Download Study Resources For Unit 1B Graphs, Equations, Linear Functions

Study Resources For
Algebra I
Unit 1B
Graphs, Equations, Linear
Functions, and Inequalities
This unit explores linear relationships
through tables, graphs, and equations.
Information compiled and written by Ellen Mangels, Cockeysville Middle School
July 2014
Algebra I
Page 1
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Topics:
Each of the topics listed below link directly to their page.











Linear Function
Analyzing Tables
o First Difference
o Second Difference
o Common Ratio
Rate of Change
Slope
o Slope Formula
o Calculate the Slope using the Slope Formula
o Rise over Run
o Positive Slope
o Negative Slope
o Zero Slope
o Undefined Slope
o The Four Types of Slope (summary)
Initial Value (y-intercept)
Slope – Intercept Form of a Linear Equation (y = mx + b)
o Graphing Equations written in Slope – Intercept Form
Evaluating Functions
o Using Substitution
o Using a Table of Values
o Using a Graph
o Evaluating Functions in a Real World Context
Solve Equations using a value of the Function
o On a Graph
o Using a Table of Values
o Solving Equations in a Real World Context
Standard Form of a Linear Equation
o Example of Standard Form in a real world context
o Graphing Standard Form Equations
 Write the Standard Form as Slope-Intercept Form
 Define x and y intercept
 Identify the x-intercept and y-intercept
o Writing Linear Equations in Standard Form
 When given the intercepts
 When given the slope and a point
 When given two points
Point-Slope Form
Literal Equations
Back to Topic List
Algebra I
Page 2
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities




