Example B
Given m =
Example C
Given m = 6.2 and b = - write the equation of the line. y = 6.2x - 
Example D
Given m = 0 and b = 2 write the equation of the line. y = 2
1
1
and b = -4 write the equation of the line. y = x - 4
2
2
In this example the initial substitution gives us y = 0x + 2. But using the
Zero Property of Multiplication 0 times 'x' is 0. And the Identity Property
of Addition tells us 0 + 2 = 2. So the equation simplifies to y = 2.
Example E
Given m = 1 and b = 0 write the equation of the line. y = x
In Example E the initial equation would have been y = 1x + 0. But using
the Identity Properties of Multiplication and Addition the equation will
simplify to y = x.
We will also have contextual problems that are similar to the process given above. In
these problems if we can identify the slope or rate of change and the y-intercept or
starting point for our line then we can write the equation. The rate of change can often
be identified by certain words or phrases. For instance the word "per" as in miles per hour
will help us locate a rate of change. There are also times when the words "each" or
"every" indicate rates of change or slopes. Below are some examples.
Example A
Tom works at "Fast Lube". He is paid $200 a week and $5 per customer
whose car he lubes. Write an equation to model how his weekly pay grows
with each customer.
The $200 is the starting point and the $5 is the rate of change (it tells us
how his pay is going to increase). We can write the equation y = 5x + 200
to model his pay. One thing we must be aware of is that the variable
'x' will be the number of customers and the variable y is his pay.
Example B
The WHS Thespian Club started the year with 12 members. The club has
grown by 3 members each month. Write an equation to model the club's
growth.
The starting point or y-intercept is 12 members and the rate of change
or slope is 3 members. The equation would be y = 3x + 12. In this example
the 'x' would represent the number of months and the 'y' would be the
number of members.
Another way in which we might see the parameters for a linear equation given to us
would be as a rate of change or slope and a point that is on the line. If we want our
answer in slope-intercept form then we must find the y-intercept. The process for
finding the y-intercept is listed as a set of steps below.
1.
2.
3.
4.
Write the slope-intercept form of a linear equation.
Substitute the slope (m) and the x and y coordinates.
Solve for b.
Write the equation.
The examples that follow use this 4-step process.
Example A
Given a line has m = -2 and the point (-1, 6) find the equation of the line.
y = mx + b
6 = -2(-1) + b
6 = 2+b
-2
-2
Step 1
Step 2
Step 3
4 = b
y = -2x + 4
Example B
Step 4
Given a line has m = 3.4 and the point (2.1, 11.7) find the line's equation.
y = mx + b
11.7 = 3.4(2.1) + b
11.70 = 7.14 + b
- 7.14
-7.14
Step 1
Step 2
Step 3
4.56 = b
y = 3.4x + 4.56
Step 4
A second form for a linear equation is called point-slope form.
The Point-Slope form of a Linear Equation is
y - y1 = m (x - x1)
where 'm ' is the slope of the line and (x1, y1) are the coordinates of the point.
When we are given the slope of a line and a point on the line and asked for the equation
in point slope form the process is somewhat different. The process for writing the
equation in point-slope form is given by the three steps below.
Step 1. Write the point-slope formula.
Step 2. Substitute the slope and the coordinates.
Step 3. Simplify if needed.
Below is an example of using the point-slope form.
Example Given a line has m = 
5
and the point (1/6, -4) find the line's equation
3
in point-slope form.
y - y1 = m (x - x1)
Step 1
5
1
y - (-4) =  (x - )
3
6
Step 2
5
1
y + 4 =  (x - )
3
6
Step 3
Exercises Unit 4 Section 2
1. In the equation y = m x + b
a. What is ‘m’ ?
b. What is ‘b’ ?
c. What is the name of this equation?
Given m and b write the equation of the line they determine.
2. m = 11 b = 7
3. m = 4, b = -2
4. m = -6 b = 5
5. m =
2
3
b = -4
6. m = 1 b = 7
7. m = -1, b =
8. m =
4
5
b=-
2
7
11. m = 5.5, b = -.4
9. m =
1
8
b=0
1
3
10. m = 0 b = 5
Given the graph find the slope, y-intercept and equation of the line.
12.
13.
m = _______
b = _______
equation is ____________
14.
m = _______
b = _______
equation is _________
m = _______
b = _______
equation is __________
Given the point and the slope find the equation of the line they determine in
Slope-Intercept form.
15. m = 2 (3, 1)
16.
m = -3 (4, -5)
18. m = 0 (2, 3)
19. m = -
17. m =
2
(-3, 5.5)
3
20.
1
(4, -3)
2
3
4
m=
(1, 1)
Given the slope and point find the equation of the line in Point-Slope form.
21. m = -5 (-1, 6)
22. m = 4.5 (2, -1.5)
1
(0, 3)
2
24. m = -4 (.5, -1.5)
23. m = -
Graph the following equations.
3
x-5
2
25.
y=6-x
26. y = -3x + 1
27.
y=
28.
y = -1
29. x = -1
30.
y = -4x + 5
3
Isolate y (in the space provided) and then use m & b to graph the equations.
31.
.5y = x - 3
32. -y + 3 = x + 4
33.
3y - 2x = -6
Unit 4 Section 3
Objectives

The student will write equations for linear functions in slope-intercept
form given two points.

The student will write equations for linear functions in point-slope
form given two points.
From the last section we should know how to find the equation of a linear function when
we are given the slope and a point. We need to examine the process of finding the
equation when we are given two points. We can write our equations in both point-slope
and slope-intercept forms. Both of these require that we find the slope first. From
section 1 in this unit we know that there is a formula for finding the slope of a line that
passes through two points. That formula is:
m
y 2  y1
x2  x1
So whenever we start with two points as our given we must first find the slope. Then
we can apply the steps we learned in the previous section. Below are examples of
starting with two points and finding the equation in slope-intercept form.
Example A
Given two points (4, -2) (8, 0) find the equation of the line they determine
in slope-intercept form.
First we find the slope
m
y 2  y1
0  (2) 2 1
 
=
84
4 2
x2  x1
We can now pick either of our two points. The point (4, -2) is listed first
so we will use it to find b.
y = mx + b
1
-2 =
(4) + b
2
-2 = 2 + b
-2
-2
-4 = b
Now that we have found both m =
1
and b = -4 we can write our equation.
2
So our equation is y =
Example B
1
x-4
2
Given the table to the right is a linear function, find the
equation of the line that passes through all the points.
Since all of the points are on the same line, the
slope between any two points we can pick will always be
the same. So we can pick any two points we like.
Let’s try the first two points:
(-1, 7) and (1, 3)
x
y
-1
1
7
3
2
1
4 -3
5 -5
We begin by finding the slope.
m
y 2  y1
37
4
=

 2
1  ( 1)
2
x2  x1
We can now pick either of our two points. The point (1, 3) has easier
numbers to work with, so we will use it to find b.
y = mx + b
3 = -2(1) + b
3 = -2 + b
+2
+2
5 = b
Now that we have found both m = -2 and b = 5 we can write our
equation. So our equation is y = -2x + 5
Finding the equation of a line when given two points in point-slope form also begins by
finding the slope and applying the previously learned techniques. The examples below
show us how to do this process.
Example A
Given two points (5, -3) (2, 1) find the equation of the line they determine
in point-slope form.
First we find the slope
m
4
y 2  y1
1  (3)
4


=
25
3
3
x2  x1
We can now pick either of our two points. The point (5, -3) is listed first
so we will substitute this point into point-slope equation
y - y1 = m (x - x1)
4
y - (-3) = - (x - 5)
3
y+3 =-
4
(x - 5)
3
And so we have our equation.
Example B
Given the table to the right is a linear function, find the
equation of the line that passes through all the points.
Since all of the points are on the same line, the slope
between any two points we can pick will always be
the same. So we can pick any two points we like.
Let’s try the first two points:
x
y
2.5
1.5
-4
-2.5
0.5
-1
-0.5 0.5
-1.5
(2.5, -4) and (1.5, -2.5)
2
We begin by finding the slope.
m
1.5
y 2  y1
 2.5  (4)

  1.5
=
1.5  2.5
1
x2  x1
We can now pick either of our two points. The point (2.5, -4) is listed first
so we will substitute this point into point-slope equation
y - y1 = m (x - x1)
y - (-4) = -1.5(x - 2.5)
y + 4 = -1.5(x - 2.5)
And so we have our equation.
Exercises Unit 4 Section 3
1. List the formula for the slope of a line passing through two points.
Using the formula above, find the slope of the line passing through the two points.
Show your work.
2. (3, 2) (-2, 12)
3. (-7, 3) (2, -3)
4. (2, 0) (-2, 3)
The rosters below represent lines. Find the slope of the line that passes through
the points. (You can choose any two points.) Show your work.
5. { (-2, 5), (0, 1), (1, -1), (2, -3) }
6. { (4, -2), (6, 1), (10, 7), (-2, -11) }
Given the tables below represent lines. Find the slope of the line that passes through
the points. Show your work.
7.
x
y
4
8.
x
y
-12
.2
1.4
1
3
.7 1.5
3
-7
1.2 1.6
0
8
1.7 1.7
Find the equation of the line that would pass through the two given points. The
equation should be in slope-intercept form. Show your work.
9. (3, 2) (5, -2)
12. (2,
1
1
) (4, - )
2
2
15. (-1, 3) (1, -3)
10. ( -1, .5) ( 3, 2.5)
11. (-7, 6) (-5, 0)
13.
14. (3, 6) (-4, 6)
16.
(9, -2) (4, 3)
(-3, 1) (6, 7)
17. (8, 1) (8, 3)
Find the equation of the line that would pass through the two given points. List your
answer in point-slope form.
18. (8, -1) (4, 11)
19. ( -4, 0) ( 4, 2)
Given m and b write the equation of the line they determine.
20. m = 3, b = -7
21. m = -1.2 b = 3.5
22. m = -1, b =
23. m = 0 b = 9
24. m = -2, b = 0
25. m =
1
4
b=-
1
2
5
3
Given the point and the slope find the equation of the line in slope-intercept form.
( Show Your Work )
26. m = 3 (4, 1)
27.
m = -2 (-5, 2)
Graph the Following.
28.
y=x
29. y = -
3
x+5
4
30.
y + 2x = 4
31. What is the equation of the line in slope-intercept form that passes through the
point (2, 3) and has a slope m = -2
a) y = 2x - 1
b) y = -2x + 7
c) y = 3x -2
d) y = -2x + 1
32. What is the equation of the line in point slope form that passes through the
point ( .5, 1.4) and has a slope of m = -.3
a) y - 1.4 = -.3(x - .5)
b) y - .5 = -.3(x - 1.4)
c) x - .3 = 1.4(y - .5)
d) x + .3 = .5(y - 1.4)
33. Given the graph of a line to the right. The equation of a line with
the same y-intercept but passing through the point ( 4, 2 ) is
a) y = 4x - 2
c) y =
2
x-2
3
10
b) y = -x - 4
d) y =
3
x-4
2
-10
10
-10
Unit 4 Section 4
Objectives