Inequalities
o Inequalities on a Number Line
o Interval Notation
o Compound Inequalities
 Union of Two Solution Sets
 Intersection of Two Solution Sets
o Solving Inequalities
 Multiplying or Dividing by a Negative Number
Absolute Value Function
o Linear Absolute Value Function
o Transformations of Linear Absolute Value Functions
o Solving Absolute Value Equations
 Multistep
 With Graphs
o Absolute Value Inequalities
Quadratic Functions
o Solving Quadratic Functions using Graphs
o Solving Quadratic Inequalities
Exponential Functions
o Solving Exponential Functions using Graphs
o Solving Exponential Inequalities
Algebra I
Page 3
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Linear Function
Back to Topic List
An example of a linear relationship
Mary is selling lemonade.
She has fixed costs of $5 for the equipment
she needs to make the lemonade
(pitcher, knife, spoon, etc.)
She has variable costs of $0.75 per cup of
lemonade for the ingredients needed to make
the lemonade (lemons, water, sugar, ice).
Let x = the number of cups of lemonade
Let y = the total cost of making the lemonade
y = 0.75 x + 5
x
Independent Variable
# of cups of lemonade
0
1
2
3
y
Dependent Variable
Total cost of making
the lemonade
$ 5.00
$ 5.75
$ 6.50
$ 7.25
Is this linear relation also a linear function? (yes)
Does it pass the vertical line test? (yes)
Algebra I
Page 4
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
First Differences
Back to Topic List
A look at how to calculate First Differences: http://www.youtube.com/watch?v=Cr1LkgO7HnA
+ 1
+ 1
+ 1
+ 1
x
y
0
0
+ 30
1
30
+ 30
2
60
3
90
4
120
+ 30
The first
difference
is 30.
+ 30
The first difference is constant.
Therefore, the table of values represents a linear function of the form y = mx + b.
+ 1
+ 1
+ 1
+ 1
x
y
0
0
+ 1
1
1
+ 3
2
4
3
9
4
16
+ 5
+ 7
The first difference is not constant.
Therefore, the table of values does not represent a linear function.
Algebra I
Page 5
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Second Differences
Back to Topic List
+ 1
+ 1
+ 1
x
y
0
0
1
1
2
4
3
+ 1
4
9
16
The second
difference is 2.
+ 1
+ 3
+ 5
+ 7
+ 2
+ 2
+ 2
The second difference is constant.
Therefore, the table of values represents a quadratic function of the form y = ax2 + bx + c
(Source: http://www.bbc.co.uk/schools/gcsebitesize/maths/algebra/transformationhirev1.shtml)
Algebra I
Page 6
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Common Ratio
Back to Topic List
+ 1
+ 1
+ 1
+ 1
x
y
0
1
1
2
2
4
3
8
4
16
+ 1
+ 2
+ 4
+ 1
+ 2
+ 4
+ 8
There is no common first difference.
There is no common second difference.
+ 1
+ 1
+ 1
+ 1
x
y
0
1
1
2
2
4
3
8
4
16
X 2
X 2
X 2
X 2
There is a common ratio.
Therefore, the table of values represents an exponential function of the form y = ax
(Source: https://people.richland.edu/james/lecture/m116/logs/exponential.html)
Algebra I
Page 7
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Rate of Change
Back to Topic List
The Rate of Change of a
Linear Function
= the Slope of the Line
The Rate of Change shows how quickly a linear function is increasing or decreasing.
Example:
Todd had 5 gallons of gasoline in his motorbike. After driving 100 miles, he had 3 gallons left.
The gasoline in the tank decreases 1 gallon for
every 50 miles that Todd drives.
The rate of change is -1 / 50.
Since -1 ÷ 50 = - 0.02, we know that Todd's bike is
burning 0.02 gallons of gasoline for every mile
that he travels. The negative value of the slope
tells us that the amount of gasoline in the tank is
decreasing.
(Source: http://www.regentsprep.org/regents/math/algebra/ac1/rate.htm)
Algebra I
Page 8
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Back to Topic List
(Source: Glencoe, Algebra I, page 258-259)
Video – Interpret The Rate of Change (Slope)
Within The Context of Everyday Life (CC.8.F.4 & CC.9-12.S.ID.7)
http://www.youtube.com/watch?v=3GhqweMiYnY&list=PL16DEDA7C66C8B23E
Algebra I
Page 9
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Slope
Back to Topic List
Slope =
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦
Slope Formula:
𝑦 −𝑦
𝑚 = 𝑥2 − 𝑥1
2
1
(Source: Glencoe, Algebra I, page 256)
Algebra I
Page 10
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
(Source: Glencoe, Algebra I, page 256)
Video Demonstration: Calculate the Slope of a Line when given Two Points
http://www.virtualnerd.com/algebra-1/linear-equation-analysis/slope-rate-of-change/slope-examples/slope-from-two-points
Calculate the Slope using the Slope Formula
Find the slope of the line
passing through the points
(−3, −5) and (2, 1).
(Source: http://catalog.flatworldknowledge.com/bookhub/reader/128?e=fwk-redden-ch03_s04)
Algebra I
Page 11
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Slope = Rise over Run
Back to Topic List
(Source: http://curriculum.nismed.upd.edu.ph/2012/08/what-is-a-slope)
Slope =
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Example of Slope
(Source: http://education-portal.com/academy/lesson/point-slope-form-definition-lesson-quiz.html#lesson)
Algebra I
Page 12
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Positive Slope
Back to Topic List
Imagine you are climbing a
steep cliff. If the cliff has
some incline to it (not
perfectly vertical), then you
will travel 'forward' while
you climb.
In the picture, the climber
ascends 8 meters while
she travels forward 5
meters.
If you think of slope, or
steepness, as 'rise over
run', then the slope of her
path is:
Slope = 8/5
(Source: http://education-portal.com/academy/lesson/undefined-slope-definition-examples-quiz.html#lesson)
(Source: Glencoe, Algebra I, page 257)
Algebra I
Page 13
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Negative Slope
Back to Topic List
Imagine you are riding a bike down a steep hill.
You will travel forward while you descend.
In the picture, the rider descends 5 meters while
he travels forward 8 meters.
Since the rider is descending, we will think of
the “rise” as -5.
5 meters
If you think of slope, or steepness, as 'rise over
run', then the slope of his path is:
Slope = -5/8
8 meters
Sometimes we think of a negative slope as “fall
over run” to remind us that the slope will be
negative.
(Source: Glencoe, Algebra I, page 257)
Algebra I
Page 14
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Zero Slope
Back to Topic List
The sail boat is traveling
horizontally.
The line generated by the
path of the boat would
have a slope of zero.
Lots of run, but zero rise.
(Source: Glencoe, Algebra I, page 257)
Algebra I
Page 15
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Undefined Slope
Back to Topic List
Here is an example of a
nearly vertical cliff.
If you tried to climb this,
your rise would be 400 feet
and your run would be
zero.
(Source: Glencoe, Algebra I, page 257)
Algebra I
Page 16
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
The Four Types of Slope
Back to Topic List
(Source: Glencoe, Algebra I, page 258)
(Source: http://www.scs.sk.ca/cyber/elem/learningcommunity/math/math30aa/curr_content/matha30rev1/lesson3-4/lesson3-4.htm)
Algebra I
Page 17
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Back to Topic List
(Source: Glencoe, Algebra I, page 258)
Algebra I
Page 18
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Initial Value (y-intercept)
The y-intercept is where a
the Y axis of a graph.
Back to Topic List
The Initial Value of a
Linear Function
= the y-intercept
of the Line
straight line crosses
(Source: http://www.mathsisfun.com/definitions/y-intercept.html)
Video – Explains the meaning of Rate of Change
and Initial Value in the context of a situation.
https://learnzillion.com/lessons/3508-interpret-linear-relationships-in-word-problems
Algebra I
Page 19
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Back to Topic List
Rate of Change and Y-Intercept in a Table
x
y
+ 1
0
3
+ 2
+ 1
1
5
+ 2
+ 1
2
7
+ 2
+ 1
3
9
+ 2
4
11
The first
difference
is 2.
The first difference is constant. Therefore this is a linear function.
The first difference is 2. The rate of change is 2. The slope of the line for this relationship is 2.
When the
x value is
zero…
x
y
0
3
1
5
2
7
3
9
4
11
… then the y
value is the
initial value.
When x-value in the table is zero, then its y-value is the initial value of the function.
The initial value of the function is the y-intercept of the graph.
The yintercept
is 3.
Run = 1
(0, 3)
Rise = 2
2
Slope = 1 = 2
Algebra I
Page 20
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Slope – Intercept Form of a Linear Equation (y = mx + b)
Back to Topic List
(Source: http://www.mathsisfun.com/equation_of_line.html)
Interactive Activity: Explore how changing m and b affects the equation and the line.
http://www.mathsisfun.com/data/straight_line_graph.html
Algebra I
Page 21
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Graphing Equations written in Slope – Intercept Form
(y = mx + b)
Back to Topic List
Example: y = 3x +5
1. Identify the slope: 3
2. Identify the y-intercept: 5
3. Plot the y-intercept on the graph (0,5)
4. Starting at the y-intercept, use the slope to plot another point.
Slope =
3
1
Rise = 3
Run = 1
Algebra I
Page 22
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
5. Draw the line through the points.
Hint: It is a good idea to plot more than two points to make sure your line is correct.
Option: If you can’t use the rise and run as written because you have run out of space
on the graph paper, you can use the opposites.
Example:
2
−2
5
−5
Slope = =
Slope =
−4
7
=
4
−7
Slope = - 5 =
−5
1
=
5
−1
Rap Song: Graph http://www.youtube.com/watch?v=jGJrH49Z2ZA
Algebra I
Page 23
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Evaluating Functions
Back to Topic List
Use substitution to evaluate the function.
To evaluate a function f (x), read as f of x, substitute the variable x with the given number.
Example: Find f (3) when the function is f (x) = -5x – 8
f (3) = -5(3) – 8 = -15 – 8 = -23
therefore f(3) = -23
Example: Find f(-2) when the function is f(x) = -5x – 8
f (-2) = -5(-2) – 8 = 10 – 8 = 2
therefore f(-2) = 2
Video – Evaluating Functions:
https://www.khanacademy.org/math/algebra/algebra-functions/graphing_functions/v/evaluating-functions
Video – Evaluating Functions (CC.9-12.F.IF.1)
http://www.youtube.com/watch?v=RZ8yLN46_YQ
Use a table of values to evaluate the function.
x
y
Example: Find f (-2) when the function is f (x) = -5x – 8
Type -5x – 8 into the Y= window of your graphing calculator.
Look at the Table of Values (2nd GRAPH).
-3
7
-2
2
-1
-3
0
-8
1
-13
2
-18
3
-23
4
-28
f (-2) = 2
Find f (0). f (0) = -8
This is the y-intercept of the function on its graph.
f (3) = -23
Rate of
Change = -5
Algebra I
Page 24
Notice the y values in the table decrease by 5
each time the x value increases by 1.
Notice that the coefficient of the x variable in
the equation is also -5.
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Use a graph to evaluate the function.
Back to Topic List
Example: Find f (-5)
(Source: http://2012books.lardbucket.org/books/advanced-algebra/s05-01-relations-graphs-and-functions.html)
In this example, f (-5) = 3.
Look at the x-value (in this case -5) and find the corresponding y-value on the graph of the
function (in this case it is 3).
Note: this is an example of a linear absolute value function.
Compare evaluating the function using a graph to the substitution method.
𝑓(𝑥) = |𝑥| – 2
𝑓(−5) = |−5| – 2
𝑓(−5) = 5 – 2 = 3
Algebra I
Page 25
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Evaluating Functions in a Real World Context
Back to Topic List
You want to rent a car. Cory’s Car Rental charges $60 to rent a car for a day plus
$0.25 per mile you drive the car.
This situation can be represented as a function:
f (m) = 0.25 m + 60
where m represents the number of miles driven
How much will it cost you to rent a car for a day and drive 12 miles?
f (12) = 0.25 (12) + 60
f (12) = 0.25 (12) + 60
f (12) = 3 + 60 = 63
It will cost you $63.00
How much will it cost you to rent a car for a day and drive 350 miles?
f (350) = 0.25 (350) + 60
f (350) = 0.25 (350) + 60
f (350) = 87.5 + 60 = 147.5
It will cost you $147.50
Although it is likely that you would have rented a car for more than one day,
that extra fact would make this problem a bit more complicated than we want to look at just now.
Algebra I
Page 26
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Solve Equations using a value of the Function
Back to Topic List
What if you wanted to go in the opposite direction? Instead of finding a value of the function
when you are given a value for x, you want to find a value for x when you are given a value for
the function.
Using Graphs
Example: If f (x) = -4, use the graph to find the value for x.
(Source: http://2012books.lardbucket.org/books/advanced-algebra/s05-01-relations-graphs-and-functions.html)
x = -5
and
x = 15
Video – Solving Equations Using a Graph:
http://www.youtube.com/watch?v=NtvyMlfPqWs&feature=player_embedded
Algebra I
Page 27
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Using Tables
Back to Topic List
Example: If f (x) = 3x + 2 and f (x) = 14, use the table to find the value for x.