The student will convert between slope-intercept, point-slope, and
standard form for a linear equation.
Linear equations can be written in three forms. These three forms are:
Slope-Intercept Form
y = mx + b
Point-Slope Form
y - y1 = m (x - x1)
Standard Form
Ax + By = C
We have used the Slope-Intercept and Point-Slope forms in the last section. Standard
form is useful when we must find the x and y intercepts. In Standard form the A,B and
C are the coefficients and constants while we still use x and y as our normal variables.
The equation of a line can be represented in any or all of the above forms. Below are
examples of converting Slope-Intercept Form into Standard form.
Example A
Given the equation y = 2x + 7 convert the equation to standard form.
y = 2x + 7
-2x = -2x
-2x + y = 7
Our equation is -2x + y = 7
Example B
Given the equation y =
3
x-6
2
3
2 . y = ( x - 6) . 2
2
2y = 3x - 12
-3x = -3x
y =
-3x + 2y = -12
In standard form the 'x' and 'y' are on the
same side of the equation. So we must
subtract '2x' from both sides. In standard
form the 'x' term should always come first so
we put the '-2x' as the first term.
3
x - 6 convert the equation to standard form.
2
In this example we start by using the
Multiplicative Property of Equality to clear the
denominator. Then the Distributive Property
is used to remove the parentheses. Finally we
subtract the '3x' from both sides to get the
x and y terms on the left hand side of the
equation.
Our equation is -3x + 2y = -12
We will also at times need to convert Standard form into Slope-Intercept form. Below are
some examples of this process.
Example A
Given the equation 4x + y = 3 convert the equation into slope-intercept form.
4x + y = 3
-4x
-4x
In slope-intercept form the 'y' variable is
isolated. So we must use the Subtraction
Property of Equality and subtract the '4x'
from both sides to get 'y' by itself.
y = -4x + 3
Our equation is y = -4x + 3
Example B Given the equation -5x + 3y = 2 convert the equation into slope-intercept form.
-5x + 3y = 2
+5x
+5x
3y = 5x + 2
3
Our equation is y =
Example C
3
3
In this example we start by adding '5x' to
both sides with the Addition Property of
Equality. Then we divide everything on both
sides of the equation by 3 in order to get
‘y’ by itself.
5
2
x+
3
3
Given the equation 9x - y = -4 convert the equation into slope-intercept form.
9x - y = -4
-9x
-9x
-1y = -9x - 4
-1
-1 -1
Our equation is y = 9x + 4
In this example we start by subtracting '9x'
from both sides with the Subtraction Property
of Equality. We also inserted the understood
coefficient of '1' with y to make sure we
remembered to use the Division Property of
Equality and divide everything by '-1'.
Another conversion between forms that we will need to do is to change from Point-Slope
form to Slope-Intercept form. Examples of this conversion follow on the next page.
Example A
Given y - 4 = 5(x - 1) convert the equation into slope-intercept form.
y - 4 = 5(x - 1)
y - 4 = 5x - 5
+4
+4
y = 5x - 1
Our equation is y = 5x - 1
Example B
In converting to slope-intercept form we
must again isolate y. Before we can
start to isolate y we must get rid of the
parentheses and so we use the
Distributive Property. Then we use the
Addition Property of Equality by adding
4 to each side to isolate y.
2
Given y + 3 = - (x - 4) convert the equation into slope-intercept form.
3
2
(x - 4)
3
2
3(y + 3) = 3 . - (x - 4)
3
3(y + 3) = -2(x - 4)
y+3 =-
3y + 9 = -2x + 8
-9
-9
One method for isolating y can be to first
clear the denominator. To do this we use
the Multiplication Property of Equality and
multiply by '3' on both sides. We then use
the Distributive Property on both sides. We
subtract '9' from both sides and then
divide everything by 3 on both sides.
3y = -2x - 1
3
3
3
Our equation is y = -
2
1
x3
3
Converting from Point-Slope form to Standard form is similar. The example below
shows this conversion.
Example A
Given y - 3 =
7
(x + 2) convert the equation into slope-intercept form.
4
7
(x + 2)
4
7
4(y - 3) = 4 . (x + 2)
4
4(y - 3) = 7(x + 2)
y-3 =
4y - 12 = 7x + 14
+12
+12
4y = 7x + 26
-7x -7x
The most efficient method for dealing with
the fractional slope is to clear the
denominator. To do this we multiply by 4
on both sides. After this we use the
Distributive Property on both sides. Then
we use the Addition and Subtraction
Properties of Equality to get the equation
into standard form.
-7x + 4y = 26
Our equation is
-7x + 4y = 26
Exercises Unit 4 Section 4
1. List the equation for the slope of a line passing through two points (x1, y1) and (x2, y2).
2. List the three forms of a linear equation.
Convert the equations in slope-intercept form to standard form.
3. y = 3x + 2
6. y = -
1
x + 10
2
4. y = -4x - 1
7. y =
5. y =
5
x
2
2
x+4
3
8. y = .7x + 5
Convert the equations from point-slope form into slope-intercept form and
standard form.
1
(x - 4)
3
9. y - 6 = -2(x + 4)
10. y + 1 =
12. y + 4 = 6(x + 1)
13. y - 9 = 3(x - 3)
11. y - .5 =
14. y -
4
(x + 6)
3
1
1
= 2(x + )
2
3
Convert the equations from standard form to slope-intercept form
15. 2x + y = 15
16. x + 6y = 3
17. -4x - 2y = -12
18. 3x - y = 12
19. 5x + 2y = 0
20. x - y = -1
Using the formula, find the slope of the line passing through the two points.
21. (8, 3) (10, 2)
22. (-4, 1) (2, 7)
23. (2, 3) (2, -6)
The roster and table below represent lines. Find the slope of the line that passes
through the points.
24. { (4, 6), (5, 9), (6, 12), (7, 15) }
25.
x
y
-1
0
4
3.5
1
3
2 2.5
Find the equation of the line that would pass through the two given points. List your
answer in the slope-intercept form of a linear equation.
26. (-3, 7) (5, -9)
27. ( -1, 8
1
1
) ( 1, 9 )
2
2
Find the equation of the line that would pass through the two given points. List your
answer in point-slope form of a linear equation.
28. (1, 3) (5, 2)
29. ( -2, 1.5) ( 0, 3.5)
Given m and b write the equation of the line they determine.
30. m = -8, b = 3
31. m = -2.5 b = 3.1
32. m = 1 b =
2
3
Given the graph find the slope, y-intercept and equation of the line.
33.
34.
35.
Given the point and the slope find the equation of the line in slope-intercept form.
36. m = 2 (5, -3)
37.
m=-
5
2
(-4, 1)
Graph the Following
38.
y = -x
39. y = -
4
x+6
3
40.
4y - 2x = 8
41. What is the equation of the line in standard form that passes through the
points (1, 3) and (5, 1)
a) x + 2y = 5
b) 2x + y = 7
c) -x + 2y = 5
d) x + 2y = 7
42. What is the equation of the line in standard form that passes through the
point ( .5, 2.2) and has a slope of m = .4
a) 2y + y = -2
b) 2x -5y = 10
c) 5y + 2x = 10
d) -2x + 5y = 10
43. Given the graph of a line to the right. The equation of a line with
the same y-intercept but passing through the point ( 4, 2 ) is
a) y = -x + 6
c) y =
2
x+3
3
b) y = -
d) y =
10
3
x-6
2
3
x+6
2
-10
10
-10
Unit 4 Section 5
Objectives

The student will find the x and y intercepts of a linear function.
The y-intercept is a point at which the graph intersects or crosses on the y-axis. This
value or point is important for two reasons. First we can consider it to be the starting
point for our graphs. Second it is the location at which the x-values change from
negative to positive. The change in sign is critical in some of our applications. The
x-intercept is also important as the location where the graph goes from positive to
negative. The graph below illustrates these ideas.
Example
profit/loss
The graph to the left predicts a
business’s profit or loss for the first
five months after it opens. The yintercept predicts the business will
start at -$3000. The x-intercept
occurs at 2 months and this means
we go from a loss to a profit at 2
months. This would be a very
important point to a new business.
$4000
$2000
$0
1
-$2000
2
3
months
4
5
-$4000
Finding the x and y intercepts from a graph is done by inspection. Finding the x and y
intercepts from an equation requires a different approach. The graphs below illustrate
the ideas we will be using.
Example A
Example B
(0,6)
(-2,0)
(-6,0)
(3,0)
(5,0)
(0,2)
(0,-1)
(0,-7)
In example A each of these lines have
points at the coordinates of the y-intercept.
The x coordinate in all these points is 0.
In example B each of the lines have points
at the coordinates of the x-intercept.
The y coordinate in all these points is 0.
The ideas and methods we can draw from these two graphs are:
To find the y-intercept we must set the x variable in the equation to 0 and solve
for the y-value.
To find the x-intercept we must set the y variable in the equation to 0 and solve
for the x-value.
The examples below show us using these methods with equations in various forms.
Example A
Given the equation y = 3x – 7 find the x and y intercepts.
To find the y-intercept set x = 0.
y
y
y
y
=
=
=
=
3x – 7
3(0) – 7
0–7
-7
The y-intercept is -7
To find the x-intercept set y = 0.
y = 3x – 7
0 = 3x - 7
+7
+7
7 = 3x
3
3
7
=x
3
The x-intercept is
Example B
7
3
Given the equation 4x – 3y = 24 find the x and y intercepts.
To find the y-intercept set x = 0.
4x – 3y = 24
4(0) – 3y = 24
0 - 3y = 24
-3y = 24
-3
-3
y = -8
The y-intercept is -8
To find the x-intercept set y = 0.
4x – 3y = 24
4x – 3(0) = 24
4x – 0 = 24
4x = 24
4
4
x=6
The x-intercept is 6
Example C
Given the equation y – 9 = 3(x + 2) find the x and y intercepts.
To find the y-intercept set x = 0.
To find the x-intercept set y = 0.
y – 9 = 3(x + 2)
y – 9 = 3(0 + 2)
y – 9 = 3(2)
y–9= 6
+9 +9
y = 15
The y-intercept is 15
y–9
0–9
–9
-9
-6
-15
3
= 3(x + 2)
= 3(x + 2)
= 3(x + 2)
= 3x + 6
-6
= 3x
3
-5 = x
The y-intercept is -5
Exercises Unit 4 Section 5
1. What is the y-intercept?
2. What is the x-intercept?
3. How do we find the x-intercept?
4. How do we find the y-intercept?
Find the x and y intercepts for each equation.
5. y = 3x - 12
6. 2x + 5y = 10
8. -x + 3y = 4
8. y - 2 =
10. y =
3
x +9
2
1
(x + 4)
2
11. 3y + 5x = -6
7. y = 1.5x - 6
9. y + 3 = .2(x + 1)
12. y – 3 = 3(x + 5)
13. List the formula for the slope of a line through the points (x1, y1) and (x1, y1).
14. y - y1 = m (x - x1) is called the ________________________________ form.
15.
y = m x + b is called the _________________________ form.
16.
Ax + By = C is called the _________________________ form.
Convert the equations into standard form.
17. y = -4x + 9
19. y - 7 = 2(x - 1)
1
x-1
2
5
20. y + 2 = (x + 6)
3
18. y = -
Convert the equations to slope-intercept form
21. 2x - 5y = 15
22. x + 6y = 3
23. y - 8 = 2(x + 3)
Using the formula, find the slope of the line passing through the two points.
24. (7, 4) (10, 2)
25. (-5, 1) (4, -5)
26. (5, -6) (2, -6)
The roster and table below represent lines. Find the slope of the line that passes
through the points.
27. { (-4, 5), (-2, 4), (2, 2), (4, 1) }
28.
x
y
-2
0
1.3
2
2
2.7
4 3.4
Find the equation of the line that would pass through the two given points. List
your answers in standard form.
1
1
29. (-5, 3) (1, -9)
30. ( -1, ) ( 2, -2 )
2
2
Find the equation of the line that would pass through the two given points. List
your answers in point-slope form.
31. (3, 4) (5, 10)
32. ( -1, 7.5) ( 1, 1.5)
Find the equation of the line that would pass through the two given points. List
your answers in slope-intercept form.
33. (-2, -1) (0, 0)
34. ( 0, 1) ( 4, 2)
Given the point and the slope find the equation of the line in slope-intercept form.
35. m = 3 (7, -2)
36.
m=-
5
4
(4, 6)
Graph the Following
37.
y = -2
38. x = 5
40. If f(x) = 3x - 10 then f(x + 2) is
a) 3x - 4
b) 3x - 16
c) 3x - 8
d) 3x - 20
39.
y = - 6x
41. Given the points (a, b) and (-a, -b) the equation of the line passing through the
points is
a) y =
b
x
a
b)
c) y =
a
x-a+b
b
d) y =
x=
b
y+a-b
a
a
x+a-b
b
42. What is the equation of the line in standard form that passes through the
1
point ( 3, .5) and has a slope of m = 2
a) 2y + x = 6.5
b) x + 2y = 4
c) x - 2y = 2
d) 3x + 2y = 4
Unit 4 Section 6
Objectives

The student will write linear equations from contextual situations.

The student will be able to solve contextual problems with linear
equations they create.
Linear equations can model many situations we see in everyday life. The ability to "model"
situations is quite important in mathematics. When an equation models a situation that
means it can generate the data points and allow us to predict new points. Equations
that run on computers model weather patterns and help predict hurricanes, storms, and
other weather patterns. Equations can be used to model TV viewer preferences and
help networks select shows and schedules. Sports teams use statistics and equations
to model their team and find ways to improve. Simulations using equations can model
a city's growth patterns and help planners make critical decisions. Equations that run
on computers model the moves of athletes and make many of our sports video games
possible.
In this section we will use our ability to create equations to model many situations and
make predictions using these equations. If we are trying to model a contextual situation
with a linear equation we need to find the rate of change (slope) which is m and the
starting point (y-intercept) which is b. The rate of change can often be identified by
certain words or phrases. For instance the word "per" as in miles per hour will help us
locate a rate of change. There are also times when the words "each" or "every" indicate
rates of change or slopes. Below are some examples.
Example A
Miguel works as a satellite dish installer for an internet provider. He is
paid $300 a week plus $25 for each installation.
Part a. Write an equation that models how his
pay grows with each customer he does
an installation for.
m = $25 and b = $300
so the equation is y = 25x + 300
Where x is the number of installations
and y the amount of his weekly pay.
In the problem above the
word "each" helps us find
the slope. Miguel would
be paid at least $300 and
so this is the starting point.
His pay can only go up
from
the $300.
Example A (continued)
Part b. Use the equation from Part 'a' to
find Miguel's pay if he installs
14 dishes in one week.
y = 25x + 300
y = 25(14) + 300
y = 350 + 300
y = 650
In Part b we are given the
number of dishes he installs,
which is the 'x' variable in
the equation. So we
substitute the 14 for the 'x'
and simplify to find the
amount of money he is paid.
Miguel's pay would be $650
Part c. If Miguel wants to earn $900 in
a week, find the number of
installations he must do.
y = 25x + 300
900 = 25x + 300
-300
-300
600 = 25x
25
25
In Part c we are given the
amount of money Miguel
wants to be paid. the 'y'
variable is the amount of his
pay. So we substitute the
#900 for the y variable and
then solve the equation to
get our answer.
24 = x
Miguel would need to install 24 dishes
In parts 'b' and 'c' of our Example A, it is critical to remember what each variable
represents. After we write our equation we should always decide what quantities
'x' and 'y' stand for.
In our next example we will be asked to do many of the same things. However, we
will start with two points. We have learned previously how to find an equation from
two points. The first thing we must do in a problem like example B that follows
is find the coordinates of each point.
Example B
Sarah owns a hot air balloon and gives customers rides in the balloon
for $300 to parties of 3 people. On a recent flight the balloon was at
a height of 1400 ft. as she started to descend. After 3 minutes the
balloon was at 1070 ft.
Part a. Write an equation to model the
balloon's descent.
The two points are:
(0, 1400) and (3, 1070)
We must pick out the important
values. The problem is about
the time and height not the
money or people so we ignore the
$300 and the 3 people. When
the action or descent started, the
time is considered 0 and the
height 1400. The other point is a
later time and height.
Example B Part a. (continued)
Now we find the slope
m
y 2  y1
1070  1400  330