Type 3x + 2 into the Y= window of your graphing calculator.

Look at the table of values (2nd GRAPH).

Find 14 in the y column.

Determine the corresponding x value.
x = 4 when f (x) = 14
You just used a table of values to solve the equation:
3x + 2 = 14
x
y
-3
-7
-2
-4
-1
-1
0
2
1
5
2
8
3
11
4
14
f (x) = 3x + 2
f (x) = 14
Here is the same problem on a graph.



Look at the y-value 14.
Find the corresponding x-value
on the graph.
x=4
x=4
Algebra I
Page 28
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Solving Equations in a Real World Context
Back to Topic List
You want to rent a car. Cory’s Car Rental charges $60 to rent a car for a day plus
$0.25 per mile you drive the car.
This situation can be represented as a function:
f (m) = 0.25 m + 60
where m represents the number of miles driven
You spent $124.75 to rent a car for a day. How many miles did you drive?
f (m) = 124.75
124.75 = 0.25 m + 60
Method 1 Solve the equation:
124.75 = 0.25 m + 60
- 60
- 60
Subtract 60 from both sides of the equation.
64.75 = 0.25 m
0.25
0.25
Divide both sides of the equation by 0.25
259 = m
You drove 259 miles in the rental car.
Method 2 Using Tables
Example: If f (m) = 0.25 m + 60 and f (m) = 124.75, use the table to find the value for m.

Type 0.25 x + 60 into the Y= window of your graphing calculator.

Look at the table of values (2nd GRAPH).

Find 124.75 in the y column.

Determine the corresponding x value.
x = 259 when f (x) = 124.75
Note: To avoid using time scrolling down the table on the graphing calculator,
you can choose the value to start the table using TBLSET (2nd WINDOW).
Algebra I
Page 29
x
y
256
124
257
124.25
258
124.50
259
124.75
260
125
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Standard Form of a Linear Equation
Back to Topic List
Ax + By = C
(Source: Glencoe, Algebra I, page 218)
Compare Standard Form (Ax + By + C) to Slope-Intercept Form (y = mx + b)
Although both forms use the letter b, the Standard Form uses a capital B.
Lower case b does not have the same value as capital B in these two equations.
Algebra I
Page 30
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Example of Standard Form in a real world context
Back to Topic List
Ben and Zach went to the Raven’s game and bought several sodas and hot dogs.
They spent a total of $57.
The sodas cost $2.50 each and the hot dogs cost $3 each.
What possible combinations of sodas and hot dogs could they have purchased?
Solution:
There are an infinite number of possible values that could satisfy the equation:
2.50x + 3y = 57
where x represents the number of sodas and y represents the number of hot dogs.
We can only consider whole number values for x and y for this problem.
It wouldn’t make sense to buy 1.4 hot dogs or -3.6 sodas.
Method 1: Guess and Check
Pick values for x and y until you find a combination that gives you a total of $57.
Method 2: Solve the equation for y and look at the table of values.
2.50x + 3y = 57
- 2.50x
- 2.50x
Subtract 2.50x from both sides of the equation.
3y = 57 – 2.50x
3
3
Divide both sides of the equation by 3
y = (57 – 2.50x) / 3
Type this equation into Y=
There are only four pairs of values that make sense for this situation.
(0, 19)
They bought 0 sodas and 19 hot dogs.
(6, 14)
They bought 6 sodas and 14 hot dogs.
(12, 9)
They bought 12 sodas and 9 hot dogs.
(18, 4)
They bought 18 sodas and 4 hot dogs.
Algebra I
Page 31
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Graphing Standard Form Equations
Back to Topic List
Write the Standard Form as Slope-Intercept Form
(Source: Glencoe, Algebra I, page 219-220)
Algebra I
Page 32
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Graphing Standard Form Equations
Back to Topic List
Definition of x and y intercept
What is an x-intercept?
What is a y-intercept?
Simply put, these are the points on the x and y axis
where the linear function crosses the x and y axis.
(Source: http://equizshow.com/print/282)
Algebra I
Page 33
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Graphing Standard Form Equations
Back to Topic List
Identify the x-intercept and y-intercept
(Source: Glencoe, Algebra I, page 220-221)
Video – Graphing Standard Form Equations when given
the x-intercept and the y-intercept:
http://www.virtualnerd.com/algebra-1/linear-equation-analysis/intercept/intercept-examples/x-y-intercepts-graphstandard-line
Algebra I
Page 34
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Writing Linear Equations in Standard Form
Back to Topic List
When given the intercepts
Example: Given the x-intercept is 3 and the y-intercept is -4,
write the equation of the line in Standard Form