 110
=
30
3
x2  x1
We now use the point (0, 1400) and the slope to find the y-intercept.
So we substitute the point and slope into slope-intercept equation and
solve for b.
y
1400
1400
1400
=
=
=
=
mx + b
-110(0) + b
0+b
b
Our equation then is y = -110x + 1400
In the equation ‘x’ is the time in minutes
and ‘y’ is the height or altitude.
Part b. Use the equation from Part 'a' to
find the balloon's height after 5 min.
y = -110x + 1400
y = -110(5) + 1400
y = -550 + 1400
y = 850
If 'x' is the time, then 5
min. will be substituted
for 'x'.
The balloon’s altitude would be 850 ft.
Part c. How many minutes will it take for
the balloon to reach the ground?
(The height or altitude would be 0.)
y = -110x + 1400
0 = -110x + 1400
-1400
-1400
-1400 = -110x
-110
To reach the ground means
the height will be 0 ft. Since
the 'y' is the height we
substitute the 0 for the 'y'.
-110
12.72 = x
The balloon will take 12.72 minutes to descend.
Exercises Unit 4 Section 6
1. In a contextual problem, what words can tells us we have a rate of change or slope?
2. In the slope-intercept equation y = mx + b
a. the slope 'm' can also be called the _________________________
b. the y-intercept 'b' can also be called the _______________________
3. Find the rate of change ( slope ) 'm' and the starting point 'b' (y-intercept) in the
following problems and write the equation for the line that models the situation.
a. A plane is at an altitude of 1200 ft. If the plane is climbing at a rate of
200 ft per minute write an equation to model this situation.
m = _________ b = ___________ equation ______________________
b. The temperature at 8:00 AM is 75o. The temperature is rising at a rate of
3o per hour. Write an equation to model this situation.
m = _________ b = ___________ equation ______________________
c. Tim is a sprinter training for track season. His time in the 100m sprint is
currently 12.2 seconds. He plans to lower his time by .1 of second per
month. Write an equation to model this situation.
m = _________ b = ___________ equation ______________________
d. Human hair grows at .5 inches per month. If Mary's hair is 8 inches long,
write an equation that will model the hair growth in this situation.
m = _________ b = ___________ equation ______________________
e. The Mercury Bicycle Messenger Service delivers packages in San Francisco.
They have 24 messengers. The company's charges are calculated as follows:
the basic charge is $30 and then a $5 charge per ounce is added to get the
final cost. Write an equation to model the charges for delivering a package.
m = _________ b = ___________ equation ______________________
f. Juan works at the "Car Stereo Palace". He installs car stereos. He is paid a
weekly salary of $200, He is also paid $25 for every stereo he installs.
Write an equation to model Juan's weekly pay.
m = _________ b = ___________ equation ______________________
4. The SADD club had 8 members at the start of the year. Every meeting for the last
month the membership has grown by 2 members per week.
a. Write an equation to model this growth.
b. Use this equation to find how big the club will be after 8 weeks of school.
c. Use the equation to determine in which week the club will have 44 members.
5. Alejandra is saving for a vacation. She put $50 into a savings account. Her plan is
to add $20 a week to the account.
a. Write an equation to model the money in the account.
b. Use the equation to find how much will be in the account in 7 weeks.
c. If the cruise she will take costs $480 how many weeks will it take her
to save at least this amount?
6. Sam is buying a car. His payments will be $245 per month. The down payment
was $2000.
a. Write an equation to model how much money he will pay for the car
over time.
b. Use the equation to find out how many months he will pay in order to
reach $10,000.
c. Use the equation to find how much money he will pay in the first year.
7. Mike is selling tickets for a raffle. Each tickets costs $6.
a. Write an equation that will model how the money from the sales will
grow with each ticket sold.
b. Use the equation to find when the total amount of money will be
at least $400.
c. Use the equation to find the amount of money Mike will have when
he sells 25 tickets.
8. Find the equation of the line passing through the two points (2, 5) (4, 9) in slope-intercept form.
Read each problem carefully, find the two data points described in the problem, and
then write an equation to model the situation in slope-intercept form. Show your work.
9. After three weeks of lifting Jim's bench press was 110 lbs. In his fifth week his
bench press was 120 lbs. Write an equation to model how his strength is growing.
(
,
) (
,
)
m = __________
equation ____________________
10. Sarah is the pilot of a passenger plane. To land the plan smoothly for the
passengers the plane should lose altitude in a linear fashion. Five minutes
into the landing pattern the plane is at 8000 ft. Four minutes later the plane
is at 2000 ft. Write an equation to model the way the plane is losing altitude.
(
,
) (
,
)
m = __________
equation ______________________
11. A water storage tank is being filled at a steady rate. After 10 minutes the tank
has 420 gallons. After 15 minutes the tank has 530 gallons. Write an equation
to model how the tank is being filled.
(
,
) (
,
)
m = __________
equation _____________________
Find the asked for equation and then use it to answer the questions that follow. Show
all your work.
12. Alex owns a company that sells widgets. In January the company sold 634 widgets.
His sales are growing linearly. In April the company sold 730 widgets.
a. Write an equation to model this situation. ( Hint: the problem is giving you
two points. The months can be 1 for January and 4 for April. )
b. Use the equation to predict how many widgets they will sell in August.
c. Use the equation to decide when the company will sell at least 1200 widgets.
( Alex's goal is to double his sales. )
13. A business has a loss in April of $2000. The company does steadily better each
month and has a profit of $1200 in November. Answer the following
a. The two points from the problem are (
,
) and (
,
).
b. Graph the line these points would determine using a grid similar to the grid below.
7K
5K
3K
1K
-1K
-3K
-5K
-7K
Mar Apr May Jun Jul Aug Sep Oct Nov
c. Find the equation of the line that models this situation. ___________________
d. Use this equation to determine when the business will have a profit of $4000.
( How many months would it take to reach this level of profit? )
Unit 4 Section 7
Objectives

The student will write equations and solve problems with arithmetic
sequences.
In this section we will be using our ability to write equations to help find 'nth' terms for
the sequences or patterns we developed in Unit 1. These patterns or sequences were
indexed by the natural numbers. The natural numbers are the set of counting numbers
we learned long ago, they are:
N = { 1, 2, 3, 4, 5 … }
When we have a pattern or sequence such as bn = 3, 6, 9, 12, 15 … we can use the
natural numbers to identify the terms or elements in the pattern. For instance in the
sequence bn the first term is 3, the second term is 6 the third term is 9. These terms
can be represented by the notation b1 = 3, b2 = 6, b3 = 9, b4 = 12, and so on.
A sequence can be shown as a list of points or a table. The above example could be in
list form as { (1, 3), (2, 6), (3, 9), (4, 12), (5, 15) …} or structured as the table below.
n
bn
1
3
2
6
3
9
4
12
5
15
There are many types of sequences. We are going to study the category of sequences
or patterns we call "arithmetic".
Definition:
An arithmetic sequence is a linear equation with the domain of the natural
numbers.
The domain of a function or relation is the numbers we are allowed to use for the 'x'
variable. If our domain is the natural numbers then we are only supposed to use
the numbers 1, 2, 3, 4, 5, and so on for 'x'.
Because a sequence is a pattern that can be represented by a linear equation we
will be able to extend it by adding the same number as we go from term to term in the
sequence. We can see this when we generate sequences as tables using
linear equations.
Example A
Example B
Example C
Example D
y = 2x + 1
y = -3x + 15
bn = .5n + 1.5
v=u+3
x
y
x
y
n
bn
u
v
1
3
1
12
1
2
1
4
2
5
2
9
2
2.5
2
5
3
7
3
6
3
3
3
6
4
9
4
3
4
3.5
4
7
5
11
5
0
5
4
5
8
In the above tables we can see that the values on the left of the table were the
natural numbers. To generate the right hand column we substituted each of the
natural numbers into the equations to get the answers for the right had column.
Examples A and B used the variables 'x' and 'y' as we often do in Algebra. However
Examples C and D used other variables which may be helpful for some applications.
When other variables are used the right hand variable is always isolated on the
left side of the equation.
We can also use these tables to draw an important conclusion. In example A
the slope of the linear equation is 2. If we look at the right hand column of Example
A we can see that the values go up by 2. We can count 3, 5, 7, 9, 11! In Example B
our slope is -3 and our values in the right hand column go down by 3 so we count
by -3! In example C the slope or rate of change is .5 and to generate the right hand
column we add .5 each time or count by .5! In Example D the slope of the equation
is 1 and in the right column we count by 1! This can help us find the equation when
we start with a table as in the example below.
Example
Given the table below find the equation of the arithmetic sequence
or 'nth' term of this pattern.
x
y
1
7
2
11
3
15
4
19
5
23
In the table we can see that we are counting by 4's. If we
subtract consecutive terms we can see the difference is 4
( 11- 7 = 4, 15 - 11 = 4, 19-15 = 4 ) . Because the left hand
column with the x values is going up by 1 each time we know
the slope is m = 4. The slope-intercept form of a linear equation
can be used to find the y-intercept or b. We use m =4 and (1,7)
y = mx + b
7 = 4(1) + b
7=4+b
-4 -4
3=b
and the equation is y = 4x + 3
When the left hand column is not the natural numbers in order, we can still find the 'nth'
term of our arithmetic sequence using the same method we used to find the equation
of any table that represents a linear function, as shown below.
Example
Given the table below find the equation or 'nth' term of the arithmetic
sequence.
n
bn
4
4
7
1
8
0
10
-2
11
-3
We start by finding the slope of the linear equation. We will
use the first two points in the table, (4, 4) and (7, 1)
m
y 2  y1
1 4  3
=

 1
74
3
x2  x1
Next we substitute the slope m = -1 and the (4, 4) into the
slope-intercept form of a linear equation.
y = mx + b
4 = -1(4) + b
4 = -4 + b
+4 +4
8= b
And the equation is y = -x + 8; however, the variables listed
in the top of the table are n and bn so we should write our
final answer as:
bn = -n + 8
Often in Algebra or real world situations we will start with a description or contextual
problem and develop tables, equations and graphs to help us interpret problems and
find solutions. Below is an example of such a contextual problem and our solutions
using these tools.
Example
Jack is a guitar instructor. He is paid $35 for each hour lesson he gives
to his students.
Part a. Construct a table of values to show how his fees accumulate
with each lesson.
x
y
1
35
2
70
3
105
4
140
5
175
The x-values represent the number of lessons starting
at 1 and going up. Because he is paid #35 dollars per
lesson the y values go up by 35 each time.
Example Part a. (continued)
x
y
0
0
1
35
2
70
3
105
4
140
The table could also have been filled out with
a starting point of 0. The 0 as a starting point
would mean that if he gives zero lessons then he
earns zero dollars, and it goes up from there.
Part b. Find the equation or 'nth' term for the arithmetic sequence.
If we use the points (1, 35) and (2, 70) we can find the slope
for the linear function.
m
y 2  y1
70  35 35

 35
=
2 1
1
x2  x1
Next we use the slope-intercept form of the linear equation and
substitute the m = 35 and the point (1, 35).
y = mx + b
35 = 35(1) + b
35 = 35 + b
-35 -35
0=0+b
0= b
So our equation or 'nth' term is y = 35x
Part c.
Construct a graph of the equation.
150
100
50
-4
-3
-2
-1
-50
-100
-150
1
2
3
4
We should pay attention
to the scale on the y-axis.
It is necessary to have a
special scale because our
numbers are too big to
count by 1's on the y-axis.
counting by 50 makes
sense because we need to
go up quickly.
Example (continued)
Part d.
You can be asked questions following the three representations
with regards to numbers of lessons and amount earned. For instance:
How many lessons must Jack give in order to earn over $500?
Using our equation y = 35x we substitute the 500 for 'y' because
'y' represents the amount of money he earns.
y = 35x
500 = 35x
35
35
14.29  x
So Jack must give 15 lessons to earn more than $500.
Exercises Unit 4 Section 7
Extend the arithmetic sequences below:
1. 6, 9, 12, 15, ________, ________
2. 21, 17, 13, 9, ________, _________
3. Use the equation below with the given domain to generate an arithmetic sequence.
y = 2x + 3 with D = { 1, 2, 3, 4, 5, 6 }
4. Fill in the blanks: An arithmetic sequence is an _____________ equation with a
domain of ______________.
Convert the sequences below into tables. Use tables similar to those shown under
each problem.
5. bn = 7, 10, 13, 16, 19
n
1
2
3
4
5
bn
6. an = 30, 25, 20, 15, 10
n
an
Convert the sequences below into tables and the find the equation or 'nth' term.
Use tables similar to those under each problem.
7. bn = 2, 8, 14, 20, 24
n
8. an = 5, 7, 9, 11, 13
bn
n
an
1
2
3
4
5
The numbers in the tables below follow a linear pattern. Find the equation that
relates the variables. List your answer in slope intercept form.
9.
u
v
4
10.
t
s
12
3
6
18
8
11.
x
y
-5
1
4
5
-9
3
5
24
7
-13
5
6
12
36
9
-17
7
7
14
42
11
-21
9
8
12. Mark builds bird houses for an online company. He is paid $14 dollars for each
bird house he finishes.
a. Create a table similar to the one on the right that shows how
his income increases with each bird house he finishes.
b. Find the equation that models this situation.
Show your work.
c. What quantities do 'x' and 'y' represent in your equation?
x
y
Problem 12 (continued)
d. Graph the equation using a set of axes similar to that given below.
$60
-4
-2
2
4
-$60
e. Explain why the graph should pass through (0, 0).
f. If Mark finishes 33 bird houses how much money is he paid? Show your work.
Explain how and why you did the substitution.
g. If Mark wants to earn $600 how many bird houses must he finish?
Show your work.
Explain how and why you did the substitution.
13. Kelly is a studio musician. She is paid $100 a day whenever she
goes into the studio. If she plays for a recording session she is
paid $30 an hour.
a. Create a table similar to the one given on the right that
shows how her income increases with each hour she plays.
b. Find the equation that models this situation.
c. What quantities do 'x' and 'y' represent in your equation?
d. Graph the equation using a set
of axes similar to the grid to
the right.
-4
$200
-2
2
-$200
4
x
y
e. Explain why the graph goes through the point (0,100).
f. How many hours must she play for to earn $280? Show your work.
Explain how and why you did the substitution.
g. If Kelly plays for 4.5 hours how much does she earn? Show your work.
Explain how and why you did the substitution.
Write an equation to model each situation.
14. A hot air balloon is rising at a steady rate of 4 ft per second. It started
from here in Wickenburg with an altitude of 1500 feet. Write an
equation that will track/model the rising altitude of the balloon.
15. James owes a friend $360. He is paying his friend back $45 a week and his
debt is going down. Write an equation that will model how the amount he
owes is going down.
16. The Fox cable company charges $79 as an installation and set-up fee. The
basic package has a charge of $39.95 per month. Write an equation to
model the growth of the cost of using this service.
Read the problems carefully, find the two data points described in the problem, and
then write an equation to model the situation in slope-intercept form.
17. After two weeks of practice Al could throw the discus 60 ft. As he practiced
he got better in a linear fashion. After five weeks he could throw the discus
67.5 ft. Write an equation to model how his length is increasing.
(
,
) (
,
)
m = __________
equation __________________
18. A water storage tank is being drained at a steady rate. After 8 minutes the
tank has 1240 gallons. After 12 minutes the tank has 1120 gallons.
a. Write an equation to model how the tank is being drained.
(
,
) (
,
)
m = __________
equation __________________
b. How much water was in the tank before the water drained?
Explain how you found your answer!
c. How many minutes will it take to empty the tank?
Explain how you found your answer!
Unit 4 Section 8
Objectives