Use the intercepts to find the slope.
If the x-intercept is 3, it can be written as the point (3, 0)
If the y-intercept is -4, it can be written as the point (0, -4)
𝑦 −𝑦
𝑚 = 𝑥2 − 𝑥1
Apply the slope formula:
2
𝑚=

4
4
4
3
4x – 12
=
-
4x
Subtract 4x from both sides of the equation
-4x + 3y = -12
This is not quite Standard Form.
Standard Form should have a positive coefficient for x.
(-1)(-4x) + (-1)(3y) = (-1)(-12)
4x – 3y = 12
Algebra I
−3
=
Multiply both sides of the equation by 3 to eliminate the fraction
3
4x
−4
Rewrite Slope-Intercept form as Standard Form
(3)y = (3) x – (3)4
-
0−3
=
Slope-Intercept Form: y = mx + b
3
3y
−4 − 0
Use the slope and y-intercept to write the equation in Slope-Intercept Form
y= x–4

1
Multiply both sides of the equation by -1 to make
the coefficient of x a positive number.
Standard Form
Page 35
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Writing Linear Equations in Standard Form
Back to Topic List
When given the slope and a point
1
Example: Given the slope of the line is 2 and the line passes through the point (-6, 5)

Substitute the slope and the point into the Slope-Intercept form of the equation
and solve for b
y = mx + b
1
5 = 2 (-6) + b
5 = -3 + b
Add 3 to both sides of the equation.
8=b

The equation written in Slope-Intercept form is
1
y=2x+8

Rewrite the equation in Standard Form
1
(2) y = (2) 2 x + (2) 8
Multiply both sides of the equation by 2
2y = x + 16
Subtract x from both sides of the equation
-x + 2y = 16
This is not quite Standard Form.
Standard Form should have a positive coefficient for x.
(-1)(-x) + (-1)(2y) = (-1)(16)
x – 2y = -16
Algebra I
Page 36
Multiply both sides of the equation by -1 to make
the coefficient of x a positive number.
Standard Form
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Writing Linear Equations in Standard Form
Back to Topic List
When given two points
Example: Given the points (-2, 13) and (4, -5), write the equation of the line in Standard Form

Use the points to find the slope.
𝑦 −𝑦
𝑚 = 𝑥2 − 𝑥1
Apply the slope formula:
2
1
−5 − 13
𝑚 = 4 −(−2) =


−18
6
= -3
Substitute the slope and a point into the Slope-Intercept form of the equation
and solve for b. It doesn’t matter which point you choose.
y = mx + b
y = mx + b
13 = (-3) (-2) + b
-5 = (-3) (4) + b
13 = 6 + b
-5 = -12 + b
7=b
7=b
The equation written in Slope-Intercept form is
y = -3 x + 7

Rewrite the equation in Standard Form
y = -3 x + 7
Algebra I
3x + y = 7
Add 3x to both sides of the equation
3x + y = 7
Standard Form
Page 37
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Point-Slope Form
Back to Topic List
(Source: Glencoe, Algebra I, page 286)
(Source: Glencoe, Algebra I, page 286)
Rewriting Point-Slope form as Standard Form
Use the equation y – 5 = -3(x + 1)
y – 5 = -3(x + 1)
Distribute the -3
y – 5 = -3x – 3
+5
y
+ 3x
+5
= -3x + 2
+ 3x
Add 5 to both sides of the equation
Notice the equation is now in Slope-Intercept Form
Add 3x to both sides of the equation
3x + y = 2
Algebra I
Page 38
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Literal Equations
Back to Topic List
Sometimes you have a formula, such as something from geometry, and you need to solve for some variable
other than the "standard" one. For instance, the formula for the perimeter P of a square with sides of length s
is P = 4s. You might need to solve this equation for s, so you can substitute in a perimeter and figure out the
side length.
This process of solving a formula for a given variable is called "solving literal equations". One of the dictionary
definitions of "literal" is "related to or being comprised of letters", and variables are sometimes referred to as
literals. So "solving literal equations" seems to be another way of saying "taking an equation with lots of
letters, and solving for one letter in particular."
At first glance, these exercises appear to be much worse than your usual solving exercises, but they really
aren't that bad. You pretty much do what you've done all along for solving linear equations and other sorts of
equations; the only substantial difference is that, due to all the variables, you won't be able to simplify your
answers as much as you're used to.
(Source: http://www.purplemath.com/modules/solvelit.htm)
Video – What is a literal equation?
http://www.virtualnerd.com/algebra-1/linear-equations-solve/isolate-variables-formulas-examples/isolate-variable/literalequation-definition
Example 1:
(Source: http://www.algebra-class.com/literal-equations.html)
Algebra I
Page 39
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Literal Equations – more examples
Back to Topic List
Example 2:
(Source: http://www.algebra-class.com/literal-equations.html)
Example 3:
(Source: Glencoe, Algebra I, page 166)
Algebra I
Page 40
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Literal Equations – even more examples
Back to Topic List
Example 4:
(Source: Glencoe, Algebra I, page 167)
Example 5:
below
(Source: Glencoe, Algebra I, page 167)
Algebra I
Page 41
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Inequalities
Back to Topic List
Inequalities on a Number Line
(Source: http://catalog.flatworldknowledge.com/bookhub/reader/128?e=fwk-redden-ch02_s07)
Notice the shaded point at 2. This is used to indicate the point is included in the solution.
The square bracket in the interval notation also indicates the 2 is included in the solution.
(Source: http://catalog.flatworldknowledge.com/bookhub/reader/128?e=fwk-redden-ch02_s07)
Notice the open circle at 2. This is used to indicate the point is not included in the solution.
The parenthesis in the interval notation also indicates the 2 is not included in the solution.
The infinity symbol in both examples, ∞, indicates the solution,
just like the number line, goes on forever to infinity.
Algebra I
Page 42
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
More Examples:
Back to Topic List
(Source: http://catalog.flatworldknowledge.com/bookhub/reader/128?e=fwk-redden-ch02_s07)
Summary of Interval Notation:
When using ≥ or ≤, use a square bracket [ with interval notation
and a closed circle on the number line
When using > or <, use a parenthesis (with interval notation
and an open circle on the number line
There is always a ( next to ∞ in interval notation.
Another example: x > 3.
(Source: http://www.shmoop.com/equations-inequalities/equivalent-inequalities.html)
Algebra I
Page 43
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Compound Inequalities on a Number Line
Back to Topic List
The Union of Two Solution Sets:
(Source: http://catalog.flatworldknowledge.com/bookhub/reader/128?e=fwk-redden-ch02_s07)
With a situation where an “or” is used, the interval notation uses U
for the union of the two solution sets.
(Source: http://catalog.flatworldknowledge.com/bookhub/reader/128?e=fwk-redden-ch02_s07)
The solution can also be written in set notation. This statement is read as
“The set of all real numbers x, such that x is less than 3 or x is greater than or equal to 6.”
The vertical line after the first x is the symbol for “such that”.
Algebra I
Page 44
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Compound Inequalities on a Number Line
Back to Topic List
The Intersection of Two Solution Sets:
(Source: http://catalog.flatworldknowledge.com/bookhub/reader/128?e=fwk-redden-ch02_s07)
Notice that in this example we want two things to happen at the same time.
Our solution has to be greater than or equal to -1 AND less than 3.
Interval Notation: [-1, 3)
(Source: http://catalog.flatworldknowledge.com/bookhub/reader/128?e=fwk-redden-ch02_s07)
Algebra I
Page 45
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
More Examples of the Intersection of Two Solution Sets
Back to Topic List
(Source: http://catalog.flatworldknowledge.com/bookhub/reader/128?e=fwk-redden-ch02_s07)
Algebra I
Page 46
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Solving Inequalities
Back to Topic List
Solution: {x | x ≤ 3}
(Source: http://cnx.org/content/m19599/latest)
Interactive Activity:
Try solving some inequalities on your own. This web site allows you to solve the
inequality and graph the solution. It gives hints if you get stuck.
http://cnx.org/content/m19599/latest
Algebra I
Page 47
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Solving Inequalities:
Back to Topic List
Multiplying or Dividing by a Negative Number
Solving inequalities is very similar to solving equations. Instead of using the Properties of Equality,
you use the Properties of Inequalities. These two sets of properties are almost identical except for
one very, very important rule:
When multiplying or dividing both sides of an inequality by a negative number, you must reverse the
inequality symbol.
Remember that rule when trying this one!
(Source: http://cnx.org/content/m19599/latest)
Algebra I
Page 48
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Video – Solving Inequalities
Back to Topic List
These videos show you step by step how to solve an inequality problem.