The student will find and apply equations for lines of best fit.
In Unit 1 we first looked at the problem solving process that is used in science,
business, sociology, and many other areas. This process consists of collecting data
as a table or roster, turning the data into a graph, and then using the graph to create
an equation and make predictions. In Unit 1 we were not able to find the equation
that would model a scatter plot. However, now in Unit 4 we have developed the
methods we need to find equations. We are going to revisit scatter plots and extend
our ability to use a scatter plots by finding equations. Below is the first example of a
scatter plot we created.
Example
The Algebra I teachers recently conducted a study. They wanted to see exactly
how missing assignments affected student grades on a unit test. The data set
below was collected. The number on the left is how many assignments were
missed during the unit by a student. The number on the right is the test score
the student earned on the unit test.
(4, 68) (1, 88) (0, 93) (3, 72) (1, 91) (0, 89) (6, 41) (7, 34)
(2, 80) (3, 76) (4, 61) (1, 86) (0, 98) (5, 52) (4, 58) (1, 42)
The data is plotted below and the line of best fit found in the diagram.
A Survey of Missing Assignments and Grades
Test Percent
100
90
80
70
60
50
40
30
0
1
2
3
4
5
6
7
Number of Missing Assignments
8
We can identify the endpoints of the line of best fit and use these to find the equation
of a line in slope-intercept form. We could actually pick any two points on the line
of best fit but the endpoints here are quite clear and easy to use. These points
are (0, 98) and (7,34).
To find our equation we must use these two points to find out slope.
m
y 2  y1
34  98  64

  9.14 (rounded off to the nearest hundredth)
=
70
7
x2  x1
We now substitute the slope m = -9.14 and the first point (0, 98) into the slope
intercept form of a linear equation.
y = mx + b
98 = -9.14(0) + b
98 = 0 + b
98 = b
So our equation is y = -9.14x + 98
In using this equation we will need to remember the quantities that 'x' and 'y'
represent. The 'x' is the number of assignments missed and the 'y' is the percent
on the unit test.
We can now use this equation to predict scores for other number of days more
accurately than before. If we want to predict the score when 8 assignments are
missed we substitute 8 for 'x' as shown below.
y
y
y
y
=
=
=
=
-9.14x + 98
-9.14(8) + 98
73.12 + 98
24.88
This means we would expect this student to earn 25% on the test.
Another question we might be asked to answer accurately would be similar to:
How many assignments could a student miss and still expect to get a 70%
on the test?
y = -9.14x + 98
70 = -9.14x + 98
-98
-98
-28 = -9.14x
-9.14
-9.14
3.06 = x (rounded to the nearest hundredth)
This means that a student who wants a C should not miss more than 3 assignments.
Exercises Unit 4 Section 8
1. Create a scatter plot for a survey's data set listed below. In the data points, the
left hand coordinate is the age of the person and the right hand coordinate is the
number of hours the student spent studying per week.
(8, 4) (10, 5.5) (11, 6) (8, 5) (9, 4.5) (10, 6) (14, 7) (13, 7) (14, 8) (8, 3)
(14, 9) (12, 7) (13, 8.5) (11, 6.5) (15, 9) (11, 7) (12, 8) (14, 8.5) (11, 5.5)
Construct a grid similar to one given below.
A Survey of Age and Hours of Studying per week
Hours
18
16
14
12
10
8
6
4
2
7
8
9
10
11
Age
12
13
14
15
a. Draw a "best fit line" for this set of data.
b. Using the endpoints from your line of best fit find its equation in
slope-intercept form.
c. Using the equation predict the number of hours a 16 year old would study.
2. Using the data points below create a scatter plot, set up your own grid.
Phictitious High School (enrollment 2000 students) has implemented a tardy
prevention plan. Students are rewarded for being to class on time with free
deserts once a week. They are tracking the progress of the plan from week to
week. The value on the left of the ordered pair is the week and the value on the
right is the number of tardies recorded during the week.
(1, 502) (2, 483) (3, 465) (4, 445) (5, 424)
a. Graph this data as a scatter plot (you can use technology if it's available).
b. Find the equation of the line of best fit. (You can use technology if it's available)
c. If a calculator is available find the coefficient of correlation. What does this
constant (the coefficient of correlation) tell us?
e. Using your equation predict the number of tardies that will be recorded
in week 10. Show your work.
f. Using the equation, when will the number of tardies reach the program's
target goal of 200. Show your work.
Given m and b write the equation of the line they determine in slope-intercept form.
3. m = 9, b = -3
5. m =
4
5
b=
4. m = -7.2 b = 4.5
3
2
6. m = 0 b = -4
Given the graph find the slope, y-intercept and equation of the line.
7.
8.
9.
Given the point and the slope find the equation of the line they determine. List
your answer in slope-intercept form. Show your work.
10. m = 3 (2, 1)
11.
12. m = 2.2 (-5, 4.5)
13.
1
(6, -5)
2
m = 0 (3, -7)
m=-
Given the slope and point find the equation of the line they determine. List your
answer in point-slope form.
14. m = 7 (-1, -4)
15. m = -.5 (  , -1)
Convert the equations into standard form. Show your work.
16. y = -3x - 2
17. y =
1
x-1
4
18. y - 3 = -(x + 2)
19. Mike owns the company "PC Tuning Inc.". The company will remove all malware
and perform any needed upgrades or optimizations required on a computer. The
charge is $40 per computer. Write an equation to model this situation.
20. A submarine is slowing as it is rising to the surface. Two minutes after it started it
was 720 ft below sea level. After 6 minutes the ship was 480 ft below sea level.
a. Use the two given points to find an equation that models the ship's depth.
b. What was the ship's depth before it started to surface?
c. What will the ship's depth be after 10 minutes?
d. At what number of minutes will the ship reach the surface?
Unit 4 Section 9
Objectives

The student will identify, formulate and apply direct variation.
There are many types of functions or equations. Usually when we change the value of an
‘x’ variable in an equation we will see a change in the ‘y’ value of an equation.
For instance in the equation below if we start with x = 4 we will produce a y-value of
8. When we change x to 5 we get a y-value of 10.
given x
y
y
we get y
=
=
=
=
4
2x
2(4)
8
If we change x to 5
with x
y
y
we get y
=
=
=
=
5
2x
2(5)
10
The x-values we were using changed by one (1) and this caused a change in the y-value
of two (2). If we continue to change the x-value by one (1) let’s see what happens.
with
x
y
y
we get y
=
=
=
=
6
2x
2(6)
12
with x
y
y
we get y
=
=
=
=
7
2x
2(7)
14
Again when we change the x-values by one (1) we see the y-values going up by two(2).
whenever ‘x’ goes up by ‘1’ then ‘y’ goes up by ‘2’. If the same change in the x-values
produces a consistent change in y-values we call this direct variation.
Definition:
Direct Variation occurs when a change in ‘x’ produces a proportional
change in ‘y’.
The key word in the definition above is the word “proportional”.
the concept of proportionality is given in the table below.
We
We
We
We
One way to illustrate
started with a value of x = 1 and it produces an answer y = 6.
doubled x to 2 and as a result the y-value doubled to 12.
tripled x to 3 and as a result the y-value tripled to 18.
quadrupled x to 4 and the y-value quadrupled to 24.
x
y
1
6
2
12
3
18
A change in ‘x’ produces a proportional change in ‘y’.
4
24
This proportionality is expressed as a linear equation in Algebra. The general equation
for dire3ct variation is:
y = kx where ‘k’ is a constant.
The ‘k’ in this equation is called the “constant of proportionality”. It is a real number.
Some examples of direct variation equations are given below.
y = 3x
f(x) = 5.5 x
y=
1
x
4
g(x) = -4x
This type of equation is actually the same as y = m x + b. The two equations are
compared below.
y = mx + b
y= kx+0
Direct variation is a linear equation where the y-intercept is zero (0) and the slope is
represented as ‘k’. This means that all the techniques we have learned for finding and
applying linear equations will work with direct variation. Because the y-intercept is
always zero there is a simpler way to find the slope or ‘k’. If we solve the general
equation of direct variation for ‘k’ we get the equation below.
y = kx
x
x
k=
y
x
This tells us that to find the constant of proportionality all we need to do is divide
the y coordinate or value by the x coordinate or value. We can then write the equation
using the general form. Below are some sample problems involving direct variation.
Example A
Given ‘y’ varies directly as ‘x’. If when x = 4 then y = 14 find the
constant proportionality k.
Since k =
Example B
y
x
we can find ‘k’ by substitution k =
14
= 3.5
4
Given ‘y’ varies directly as ‘x’. If when x = 5 then y = 4 find the
constant proportionality ‘k’ and the equation for the function.
Since k =
y
x
we can find ‘k’ by substitution k =
so our function is y =
4
x
5
4
5
Example C
The number of feet an object measures varies directly as number
of inches. Find an equation that will represent this relationship.
In general terms we would say:
‘y’ varies directly as ‘x’
In the problem we see:
# of feet varies directly as # of inches
This means the ‘y’ variable represents feet and the ‘x’ variable represents
the inches. Using the idea that 1 foot = 12 inches we find the constant
‘k’ as shown below.
Since k =
y
x
we can find ‘k’ by substitution k =
So our equation is y =
Example D
1
12
1
x
12
Mike is traveling on I-10 in West Texas. He has his cruise control on
in his car. If Mike goes 230 miles in 3 hours write an equation that
shows how his distance varies with the hours he drives.
In general terms we would say:
‘y’ varies directly as ‘x’
In the problem we see:
distance varies directly as hours
This means the ‘y’ variable represents distance and the ‘x’ variable
represents the hours. We find the constant ‘k’ as shown below.
Since k =
y
x
we can find ‘k’ by substitution k =
So our equation is y =
Example E
230
3
230
x or y = 76.6x
3
Juan works as a clerk at the Wickenburg BMC 24 Movie Theater.
He serves 34 people in two hours.
Part a. Write an equation that shows how the number of people
serves varies over the hours he works.
k=
34
= 17
2
so our equation is y = 17x
Part b. How many people will he serve in 4 hours?
Since ‘y’ represents people and ‘x’ represents hours we
substitute the 4 for the ‘x’.
y = 17x
y = 17(4)
y = 68 people
Example E (continued)
Part c. How long would it take for Juan to serve 25 people?
Since ‘y’ represents people and ‘x’ represents hours we
substitute the 25 for the ‘y’.
y = 17x
25 = 17x
17
17
1.47 = x
(rounded to the nearest hundredth)
So it will take almost 1.5 hours to serve 25 people.
Exercises Unit 4 Section 9
Fill in the blanks or complete the sentence.
1. If 'x' and 'y' vary directly then as 'x' goes up ______________________________.
2. The general equation for direct variation is ______________.
3. In the general equation the 'k' is called the _______________________________.
4. In the general equation 'k' also functions as the ______________ of a line.
Write equations that model the given situations.
5. The distance Paul can hike varies directly with the number of hours he walks.
If Paul can cover 3 miles in an hour, write the equation to model this
situation.
6. Jim is a vendor for the Diamondbacks. He sells soda during the games.
He makes $1.20 for every drink he sells. Write an equation to model how
his earnings grow with each drink he sells.
7. Alejandro runs a bicycle rental business on Pacific Beach. He charges
6.25 per hour to rent a bicycle. Write an equation that models how
the charges grow with each hour.
8. A water tank is being filled at a constant rate 'z'. If we allow the water
to flow into the tank for a length of time 't'. Write an equation that will
model the amount of water ‘w’ in the tank at any time.
9. The volleyball team is holding a car wash. Terry can wash 8 cars in one hour.
a. Write an equation that models how the number of cars Terry will wash varies
as the number of hours increases.
b. Use the equation to find the number of cars Terry washes in four and half hours.
c. Use the equation to determine how long it will take Terry to wash 27 cars.
d. Sarah can do 7 cars an hour and Sally can do 7.5. Write an equation that
models the number of cars Terry, Sarah and Sally can wash as the number
of hours increases. _______________________
Find the constant of proportionality and equation (if asked for) in each problem.
10. Given 'y' varies directly as 'x'. When x = 12 then y = 30 what is the
constant of proportionality?
11. The number of centimeters in an inch varies directly. In 10 inches there
are 25 centimeters. What is the constant of proportionality when 'x' is
the number of inches and 'y' is the number of centimeters?
12. The circumference of a circle varies directly as the diameter. If a circle has
a circumference of 25.1327 ft and a diameter of 8 ft, find the equation that
models this situation when 'y' is the circumference and 'x' is the diameter.
13. A train is traveling at a constant rate of speed. If it goes 205 miles in 3 hours
write an equation to model how the distance varies with the hours it
travels.
14. Jose just started his first job. During his first week he worked 28 hours. When he
got his paycheck the gross pay was for $231.00.
a. Write an equation to model how his pay grows with each hour.
b. How much would he make in 40 hours?
c. He needs to have a gross pay of $275 a week to make his car payment.
How many hours does he need to work?
Find the slope of the line passing through the listed points. (Show your work)
15. (5, -1) (7, -5)
16. (-1.5, 3) (-1, 3.25)
_______
17. Given the roster { (-2, 4), (2, 4), (6, 4), (10, 4) } is linear, the slope of the line
that passes through the points is:
a) -1
b) 1
c) 0
d) no slope
18. Given the table to the right represents a linear function,
the slope of the line that passes through the points is:
x
y
4
3
3
-2
-9
c) .3
1
0
d) -.3
0
-3
a) -3
b)
19. Mike and Tom are putting a fence around a property. In the first two hours
they put up 64 ft of fence. After 5 hours they had finished 170 ft of fence.
a. List the two data points in this problem.
b. Find the rate of change or slope and list the units. (round your answer
to the nearest tenth.)
20. In the general equation y = m x + b
a. What is “m” ? _______________________
b. What is “b” ? _______________________
Given m and b write the equation of the line they determine.
21. m =
5
, b = -4
3
22. m = 1.4 b = 0
Given the graph find the equation of the line in slope-intercept form.
23.
24.
Given the point and the slope find the equation of the line they determine. List
your answer in slope-intercept form.
25. m = -3 (-1, 6)
26. m = -
1
(-4, 6)
2
Given the slope and point find the equation of the line in point-slope form.
27. m = -5 (-1, 1.8)
28. m =
3
3
(1, )
2
4
Find the equation of the line that would pass through the two given points.
List your answer in slope-intercept form.
29. (-6, 1) (3, 4)
30. ( 2, .5) ( -3, 3.5) __________________
31. Find the equation of the line that would pass through the two given points.
List your answer in point-slope form.
(3, -1) (5, 6)
32. Convert the equation y+3 =5x to standard form.
33. Convert the equation y + 2 = 3(x - 1) to slope-intercept form.
34. Find the x and y intercepts for the equation 4x - y = 9
35. Sandy is saving to buy a guitar. She put $80 into a savings account. Her plan is
to add $20 a week to the account.
a. Write an equation to model this situation.
b. Use the equation to find how much will be in the account in 9 weeks.
c. If the guitar she wants costs $450 how many weeks will it take her to
save at least this amount.
Unit 5
Systems of Equations or Inequalities
In analyzing data and problem solving in general there will be times when we will
put two lines on one set of coordinate axes. Below we see a graph of the monthly
expenses and income for Great Western Jeep Tours. At the beginning of each month
they must pay a number of bills or fixed expenses. These bills include vehicle payments,
property rental, insurance, utilities and other expenses. During the month they will have
income for each tour they give to customers. During the month they must pay their
employees for each trip and fuel for each tour. At the start of each month the business
will have no income, but they will have to pay the fixed expenses.
Great Western Jeep Tours Income and Expenses
$21000
$18000
$15000
Expenses
$12000
Income
$9000
$6000
$3000
$0
0
10
20
30
40
Tours
50
60
70
The solid line in the graph represents the income for the month. The income goes up
with each tour the company gives to a customer. The dashed line represents the
expenses the company has during the month. The expenses start at over $6000 after
paying all the fixed expenses. The expenses then go up when the company pays for
the fuel and the wages for the guides.
One of the most important points on the graph is where the two lines intersect or cross.
The place where the two lines cross is a point that makes both equations true. It is
also the point where the two equations have the same y-value. The y-values of the
solid lines are the income. The y-values of the dashed line are the expenses. When
the y-values are equal the income and the expenses are equal. To the left of this
point the graphs show the business losing money to the right they are making money.
This is the point at which the business breaks even. The business must have give
about 23 tours to break even. Finding this point is important to any business. In this
unit we will be finding a place where functions share a point.
Unit 5
Vocabulary and Concepts
Parallel Lines
Lines that are in the same plane and do not intersect are
called parallel lines.
Perpendicular Lines
Lines that intersect and create right angles are called
perpendicular lines.
System
A system is two equations or inequalities.
Key Concepts for Systems of Equations or Inequalities
Points on a Line
A point is on a line if its coordinates, when substituted into the
equation, makes it true.
Point of Intersection
A point of intersection is the place where two lines cross.
This point is the solution to the system and its coordinates
will make both equations true.
Method of Substitution
A way to solve a system of equations by replacing one
of the variables in an equation by the expression it is
equal to from a second equation.
Method of Elimination
A way to solve a system of equations by adding or
subtracting two equations.
Unit 5 Section 1
Objective