One Step Inequalities involving Addition
https://www.khanacademy.org/math/algebra/linear_inequalities/inequalities/v/one-step-inequalities

Two Step Inequalities
http://www.khanacademy.org/math/algebra/linear_inequalities/inequalities/v/solving-inequalities

Compound Inequalities
https://www.khanacademy.org/math/algebra/linear_inequalities/
compound_absolute_value_inequali/v/compound-inequalities
Algebra I
Page 49
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Absolute Value Functions
Back to Topic List
A Linear Absolute Value Function can be viewed as a piecewise function of two linear
functions.
Consider the function f(x) = x
(which can also be written as the equation y = x)
Now consider the function f(x) = -x
(which can also be written as the equation y = -x)
What if you only used the positive x values for y = x
and only used the negative x values for y = -x?
Algebra I
Page 50
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Now put the two functions together.
Back to Topic List
(Source: http://catalog.flatworldknowledge.com/bookhub/10121?e=fwk-redden-ch02_s04)
Algebra I
Page 51
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
The Linear Absolute Value Function
Back to Topic List
Notice that the graph only has positive y values.
This is because the absolute value of a number is always positive.
(Source: http://www.chilimath.com/algebra/intermediate/abs/graph-absolute-value-functions.html)
Algebra I
Page 52
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Transformations of Linear Absolute Value Functions
Back to Topic List
When other values are part of the
absolute value equation, the graph
can shift vertically and horizontally
from the parent function.
(Source: http://www.chilimath.com/algebra/intermediate/abs/graph-absolute-value-functions.html)
Algebra I
Page 53
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Solving Absolute Value Equations
Back to Topic List
Consider the equation x = 3. There is only one solution to this equation: 3.
Now compare that equation to the equation |x| = 3
What values will make this equation true?
 Well obviously 3 will work:
 How about -3?
|3| = 3
|-3| = 3
So this equation |x| = 3 has two solutions. x = 3 or x = -3
Consider the equation |x| = -3
What values will make this equation true? Nothing!
When you take the absolute value of a number, the answer is always positive.
There is no number for x that will satisfy the equation |x| = -3
So how do you answer the question: Solve the equation |x| = -3.
Your answer is No Solution.
Here is a special case: Solve the equation |x| = 0.
Your answer is x = 0. There is only one solution.
Algebra I
Page 54
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Solving Multistep Absolute Value Equations
Back to Topic List
Example 1: | x + 3 | = 8
We know from the previous example of |x| = 3 that there are two solutions (3 and -3),
so we are going to set up our work to find both solutions.
Case 1
Case 2
x+3=8
x + 3 = -8
x=5
x = -11
Test your solutions:
|5 + 3| = |8| = 8
|-11 + 3| = |-8| = 8
Steps:
1.
2.
3.
4.
5.
Write the equation twice
Remove the absolute value symbols
Set the second equation equal to the opposite of the right side of the equation
Solve both equations
Check your solutions
Example 2: | x – 4 | = -13
STOP! Don’t even bother to try to solve this one! The absolute value of any number
is always postive. There is no number in this world that will give you a negative
value when you take the absolute value of that number.
Solution: No Solution
Video – Solving Absolute Value Equations
This videos show you step by step how to solve an absolute value equation.
http://www.youtube.com/watch?v=G5TW6Lq2t9c
Algebra I
Page 55
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Solving Multistep Absolute Value Equations
Back to Topic List
Example 3: | x – 6 | – 21 = -13
At first glance this looks like another one of those impossible No Solution kind of
questions. But wait! Check out the minus 21 on the left. Take care of that before you
start to use the absolute value equation steps.
| x – 6 | – 21 = -13
Add 21 to both sides of the equation
|x–6| =8
Now this equation can be solved!
Case 1
Case 2
x–6=8
x – 6 = -8
x = 14
x = -2
Test your solutions:
| 14 – 6 | – 21 = -13
| 8 | – 21 = -13
8 – 21 = -13
-13 = -13
Algebra I
Page 56
| -2 – 6 | – 21 = -13
| -8 | – 21 = -13
8 – 21 = -13
-13 = -13
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Solving Multistep Absolute Value Equations
Back to Topic List
Example 4:
When solving equations with several steps, remember to follow the order of operations
in reverse.
If we already knew the value of x, we would simplify the expression using order of
operations:
1.
2.
3.
4.
Grouping symbols
Powers (there are no exponents in this problem, so we would have skipped this step)
Multiply or Divide
Add or Subtract
Since we don’t know the value for x, we have to work backwards to find x.
5 | x + 7 | + 9 = 39
5 | x + 7 | = 30
|x+7| =6
Subtract 9 from both sides of the equation
Divide both sides of the equation by 5
Now set up the two cases for the absolute value
Case 1
Case 2
x+7=6
x + 7 = -6
x = -1
x = -13
Test your solutions:
5 | -1 + 7 |
5|6|
5(6)
30
Algebra I
+ 9 = 39
+ 9 = 39
+ 9 = 39
+ 9 = 39
39 = 39
Page 57
5 | -13 + 7 |
5 | -6 |
5(6)
30
+ 9 = 39
+ 9 = 39
+ 9 = 39
+ 9 = 39
39 = 39
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Solving Absolute Value Equations Using Graphs
Back to Topic List
Example 1: Solve: |x| = 3