The student will solve systems of equations graphically.
A system of equations consists of two or more functions. In Algebra I we will study
systems of two linear equations. Before we examine two equations we must understand
how we can tell when a point is one a line and when it is not. If we were to graph the
equation y = 2x - 4 as illustrated below we would find the slope and intercept, then graph
the line. In the diagram below point A(3, 2) is on the line and point B(1, 6) is not on the
line.
y = 2x – 4
If we substitute point A into the
equation we get
y = 2x – 4
2 = 2(3) - 4
2=6–4
B
2=2
which is true!
A
If we substitute point B into the
equation we get
y = 2x – 4
6 = 2(1) - 4
6=2–4
6 = -2
which is false!
The conclusion or conjecture we can draw from this is that “If a point is on the line
then the point will make the equation true.” We can use this conjecture to determine
when a point is on a line. The examples below illustrate this idea.
Example A
Given the equation of a line y = .5x + 3.5 is (4, 2) on the line?
We substitute the point into the equation y = .5x + 3.5
2 = .5(4) + 3.5
2 = 2 + 3.2
2 = 5.5
This is false and the point is NOT on the line.
Example B
Given the equation of a line y = -4x + 1 is (-1, 5) on the line?
We substitute the point into the equation y = -4x + 1
5 = -4(-1) + 1
5= 4+1
5=5
This is true and the point is on the line.
When we examine a system of equations we will be looking for the point at which
the two lines intersect or cross. As we saw in the “Unit 5 Introduction” this is an
important point in applications. The point at which two lines intersect is the only
point that will make both equations true. We can illustrate this with the example
below.
The graph below has the lines y = x + 1 and y = -2x + 4. We will test the points
A(-1, 0), B(1, 2), and C(4, 1).
B
When we substitute point A into the
equations we get:
A
C
y=x+1
0 = -1 + 1
0=0
y = -2x + 4
0 = -2(-1) + 4
0=2+4
0=6
Point A only makes one of the equations
true and it is NOT on both lines.
When we substitute point B into the
equations we get:
y=x+1
2=1+1
2=2
y = -2x + 4
2 = -2(1) + 4
2 = -2 + 4
2=2
Point B makes both of the equations
true and it is on both lines. It is the
point where the lines intersect.
When we substitute point C into the
equations we get:
y=x+1
1=4+1
1=5
y = -2x + 4
1 = -2(4) + 4
1 = -8 + 4
1 = -4
Point C makes both of the equations
false and it is Not on either line.
The conclusion we can draw from the example above is:
If a point makes both equations true then the point is where the two lines intersect.
There are several methods for finding the point of intersection. We will begin by
examining the graphical method. This method is similar to what we have seen
in the example above. This consists of graphing the two lines and finding the
point of intersection by inspection. The point can be checked by substituting
the coordinates into the original equations. Examples of this process are on the
following page.
Example A
Given the equations y = 3x – 6 and y = -x + 2 find the point of
intersection.
In order to graph the two lines we need to find their slopes and y-intercepts.
The first equation y = 3x – 6 has m = 3 and b = -6. The second equation
y = -x + 2 has m = -1 and b = 2. They two lines are graphed below.
The point of intersection is where
the two points cross.
The coordinates of the point are
(2, 0)
We can check our answer by substituting the point into both equations.
y = 3x – 6
0 = 3(2) – 6
0=6–6
0=0
y = -x + 2
0 = -(2) + 2
0 = -2 + 2
0=0
The point makes both equations true and so is the solution to the system.
Example B
Given the equations y = 2x + 1 and y = -.5x - 3 find the point of
intersection. In order to graph the two lines we need to find their slopes
and y-intercepts.
The first equation y = 2x + 1 has m = 2 and b = 1. The second equation
y = -.5x - 3 has m = -.5 and b = -3. They two lines are graphed below.
The point of intersection is where
the two points cross.
In this example we must estimate
coordinates of the point
(-1.3, -2.1)
Since this is an estimate checking it will not result in equations that
are necessarily true.
Exercises Unit 5 Section 1
Set A
1. The point where two lines intersect will make the two equations ______________.
For each given system of equations check to see if the listed point is the solution.
Answer yes or no and show your work.
2.
y = 3x - 4
y = 2x - 3
(1, -1)
5.
2y + 4x = 0
y - x = 3
(-1, 2)
3.
y = -x + 4
y = 3x + 8
4.
y = .5x + 3
y = 3x - 2
(2, 2)
2
x+1
3
x + y = 11
6. y =
(2, 4)
7.
x=y+2
3y = x - 5
(6, 5)
(3, 1)
Graph each pair of lines and find the point of intersection.
8.
y=x-1
y = -2x + 8
11.
y = -5x + 8
y = 3x - 8
14.
y = -x - 3
y= x+1
1
x-4
2
y= x+5
9. y = -
2
x-1
3
y = -2x - 1
10.
y=
1
x+1
2
y = 2x + 4
13.
y= 2
1
x+4
2
1
y= - x-2
2
16.
12. y =
15. y =
x = -3
y = 2.5
y =
1
x
2
17. Given the data set below find the mean, median, mode, and range.
12, 14, 11, 17, 13, 14, 15, 9, 10, 14, 11
18. The data set below represents the ages of people in a bowling league. Create
a stem and leaf plot to represent the data.
28, 33, 42, 25, 36, 31, 33, 45, 40, 46, 27, 38, 51, 33, 49
Exercises Unit 5 Section 1
Set B
1. When a single point can make two different equations true we know
this is a point at which the two lines __________________.
For each given system of equations check to see if the listed point is the solution.
Answer yes or no and show your work.
2.
y = 5x - 7
y = 2x + 2
3.
y = -2x + 4
y = x - 10
(3, 8)
5.
4.
y = -x + 4
y = 4x - 2
(5, -6)
(-2, 6)
1
x+3
2
x - y = -4
2y + .5x = 0
6. y =
y - 2x = 4
(8, -2)
7.
x = 2y
-y = x - 10
(-2, 2)
(18, -9)
Graph each pair of lines and find the point of intersection.
8.
3
x+1
2
y= x+3
y = 2x - 4
9. y =
y = -3x + 6
11.
1
x
3
y = 2x
y = -3x + 4
12. y =
y= x+4
14.
y+x=0
2y = 4x + 6
1
x-1
2
y= x - 4
10.
y=-
13.
y = -4
x=1
15. y = x + 4
y= x-2
16.
y = 2x + 1
2y = 4x + 2
16. The solution to the equation -2(x – 4) + x = x – 6 is
a. 1
b. 7
c. -1
17. The solution to the inequality
a. x > -2
b. x < 
8
3
d. 14
-3x + 1 > 7 is
c. x< -2
d. x > 
8
3
Unit 5 Section 2
Objective

The student will define, recognize and find parallel and perpendicular
lines.
There are two important ways lines can relate in a plane. These are called parallel and
perpendicular.
Definition
Parallel lines are in the same plane and do not intersect.
Below is an example of what these lines would look like with and without a graphing
plane.
Both of these lines
2
have slopes of
3
Because the lines go in the same direction they will never intersect. These kinds of
lines appear everywhere in our world. The sides of a window, the legs of a table,
the edges of a book, a railroad track, and many other figures or objects have parallel
lines in them. Also because the lines go in the same direction they have equal slopes.
This idea will make it possible for use to identify and draw parallel lines.
Definition
Perpendicular lines form right angles when they intersect.
Below is an example of how perpendicular lines would look.
The angles formed here are 90o and are called right angles. These angles are the
“right” angles because when we need to build walls or use rectangles or squares these
are the angles we must use. So they are the “right” angles for many purposes. When
we see angles like these graphed on the x-y plane we can see that they go in opposite
directions. There are two perpendicular lines graphed in the next diagram.
An Example of Perpendicular Lines
These two lines are perpendicular. They
go in opposite directions. This means that
one line has a positive slope and the other
has a negative slope. One of the lines is
very steep while the only slants a little.
that is why we can say they go in opposite
directions.
Another way to see how perpendicular lines are created is to start with a vertical line
and a horizontal line and rotate them as in the diagrams below.
Perpendicular Lines
Rotated 15o
Rotated 45o
If we examine the slopes in these diagrams we can draw some conclusions.
Top Diagram
m = 2 and m = -
Rotated 15o Diagram
1
2
m = 3 and m = -
1
3
Rotated 45o Diagram
m = 1 and m = -1 or -
1
1
When we look at the pairs of slopes that produce perpendicular lines we find they
are opposite in sign and they are reciprocals. This conclusion makes it possible for
us to identify pairs of perpendicular lines from their equations and to draw perpendicular
lines by counting the slopes.
There are many symbols we use in Algebra. We have symbols for the two new
relationships we now have. For parallel lines we use two vertical bars that look like parallel
lines. The symbol is ||. For perpendicular lines we use a symbol that looks like the
intersection of two perpendicular lines. The symbol is
.
The examples that follow show us how to draw || and
lines.
Example A Given the graphed line draw a || (parallel) line through the point (3,2).
Original Graph
Parallel Line
By counting we know the
original line has a slope of
m = -2
So we need to plot the point
(3,2) and then count the
slope of -2 to find points
for the new line.
Example A Given the graphed line, draw a
(-2,3).
Original Graph
(perpendicular) line through the point
By counting we know the original
line has a slope of
m=
Perpendicular Line
3
4
The opposite reciprocal is
m=-
4
3
So we need to plot the point
(-2,3) and then count the slope
to find points for the new line.
We also need to be able to recognize when lines are parallel or perpendicular when
we are given their equations. The ideas we can use to identify these relationships
are given below.
Lines are parallel when their slopes are equal.
Below are examples of how to use this fact to recognize parallel lines.
Example A
Given the equations
Example B
Given the equations
y = 3x + 1
y = 3x – 5
2x + y = 7
-4x – 2y = 10
Both equations have
a slope of m = 3.
So the two lines are
parallel.
We must first isolate y
in both equations. We
get the equations below.
y = -2x + 7
y = -2x – 5
The slopes are equal,
the lines are ||.
Example C
Given the equations
y = -5x + 1
y = 5x - 4
The slopes are not equal.
The lines are not parallel.
Lines are perpendicular when their slopes are opposite reciprocals.
Below are examples of how to use this fact to recognize parallel lines.
Example A
Given the equations
1
y= x–1
2
y = -2x + 4
The slopes are opposite
in sign. The slope of
m=
1
2
has a reciprocal of
2
m=
=2
1
Since the slopes are
opposite reciprocals
the lines are
.
Examples B
Given the equations
3
y= x+4
5
5
y=- x-2
3
The slopes are opposite
in sign. The slope of
m=
3
5
has a reciprocal of
m=
5
3
Since the slopes are
opposite reciprocals
the lines are
.
We may also be asked to use the relationships of parallel and perpendicular to solve
other kinds of problems. Some examples follow.
Example A
Find the equation of a line parallel to y = 6x – 4 passing through the
point (-1, 3).
Since the slope of the given line is m = 6 the problem becomes
something like what we did in Unit 4. We need to find the equation of
a line with a slope of m = 6 and passing through (-1, 3). So we
follow our process:
y = mx + b
3 = 6(-1) + b
3 = -6 + b
+6 +6
Write the slope-intercept form of an equation.
Substitute the coordinates and slope.
9=b
So our equation is y = 6x + 9, which is the equation of the line.
Example B
Find the equation of a line
to y =
1
x + 1 passing through (2, -4).
3
1
and the opposite reciprocal is m = -3.
3
Again we will follow the process for finding the equation of a line
with a slope of m = -3 and passing through (2, -4).
The slope of the given line is m =
y = mx + b
-4 = -3(2) + b
-4 = -6 + b
+6 +6
2=b
Write the slope-intercept form of an equation.
Substitute the coordinates and slope.
So our equation is y = -3x + 2, which is the equation of the line.
Exercises Unit 5 Section 2
Set A
1. What does it mean for two lines to be parallel?
2. What is the symbol we use for parallel lines?
3. What does it mean for two lines to be perpendicular?
4. What is the symbol we use for perpendicular lines?
Find the slope of each of the given lines then draw the asked for line through the
given point on a set of coordinate axes.
5. a parallel line through (4, 2)
6. a parallel line through (-2, 3)
7. a parallel line through (-1, -2)
8. a || line through (1, 5)
9. a perpendicular line through (0, 1)
11. a  line through (1, 2)
10. a perpendicular line through (3, 0)
12. a  line through (0, -1)
Decide from the two listed equations if the lines are parallel, perpendicular or
neither.
1
13. y = 4x + 2
14. y = x + 3
15. y = 6x + 1
2
y = 4x - 1
y = 2x + 3
y = -6x + 1
16. y = -4x + 2
y=
1
x-1
4
17. y = -
1
x+3
2
y = 2x + 3
5
19. y = - x + 2
3
3
y= x-1
5
20. y = x + 3
22.
23. y = 4
y= 2
1
y=2
y= x-7
x= 3
2
x+1
3
2
y= x-1
3
18. y =
21. y = -x + 1
y=x-8
24. y = 5x + 1
y = -5x - 1
25. 2y - 4x = 2
1
y=- x-1
2
26. y + 2x = 5
27. 2y - 5x = 2
y = 2x + 3
5y = 2x - 15
28. Find the equation of the line parallel to y =
1
x + 3 and passing through (4, -5)
2
29. Find the equation of the line parallel to y = 3x - 2 and passing through (-1, 7)
30. Find the equation of the line
31. Find the equation of the line
to y =
1
x - 1 and passing through (2, -1)
2
to y = -
2
x + 3 and passing through (-2, 3)
3
32. Find the equation of the line in slope-intercept form going through (4, -1) & (-2, -4)
Graph each pair of lines and find the point of intersection.
33.
y = 2x - 5
y= x-6
1
x+4
2
y = -x - 2
34. y =
35.
3
x-1
2
y= x+1
y=
Isolate 'y' as needed in the equations below before graphing the systems
36.
2y = -4x + 8
x+y= 1
1
x= 3
2
y= x+1
37. y -
Exercises Unit 5 Section 2
38.
y = -2
x=0
Set B
1. The point where two lines intersect will make the two equations ______________.
Check to see which if the listed point will make both equations true. Answer
yes or no and show your work.
2.
y = -4x + 5
y = 2x - 7
3. y = -.5x + 4
y = 3x - 7
(2, -3)
(2, 0)
Graph each pair of lines and find the point of intersection.
1
4
4. y = x - 4
5. y = - x - 4
2
3
y = -x + 5
y = 2x + 6
5. What does it mean for two lines to be parallel? ( List our definition. )
6. Two lines will be parallel when their slopes are ___________________.
7. What is the symbol we use for parallel lines?
8. What does it mean for two lines to be perpendicular?
9. Two lines will be perpendicular when their slopes are ________________________.
10. What is the symbol we use for perpendicular lines?
Find the slope of the given line then draw the asked for line on a set of coordinate axes.
11. a parallel line through (-1, -3)
13. a parallel line through (1, 4)
15. a perpendicular line through (2, -1)
12. a parallel line through (-1, 4)
14. a
line through (-1, 2)
16. a  line through (2, 4)
Decide from the two listed equations if the lines are parallel, perpendicular or neither.
1
17. y = 4x + 2
18. y = x + 3
19. y = -.5x + 1
3
y = -4x - 1
y = -3x + 3
y = -.5x + 4
20. y = 4x + 2
y=
1
x-1
4
2
x+2
7
7
y=- x-1
2
23. y =
26. y = 2
1
y=2
21. y = -
1
x+3
2
1
y=
x+3
2
2
22. y = - x + 5
3
2
y=- x-1
3
24. y = 3
25. y = 5
y = -1
x=5
27. y = 2x + 1
( Think carefully about problem 27 )
y = 2x + 1
28. Find the equation of the line parallel to y = -
2
x + 3 and passing through (4, 0)
3
29. Find the equation of the line parallel to y = 4x - 2 and passing through (-2, 5)
30. Find the equation of the line
31. Find the equation of the line
to y =
1
x - 1 and passing through (2, 1)
4
to y = -x + 3 and passing through (-5, 8)
32. Find the equation of the line in point-slope form passing through (4, 7)
and (-3, 5)
Unit 5 Section 3
Objectives

The student will solve systems with an isolated variable by
substitution.
In Unit 2 we learned how to solve equations with one variable in them. The equations
we have been using in the last several units all have focused on equations with two
variables, ‘x’ and ‘y’. We can solve a system of equations by making equations with
two variables into an equation with one variable. This method for finding the solution
or point of intersection is called substitution. The steps we can use for this process
are listed below.
Step
Step
Step
Step
1.
2.
3.
4.
Find the equation with an isolated variable.
Do the substitution for the isolated variable in the other equation.
Solve the resulting equation.
Substitute the first coordinate into one of the equations
and find the second coordinate.
We will use these steps to solve the systems in the examples that follow.
Example A
Given the system of equations below find the point of intersection.
y = 5x – 12
2x + y = 16
Step 1.
y = 5x – 12
The y is isolated in this equation.
Step 2.
y = 5x – 12
We take what ‘y’ is equal to and
substitute it for ‘y’
2x + y = 16
2x + (5x – 12) = 16
Step 3.
2x + (5x – 12) = 16
2x + 5x – 12 = 16
7x – 12 = 16
+12 +12
This is the result of the substitution.
Solving the remaining equation.
7x = 28
7
7
x = 4
Step 4.
y = 5x – 12
y = 5(4) – 12
y=8
Substitute to find the second coordinate.
The solution is (4, 8) for this system of equations.
Example B
Given the system of equations below find the point of intersection.
-x + 2y = -7
x = 3y + 2
Step 1.
x = 3y + 2
The x is isolated in this equation.
Step 2.
x = 3y + 2
We take what ‘x’ is equal to and
substitute it for ‘x’
-x + 2y = -7
- (3y + 2) + 2y = -7
Step 3.
- (3y + 2) + 2y = -7
-3y - 2 + 2y = -7
-1y - 2 = -7
+ 2 +2
This is the result of the substitution.
Solving the remaining equation.
Use the Distributive Property.
-1y = -5
-1
-1
y = 5
Step 4.
x= 3y + 2
x = 3(5) + 2
x = 17
Substitute to find the second coordinate.
The solution is (17, 5) for this system of equations.
Example C
Given the system of equations below find the point of intersection.
x = 4y - 1
2x - 8y = -2
Step 1.
x = 4y – 1
The x is isolated in this equation.
Step 2.
x = 4y – 1
We take what ‘x’ is equal to and
substitute it for ‘x’
2x – 8y = -2
2(4y - 1) – 8y = -2
Step 3.
2(4y - 1) – 8y = -2
8y - 2 – 8y = -2
-2 = -2
This is the result of the substitution.
Solving the remaining equation.
Use the Distributive Property.
When we get a true equation that no longer has a variable this means the
two equations represent the same line! When this happens we say the
lines are coincident. The two lines actually intersect at all points and so they
coincide completely. ( “Coincidentally” in the English language means things
happen at the same time – in math it can mean they happen at the same place. )
Example E
Given the system of equations below find the point of intersection.
y = 2x - 4
4x - 2y = 9
Step 1.
y = 2x - 4
The ‘y’ is isolated in this equation.
Step 2.
y = 2x - 4
We take what ‘y’ is equal to and
substitute it for ‘y’
4x – 2y = 9
4x – 2(2x – 4) = 9
Step 3.
This is the result of the substitution.
4x – 2(2x – 4) = 9
4x – 4x + 8 = 9
-8 = 9
Solving the remaining equation.
Use the Distributive Property.
When we get a false equation that no longer has a variable this means the
two equations represent lines that are parallel! The false equations means
the lines never intersect! The only way lines can fail to intersect is if they are
parallel.
Exercises Unit 5 Section 3
Solve the listed systems using substitution.
1.
x=y+3
3x + 2y = 14
2. -x + y = 8
x = 2y + 7
3.
y=x+4
-2x + 3y = 10
4. 5x – 2y = 20
y = -2x + 8
5. 3x + 7y = 78
y=x+4
6.
y = 7 – 6x
8x - y = 0
7. 4x – 3y = -1
y = -2x
8.
9.
y = 4x - 5
y =0
y = 6x - 4
6 = 9x + y
10. What does it mean for two lines to be parallel?
11. What is the symbol we use for parallel lines?
12. What does it mean for two lines to be perpendicular?
13. What is the symbol we use for perpendicular lines? _________
Find the slope of the line in each graph then draw the asked for line on another
set of x-y axes.
14. a parallel line through (2, 0)
15. a || line through (-2, 1)
16. a perpendicular line through (2, 3)
17. a
line through (-1, 3)
Decide from the two listed equations if the lines are parallel, perpendicular or
neither.
18.
y = 4x + 2
y = -4x - 1
21. y = -4x + 2
y=
1
x-1
4
5
24. y = - x + 2
3
3
y=- x-1
5
1
x+3
2
y = -2x + 3
20. y = 7x + 1
22. y =
1
x+3
2
1
y=
x+7
2
2
23. y = - x + 1
3
2
y= x-1
3
25. y = x - 5
26. y = x + 1
19. y =
y= x+7
y = 7x -4
y = -x - 9
27. y = 5
y = -6
28. y = 0
x= 2
29. y = 5x + 1
y = 5x + 1
30. Find the equation of the line parallel to y = 3x + 1 and passing through (2, -5)
31. Find the equation of the line parallel to y = -
1
x - 2 and passing through (7, 1)
2
32. Find the equation of the line perpendicular to y =
3
x + 4 passing through (6, -1)
2
33. Find the equation of the line perpendicular to y =
2
x + 3 passing through (4, 5)
3
Unit 5 Section 4
Objective

The student will solve systems by isolating a variable and using the
method of substitution.
In this section we are going to be solving systems of equations by substitution again.
The process will be slightly different because we will need to isolate a variable in one
of our equations before we can go through the rest of the steps. The algorithm or set
of steps we should follow is given below.
Step
Step
Step
Step
1.
2.
3.
4.
Isolate a variable in one of the equations.
Do the substitution for the isolated variable in the other equation.
Solve the resulting equation.
Substitute the first coordinate into one of the equations
and find the second coordinate.
The examples that follow show us how the process changes.
Example A
Given the system of equations below find the point of intersection.
-3x + y = 7
2x + 4y = 14
Step 1.
-3x + y = 7
Since ‘y’ is positive and has a coefficient of 1
+3x
+3x
it is easiest to isolate ‘y’ in this equation.
y = 3x + 7
Step 2.
y = 3x + 7
2x + 4y = 14
2x + 4(3x + 7) = 14
Step 3.
2x + 4(3x + 7) = 14
2x + 12x + 28 = 14
14x + 28 = 14
- 28 - 28
We take what ‘y’ is equal to and
substitute it for ‘y’
This is the result of the substitution.
Solving the remaining equation.
14x = -14
14
14
x = -1
Step 4.
y = 3x + 7
y = 3(-1) + 7
y=4
Substitute into the equation in which
isolated the variable to find the second
The solution is (-1, 4) for this system of equations.
The overall process has not changed. We merely need to isolate a variable in one of
the equations before we can do our substitution.
Exercises Unit 5 Section 4
Solve the listed systems using substitution.
1. 3x + 5y = 20
x=y+4
2. -2x + y = 7
x=y-3
3.
5x - 4y = 15
2y = 4x - 6
4. x + y = 11
x - y = 13
5. 3x + 4y = 2
x – 2y = -1
6.
2x - y = 12
x + 4y = 51
7. 4x + 6y = -7
2x - 2y = 4
8. 3x + 6y = 2
6x – 7y = 4
9.
x = y2
x + 2y2 = 75
10. The point where two lines intersect will make the two equations ______________.
Check to see which if the listed point will make both equations true.
11. y = 2x - 4
y = -3x + 11
(3, 2)
12. y = -x + 5
2y + x = 8
(2, 3)
Graph each pair of lines and find the point of intersection.
13.
y = 3x - 4
y = -x + 4
1
x+2
2
y= 4
14. y = -
Find the slope of the line in each graph then draw the asked for line on another set of axes.
15. a || line through (-2, 2)
16. a || line through (2, 4)
17. a
line through (1, 3)
18. a
line through (-2, 6)
Decide from the two listed equations if the lines are parallel, perpendicular or neither.
19.
y=
1
x+2
3
20. y = -
y = -3x - 1
22.
y = 5x + 2
1
x -5
2
y = 2x + 3
5
x+3
2
2
y=
x+7
5
23. y =
y = -5x - 1
21. y = 7x + 1
y=
1
x-4
7
24. y = - 3 x + 1
y = -3x - 5
25. y = 1.5x + 1
y = -1.5x - 1
26. y = x + 5
y= x-9
27. y = -3x + 1
3y = x - 6
( Hint: Isolate y )
28.
29.
30. 2y = 4x - 8
1
y+ x= 1
2
y= 2
x=2
y = -3
y= 6
31. Find the equation of the line parallel to y =
1
x - 1 and passing through (4, -3)
2
32. Find the equation of the line parallel to y =  4 x + 2 and passing through (5, 2)
1
33. Find the equation of the line perpendicular to y = - x + 4 passing through (-1, 4)
3
34. Find the equation of the line perpendicular to y = -5x + 3 passing through (10, -1)
Unit 5 Section 5
Objective

The student will solve systems using elimination with similar
coefficients.
We have two methods to solve systems of equations. We can solve them graphically
or by substitution. Both of these methods have advantages and disadvantages. There
is, however, one more method to solve systems that we will learn. This method is called
elimination or linear combination. Elimination is another way of changing equations
with two variables into a single equation with one variable. The algorithm or set of
steps follow.
Step 1. Determine which variable can be eliminated
by adding or subtracting.
Step 2. Add or subtract the two equations using our
Properties of Equality.
Step 3. Solve the resulting equation.
Step 4. Find the other coordinate of our solution.
Below are some examples of using this algorithm.
Example A
Given the system of equations below find the point of intersection or
solution to the system using the method of elimination.
3x + 4y = -18
3x + y = 0
When we examine the two equations we can see that the ‘x’ variable
in both equations has a coefficient of 3. If we subtract 3x – 3x we
get 0x or 0 and the ‘x’ terms will be gone. So we use the Subtraction
Property of Equality and subtract the two equations. ( The left and
right sides of an equation are equal so we are subtracting two equal
quantities. )
3x + 4y = -18
-(3x + y = 0)
3y = -18
3
3
y = -6 We have our first coordinate.
3x + y = 0 We now substitute the y coordinate
3x – 6 = 0 into one of the equations and solve.
+6 +6
3x = 6
3
3
x=2
So the point (2, -6) is our solution or point of intersection.
Example B
Given the system of equations below find the point of intersection or
solution to the system using the method of elimination.
5x + 2y = 19
2x – 2y = 9
When we examine the two equations we can see that the ‘y’ variable
in the equations has a coefficient of 2 and -2. If we -2y + 2y we
get 0y or 0 and the ‘y’ terms will be gone. So we use the Addition
Property of Equality and add the two equations.
5x + 2y = 19
2x – 2y = 9
5x + 2y = 19
+(2x – 2y = 9)
7x = 28
7
7
x =4
We have our first coordinate.
5x + 2y = 19 We now substitute the x coordinate
5(4) + 2y = 19 into one of the equations
20 + 2y = 19 and solve.
-20
-20
2y = -1
2
2
y=So the point (4, -
1
2
1
) is our solution or point of intersection.
2
Exercises Unit 5 Section 5
Solve the following by linear combination or elimination. (add or subtract the equations)
Show your work. Check your answers.
1. 4x + 2y = 12
x - 2y = 13
2. 3x - 4y = -11
3x + y = -1
3.
4. 7x - 4y = 37
-2x + 4y = -12
3x + 5y = 7
2x + 5y = 3
5. 2x + 3y = 20
2x – 4y = 6
6.
5x – 4y = 5
6x – 4y = -8
7.
9.
x + 2y = 3.5
-x + 4y = -2
8.
x - 2y = -5
3x - 2y = 10
3x – 7y = -24
2x + 7y = -16
10.
.5x + .8y = 5
.2x – .8y = -3.6
Use substitution to solve the following systems. SHOW YOUR WORK!
11.
y = 3x + 7
y = 4x + 2
12. y = 2x - 3
y - x = -6
13. y - 5x = 3
3y - 2x = 22
14.
y= x+3
y + 2x = -9
15. Given the circle graph below that represents the breakdown of participation in
activities at Washington High School. Answer the questions that follow.
Washington High Extracurricular Activities Participation
Spring Sports
Fall Sports
25%
Other
11%
8%
20%
Winter Sports
34%
12%
Choir
Band
If there are 420 students participating in Extracurricular Activities at WHS
answer the questions that follow.
a. How many students were involved in fall sports?
b. How many students were involved in all sports?
c. How many more students were in Fall sports than in Winter sports?
Unit 5 Section 6
Objective

The student will solve systems using elimination using factors to
produce like coefficients.
When the coefficients of a system are not similar we must multiply one or both of the
equations by factors that will allow us to cancel out a variable. The process is listed
below as a set of steps or algorithm.
Step 1. Multiply one or both of the equations by factors to
produce opposite coefficients and cancel out variable terms.
Step 2. Add the two equations using our Property of Equality.
Step 3. Solve the resulting equation.
Step 4. Find the other coordinate of our solution.
Example A
Solve the system below using elimination or linear combination.
3x + 5y = -1
x - 3y = -5
Step 1. Adding or subtracting the equations now will not cancel out
out a variable’s terms. However if we multiply the bottom
equation by -3 then the ‘x’ terms will add up to 0. So we
use the Multiplicative Property of Equality and do the
multiplication.
-3(x – 3y) = -5(-3) Use the Multiplication Property of Equality
-3x + 9y = 15
Use the Distributive Property
Step 2. Add the equations
3x + 5y = -1
-3x + 9y = 15
Step 3.
14y = 14
14
Solve the resulting equation.
14
y= 1
Step 4.
x – 3y = -5
x – 3(1) = -5
x – 3 = -5
+3 +3
Find the value of the x-coordinate.
x = -2
So our point of intersection is (-2, 1).
Example B
Solve the system below using elimination or linear combination.
4x – 2y = 22
5x + 3y = 0
Step 1. Adding or subtracting the equations now will not cancel out
out a variable’s terms. However if we multiply the bottom
equation by 2 and the top equation by 3 then the ‘y’ terms
will add up to 0. So we use the Multiplicative Property of
Equality and do the multiplication.
3(4x – 2y) = 22(3) Use the Multiplication Property of Equality
12x – 6y = 66
Use the Distributive Property
2(5x + 3y) = 0(2) Use the Multiplication Property of Equality
10x + 6y = 0
Use the Distributive Property
Step 2. Add the equations
12x – 6y = 66
10x + 6y = 0
Step 3.
22x = 66
22
Solve the resulting equation.
22
x= 3
Step 4.
4x – 2y = 22
4(3) – 2y = 22
12 – 2y = 22
-12
Find the value of the y-coordinate.
-12
-2y = 10
-2
-2
y = -5
So our point of intersection is (3, -5).
Exercises Unit 5 Section 5
Solve the following by linear combination or elimination. (add or subtract the equations)
Show your work. Check your answers.
1. 3x + 5y = 1
x - 6y = 8
2. 3x - 4y = -21
8x + y = 14
3.
4.
4x – 7y = -6
-2x + 4y = 3
5. 2x + 3y = 20
8x – 6y = 80
6.
4x – 2y = -4
8x – 4y = -8
7.
8.
9.
x + 9y = 38
2x – 5y = 7
9x + 2y = -4
-4x + 3y = 29
x - 2y = 11
3x - 6y = 2
11. -3x - 2y = -31
3x - y = 25
5x - 11y = -2
2x + 8y = 24
10.
.8x + .7y = 11
.2x - .3y = -2
12.
x + 7y = 24
x - 9y = -40
Solve the following by substitution. Show your work.
13.
y = 3x
y + 2x = -20
14. 2x – 7y = 11
y – x = -8
List the name of the property illustrated in each problem.
15. -x = -1
17. x
.
.
x
y = y
16. a + 0 = a
.
x
18. 1
.
z=z
1
=1
b
19. -w + w = 0
20. b
.
21. (b + c) + t = b + (c + t)
22. x(y – z) = xy – xz
Unit 5 Section 7
Objective
 The student will graph linear inequalities in two variables.
In Unit 2 we learned to solve and graph inequalities in one variable. In this section we
will solve and graph inequalities in two variables. The graphing process is very similar
to graphing a linear equation. We will need to isolate the ‘y’ variable and use the slope and
intercept to graph the line then determine the part of the graph that needs to be shaded.
The algorithm for this process follows.
Step
Step
Step
Step
Step
1.
2.
3.
4.
5.
Isolate ‘y’ if needed.
Find m and b.
Graph the line and draw it either solid or dashed.
Pick a point and test this point in the inequality.
Shade the side of the line that makes the inequality true.
The examples below show us how to use the algorithm.
Example A
Graph y > 2x – 3
In this example the ‘y’ variable is initially isolated, so we don’t need to
perform Step 1.
y > 2x -3
m = 2 and b = -3
Step 2. Find the slope and intercept.
Step 3. We graph the line.
The line should be solid because the points on the line will make the
inequality true. We know the points on the line will make the
inequality true because there is an “equal to” bar underneath the
inequality symbol. For instance the point (2, 1) is on the line.
We can prove this by substituting the point into the inequality.
y > 2x -3
1 > 2(2) – 3
1>1
Which is true.
Example A (continued)
Now that we have graphed the line we need to pick a point. The
point (0, 0) is easy to use because arithmetic with a zero is simple.
The point we will pick is (0, 0)
We will now substitute the point (0, 0) back into the inequality.
y > 2x – 3
0 > 2(0) – 3
0 > -3
Example B
Step 4. Decide which side of the line to
shade.
This is true so we should shade the
side of the line that has (0, 0) in it.
Graph -2y + x > 4
There are two differences between example A and example B. The
‘y’ variable is not isolated here and the “equal to” bar is missing so
we are dealing with a strict inequality.
-2y + x > 4
–x –x
Step 1. Solve the inequality for y.
-2y > -x + 4
-2
-2 -2
1
y < x–2
2
When we divide by a negative value we
switch the inequality symbol.
Example B (continued)
1
x–2
2
1
m=
and b = -2
2
y <
Step 2. Find the slope and intercept.
Step 3. We graph the line.
The line should be dashed or dotted because the points on the line will
NOT make the inequality true. We know the points on the line will make
the inequality false because the points are on the line and will make
the two sides of the inequality the same. But we don’t have an “equal
to bar”. For instance the point (4, 0) is on the line. We can prove this
by substituting the point into the inequality.
1
x–2
2
1
0 < (4) – 2
2
0 <2–2
y <
0 <0
Which is false!
This is why we need to have a dashed line – points on the line do not
make the inequality true.
We will now substitute the point (0, 0) back into the inequality.
1
x–2
2
1
0 < (0) – 2
2
0 < -2
y <
Step 4. Decide which side of the line to
shade.
This is false so we should shade the side of
the line that does NOT have (0, 0) in it.
There are two important ideas we must remember:
1. When to use a solid line and when to use a dashed or dotted line.
2. How to determine which side of the line to shade
If we have a strict inequality which is > or < the line is dashed or dotted.
If the inequality is > or < the line is solid.
To decide which side to shade we pick an easy point to use, substitute the ‘x’ and
‘y’ values into the inequality. If the inequality is true shade the side that the test
point is on. If the inequality is false then shade the other side that the point is
not on.
Exercises Unit 5 Section 7
1. What are the steps we use to graph a linear inequality?
Graph the following inequalities
2.
y < 2x - 5
3. y > -x + 7
4.
5.
y>3
6. x > 1
7.
3
x-2
2
y> 4x–5
y<
3
8.
y < 4 – 2x
9. y > -
2
x
5
10. y <
x
+1
3
Isolate y and then graph.
11.
5y < -15x + 10
12. y - 3 > 2x - 9
13. -2(x – 6) > 2y + 8
15.
-2y < x - 6
16. x – y > 5x - 2
17. 3(x + 4) + x > -y +14
Graph each pair of lines and find the point of intersection.
18.
y = -2x + 4
y=
x+7
5
x–7
2
y= 3
19. y =
20. What are the advantages and disadvantages of using each of the three methods
we have; graphing, substitution, and elimination.
Unit 5 Section 8
Objective

The student will graph systems of linear inequalities in two variables.
Previously in this unit we learned to solve systems of equations by graphing. We can also
graphs systems of inequalities as well. The two processes are very similar. To
solve a system of equations we can graph both equations and look to find the point of
intersection. When we solve a system of inequalities we need to graph both of the
lines in the inequalities but we need to find the region to shade that makes both of the
inequalities true. The algorithm for this process is given below.
Step 1. Graph the two lines of inequalities using the technique
in the previous section.
Step 2. Select a test point from each of the four regions.
Step 3. Test each point in both inequalities until we find a point that
makes both inequalities true.
Step 4. Shade the region that makes both inequalities true.
We will use these steps in the examples that follow.
Example A
Given the system of inequalities below, graph the solution set.
y > 4x – 3
y < -x + 2
Using the slopes and intercepts we graph the lines that help form the
inequalities.
for y > 4x – 3
for y < -x + 2
m = 4 and b = -3
m = -1 and b = 2
The graph of the lines would be
The first line must be dashed
or dotted because it was a
strict inequality. The second
line is solid because it had the
“equal to bar”.
Step 1. Graphing the two lines of the inequalities is now complete!
There are four regions into which the lines divide the x-y plane. We must
choose a point from each of the four regions.
(0, 4)
(0, 0)
(5, 0)
Step 2. Picking a point from
each region. We
pick points with
zero’s to make the
evaluation easier.
(0, -6)
Now we need to test each of the points in the original inequalities
to see which region we need to shade.
Step 3. Testing each point in the inequalities.
Testing (0, 4)
y > 4x – 3
4 > 4(0) – 3
4>0–3
4 > -3
true
y < -x + 2
4 < -(0) + 2
4< 2
false
Testing (5, 0)
y > 4x – 3
0 > 4(5) – 3
0 > 20 – 3
0 > 17
false
y < -x + 2
0 < -(5) + 2
0 < -3
false
Testing (0, -6)
y > 4x – 3
-6 > 4(0) – 3
-6 > 0 – 3
-6 > -3
false
y < -x + 2
-6 < -(0) + 2
-6 < 2
true
Testing (0, 0)
y > 4x – 3
0 > 4(0) – 3
0>0–3
0 > -3
true
y < -x + 2
0 < -(0) + 2
0<2
true
Only one of the test points made both inequalities true. The point (0,0)
made both inequalities true. This is the region we will shade.
Step 4. Shade the appropriate region.
Example B
Given the system inequalities below graph the solution set.
-3y > 2x – 6
2y + 4 < 6x + 4
In this example we must first isolate the y variables so that we can use
the slope and intercept to graph the line.
2y + 4 < 6x + 4
–4
–4
-3y > 2x – 6
-3
-3 -3
2
y<- x+2
3
2y < 6x
2
2
Since we divided
by a negative value
we had to switch
the symbol.
y < 3x
Using the slopes and intercepts we graph the lines that help form the
inequalities.
for y < 3x
2
for y < - x + 2
3
m = 3 and b = 0
m=-
2
and b = 2
3
The graph of the lines would be
Again one line must be dashed
and the other solid. This does
not always have to happen but
it can.
Step 1. Graphing the two lines of the inequalities is now complete!
There are four regions into which the lines divide the x-y plane. We must
choose a point from each of the four regions.
(0, 4)
Step 2. Picking a point from
each region. We
(6, 0)
couldn’t pick (0,0)
since it is on one of
the lines.
(-1, 0)
(0, -3)
Now we need to test each of the points in the original inequalities
to see which region we need to shade.
Step 3. Testing each point in the inequalities. ( We can use the
inequalities in which we isolated y. )
Testing (0, 4)
Testing (6, 0)
y < 3x
4 < 3(0)
4 <0
y < 3x
0 < 3(6)
0 < 24
false
true
Testing (0, -3)
y < 3x
-3 < 3(0)
-3 < 0
true
y<-
y<-
2
x+2
3
2
0 < - (6) + 2
3
2
x+2
3
2
-3 < - (0) + 2
3
4<0+2
0 < -4 + 2
-3 < 0 + 2
4< 2
0 < -2
-3 < 2
false
false
true
2
x+2
3
2
4 < - (0) + 2
3
y<-
Testing (-1, 0)
y < 3x
0 < 3(-1)
0 < -3
false
2
x+2
3
2
0 < - (-1) + 2
3
2
0<
+2
3
2
0< 2
3
y<-
true
Only one of the test points made both inequalities true. The point (0,-3)
made both inequalities true. This is the region we will shade.
Step 4. Shade the appropriate region.
Exercises Unit 5 Section 8
1. List the algorithm we have for solving a systems of inequalities.
Graph the systems of inequalities
2. y > x – 3
y < 2x + 1
3. y > 4x – 6
y < -2x + 1
4.
y > 3x – 1
x>1
5. y < -x + 3
6. -2y < 6x – 6
2
y<
x+1
3
7.
y > 3x – 1
y < 3x - 5
8
y>1
x<0
9.
y < -1
y < 4x
2y < -3x – 4
10.
y>x+1
y < -3
y < -x + 2
12. -y < 2x - 4
13.
y+1>
Graph the following inequalities
11.
y<x-5
3
x-3
2
14. Find the slope of the line that passes through the points (2,4) and (-1, 6).
15. Find the equation of the line in slope-intercept form that has a slope of
m = 3 and passing through the point (-4, -5).
16. Find the equation of a line in point-slope form that has a slope of
m = -2 and passing through the point (-2, 7).
17. Find the equation of a line in slope-intercept form that passes through
the points (-3, 4) and (0, 10).
18. Find the equation of the line in slope-intercept form that is parallel to
y = 5x + 1 and passing through the point (2, -4).
19. If a line has a slope of m =
20. Convert the equation y =
1
find the slope of a perpendicular line.
7
3
x – 1 to standard form.
4
Unit 5 Section 9
Objective

The student will solve contextual problems with systems of equations
and inequalities.
Systems always have two equations or inequalities. When we deal with contextual
problems our first task will be write two equations or inequalities. Some steps that
may be useful in creating these equations or inequalities are given below.
Step
Step
Step
Step
Step
1.
2.
3.
4.
5.
List what the variables will represent.
List the quantities or constants we see in the problem.
Determine how to calculate the quantities.
Write the two equations or inequalities.
Solve the system.
The applications for systems in this section can be broken into several categories. The
first category we will deal with involves translation. These problems use our knowledge
of operations and the words that represent them. These problems may not require that
we use all of the steps above. Some examples of this type follow.
Example A
The sum of two numbers is 18 and the difference of the same
two numbers is 14. Find the numbers.
Step 1. The problem says to find two numbers, and in all of these
exercises we will need to have two variables - one for
each number.
x = first number
y = second number
Step 2. The constants in the problem are 18 and 14. In this
problem they are simply numbers.
Step 3. The key words are “sum” and “difference” which means
one equation will need to use addition and the other
subtraction.
Step 4. The two equations are:
x + y = 18
x – y = 14
The equation for the sum.
The equation for the difference.
Step 5. Solving the system. We can use any of three methods.
Graphing may not be practical because our numbers could
be large or they might not work out evenly. (With a
graphing calculator this method could be a good alternative.)
When we look at the system below we can see that adding
the two equations will cause the ‘y’ variable to cancel out.
We should use the method of elimination.
x + y = 18
x – y = 14
2x = 32
2
2
x = 16
x + y = 18
16 + y = 18
-16
-16
y= 2
Now we find the other number.
Our two numbers are 16 and 2.
Example B
The sum of two numbers is 74. The larger number is eight more
than twice the smaller. Find the numbers.
Step 1. The problem says to find two numbers, and in all of these
exercises we will need to have two variables - one for
each number.
x = smaller number
y = larger number
Step 2. The constant in the problem is 74.
Step 3. The key word is “sum” one equation will need to use
addition. The other equation can be found by direct
translation of the second sentence in the problem.
Step 4. The two equations are:
x + y = 74
y = 2x + 8
The equation for the sum.
The direct translation
Step 5. Solving the system. We can choose any of three methods.
When we look at the system below we can see that in the
bottom equation the ‘y’ is isolated. We can use the method
of substitution.
x + y = 74
y = 2x + 8
x + 2x + 8 = 74
3x + 8 = 74
-8 -8
3x = 66
3
3
x = 22
x + y = 74
22 + y = 74
-22
-22
y = 52
Now we find the other number.
Our two numbers are 22 and 52.
Another category of problems could be called the “together” type of problem. We follow
the same set of steps or algorithm. The example below shows us how to attack this
type of problem.
Example C
Mike and Tom are players on the Washington High School Basketball
team. In a recent game Mike scored five less than four times as many points as
Tom. Together they scored 30 points. Find the number of each person
scored.
Step 1. The problem says to find two numbers, and in all of these
exercises we will need to have two variables - one for
each number.
x = Tom’s points
y = Mike’s points
The smaller number
The larger number
Step 2. The constant in the problem is 30 points.
Step 3. The key word is “together”, so one equation we will need to
use is addition. The other equation can be found by direct
translation of the second sentence in the problem.
Example C (continued)
Step 4. The two equations are:
x + y = 30
y = 4x - 5
The equation for together.
The direct translation
Step 5. Solving the system. We can choose any of three methods.
When we look at the system below we can see that in the
bottom equation the ‘y’ is isolated. We can use the method
of substitution.
x + y = 30
y = 4x - 5
x + 4x - 5 = 30
5x – 5 = 30
+5 +5
5x = 35
5
5
x=7
x + y = 30
7 + y = 30
-7
-7
y = 23
This is Tom’s points.
Now we find Mike’s points.
So Mike scored 23 points and Tom scored 7
A third category or type of problem can be called coin problems. The example below
shows us how to use our steps for this category.
Example D
Mr. Jones wants his class to solve a problem for him. He has a small glass
jar with coins in it. The jar contains only dimes and nickels. He tells his
class the jar has $5.00 in it. He also tells his students that the total
number of coins is 71. The first student to find the numbers of nickels
and the number of dimes in the jar gets to keep the coins.
Step 1. The problem says to find two numbers, and in all of these
exercises we will need to have two variables - one for
each number.
x = the number of nickels
y = the number of dimes
Example D (continued)
Step 2. The constants in the problem are 71 which is the number of
coins and $5.00 which is the value of the coins.
Step 3. The word “total” tells us we have to add the numbers
of coins together to get the 71. To find the value of the
value of the coins we have to look at how we can find the
value of just one type of coin. If we had 2 nickels we
could multiply $0.05 times 2 and get $0.10. If we had
6 nickels we could multiply $0.05 times 6 and get $0.30.
So if we have ‘x’ nickels we would multiply $0.05 times x.
We would write this as .05x. We can do the same thing
for the dimes and we would get .10y or .1y. These two
expressions for the value must be added together to get
the total value.
Step 4. We should be able to write the equations now.
x + y = 71
.05x + .1y = 5
The total number of coins
The total value of the coins.
Step 5. Solving the system. We can choose any of three methods.
One method for solving this system would be to use
elimination and multiply the bottom equal by -20.
This will allow us to cancel out the ‘x’ terms.
x + y = 71
-20(.05x + .1y) = 5(-20)
x + y = 71
-x - 2y = -100
After multiplying by -20.
x + y = 71
-x - 2y = -100
-1y = -29
-1
-1
y = 29
x + y = 71
x + 29 = 71
-29 -29
x = 42
The number of dimes
Now we find the number of nickels.
the number of nickels
So there are 42 nickels and 29 dimes.
Example E
An insecticide must be diluted before it can be sprayed. Currently the
solution is 60% insecticide. The most efficient and safe solution is a
25% solution. The current volume of the solution is 2600 ml. What
volume of water must be added to make the solution 25%.
Step 1. This problem involves amount of a solution. We need to know
how much water needs to be added and how much solution
we will have when we are done.
x = the volume of water to be added
y = the volume of the solution after adding water
Step 2. The two quantities in the problem are the amounts of liquid
and the amount of insecticide.
Step 3. One of the equations should give us the “total” amount of
liquid. The other equation will be based on the percentages
that will help us calculate the amount of insecticide, which
must be the same before and after adding the water. Adding
water will not change the amount of insecticide in the
mixture. We use percentages in this case to multiply.
Step 4. Our equations are.
y = x + 2600
.6(2600) = .25(y)
The total amount of liquid
The amount of insecticide
We need to solve the second equation and then substitute.
Step 5.
1560 = .25y
.25
.25
6240 = y
y = x + 2600
6240 = x + 2600
-2600
-2600
3640 = x
The second equation solved
Now we substitute
So our answer is we must add 3640 ml of water.
These techniques can be used on many types of problems not just the types we have listed.
When we do the problems we need to show steps 1, 4, and 5. Steps 2 and 3
in our algorithm must be performed mentally.
Exercises Unit 5 Section 9
Solve the following problems. Show your work.
1. The sum of two numbers 51. The difference of the numbers is 9. Find the
numbers.
2. Two numbers have a sum of 45 and a difference of 13 find the numbers.
3. The sum of two numbers is 26 and the difference of the same two numbers is 4.
Find the numbers.
4. The sum of two numbers is -34 and the difference of the numbers is 8. Find
the numbers.
5. The sum of two numbers is 61. The larger number is thirteen more than twice
the smaller number. Find the numbers.
6. The difference of two numbers is 72. The larger number is sixteen more than
five times the smaller number. Find the numbers.
7. A rectangle has a perimeter of 96 cm. The length is six less than twice the width.
Find the length and width.
8. The width of a rectangle is half the length. If the perimeter 81 inches. Find the
length and width.
9. A rectangle has a perimeter of 68 m. The length is eight meters more than
the width. Find the AREA of the rectangle.
10. A rectangle has a perimeter of 56 cm. If we were to add 5 cm to the width and
subtract 5cm from its length we would have a square. Find the length and width.
11. Washington High School has a fund raiser on Halloween. They are running a
haunted house. The tickets are $5 for children and $10 for adults. At the end
of the night they know they had $1705. They also know they had 273 people
go through the haunted house. How many of each type of ticket was sold?
12. Sam is buying clothes for the start of the school year. He buys a number of pairs of
pants and some shirts. The pants cost $42 each and the shirts cost $38 dollars
each. The total cost was $316. If he bought 8 items of clothing altogether find
how many shirts he bought.
13. The local zoo is running a fund raiser. They are holding a pancake breakfast.
There is a fixed price for each breakfast but they charge extra for coffee.
One table has a total bill of $84 for 6 orders of pancakes and 4 cups of coffee.
A second table is charged $77.50 for 5 orders of pancakes and 5 cups of coffee.
Find the price of the coffee and pancakes.
14. Alicia manages a toy store. She is buying two sizes of teddy bears, large and small.
The large bears cost twice as much as the small bears. If she buys 20 small bears
and 12 large bears and the cost is $448.80 find the cost of each size bear she
bought.
15. A chemist has 2000 g of a solution. The solution is 30% acid. He wants to use the
solution to wash some metal equipment but the acid content is too high and will
damage the metal. He needs to add water to dilute the acid. If he wants the
solution to be 12% acid how much water does he need to add?
16. A chemical engineer wanted to etch glass with an acidic solution. The solution that
he had a 25% acid content. This is too weak to do the etching. He wanted to
raise the content to 40%. He added 800 g of acid to the solution and raised the
content to 40%. What was the size of the original solution?
17. A new High School just opened. The students were asked to vote on the mascot.
The choice they had was between the “Wolves” and the “Hawks”. In the voting
the total number of votes cast was 924. If the “Wolves” won by 90 votes how
many votes did each mascot received?