Construct the graph for f(x) = |x|
Construct the graph for f(x) = 3
Determine the points where these two graphs intersect
The solutions for the
equation |x| = 3
are x = -3 and x = 3
(Source: http://www.purplemath.com/modules/solveabs.htm)
Example 2: Solve: |x + 2| = 7



Construct the graph for f(x) = |x + 2|
Construct the graph for f(x) = 7
Determine the points where these two graphs intersect
The solutions for the
equation |x + 2| = 7
are x = -9 and x = 5
(Source: http://www.purplemath.com/modules/solveabs.htm)
Algebra I
Page 58
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Example 3: Solve: |2x – 3| – 4 = 3

Back to Topic List
Simplify the equation to isolate the absolute value
|2x – 3| – 4 = 3
add 4 to both sides of the equation
|2x – 3| = 7



Construct the graph for f(x) = |2x – 3|
Construct the graph for f(x) = 7
Determine the points where these two graphs intersect
The solutions for the
equation |2x – 3| – 4 = 3
are x = -2 and x = 5
(Source: http://www.purplemath.com/modules/solveabs.htm)
Steps to Solving Absolute Value Equations on the Graphing Calculator (TI-83 and TI-84)
http://mathbits.com/MathBits/TISection/Algebra2/absolutevalue.htm
Algebra I
Page 59
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Absolute Value Inequalities
Back to Topic List
Example 1: Solve: |x| ≤ 3



Construct the graph for f(x) = |x|
Construct the graph for f(x) = 3
Determine the points where these two graphs intersect
The solutions for the equation
|x| = 3
are x = -3 and x = 3
But we want to know what x values
will make the inequality true.
Think about the values of x that would make the inequality true.
Try 2 and -2
|2| ≤ 3 is true and so is |-2| ≤ 3
So we want all the x values between -3 and 3.
Solve: |x| ≤ 3
Answer: -3 ≤ x ≤ 3
Video – Solving Absolute Value Inequalities
https://www.youtube.com/watch?feature=player_embedded&v=iI_2Piwn_og
Algebra I
Page 60
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Example 2: Solve: |x + 2| > 7



Back to Topic List
Construct the graph for f(x) = |x + 2|
Construct the graph for f(x) = 7
Determine the points where these two graphs intersect
The solutions for the
equation |x + 2| = 7
are x = -9 and x = 5
But we want to know what
x values will make the
inequality true.
Notice in this example the inequality does not include the equals.
For this reason, the points will be open circles.
The value of 3 was chosen
as a test point because it
was an x value between -9
and 5. Lots of other values
could have been tested too.
Think about the values of x that would make the inequality true.
Try 3:
|3 + 2| > 7
5 is not greater than 7, so 3 is not a solution.
Try -10:
|-10 + 2| > 7
|-8| > 7
8>7
8 is greater than 7, so -10 is a solution.
|9 + 2| > 7
|11| > 7
11 > 7
11 is greater than 7, so 9 is a solution.
Try 9:
Solve: |x + 2| > 7
Answer: x < -9 or x > 5
Algebra I
Page 61
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Example 3: Solve: |2x – 3| – 4 ≥ 3

Back to Topic List
Simplify the inequality to isolate the absolute value
|2x – 3| – 4 ≥ 3
add 4 to both sides of the equation
|2x – 3| ≥ 7



Construct the graph for f(x) = |2x – 3|
Construct the graph for f(x) = 7
Determine the points where these two graphs intersect
The solutions for the
equation |2x – 3| – 4 = 3
are x = -2 and x = 5
But we want to know what x values will
make the inequality true.
Think about the values of x that would make the inequality true.
Try 1:
Try -4
Try 6
|2(1) – 3| – 4 ≥ 3
|2 – 3| – 4 ≥ 3
|-1| – 4 ≥ 3
1–4≥3
-3 ≥ 3
always go back to the original problem
-3 is not greater than 3, so 1 is not a solution.
|2(-4) – 3| – 4 ≥ 3
|-8 – 3| – 4 ≥ 3
|-11| – 4 ≥ 3
11 – 4 ≥ 3
7≥3
7 is greater than 3, so -4 is a solution.
|2(6) – 3| – 4 ≥ 3
|12 – 3| – 4 ≥ 3
|9| – 4 ≥ 3
9–4≥3
5≥3
5 is greater than 3, so 6 is a solution.
Solve: |2x – 3| – 4 ≥ 3
Answer: x ≤ -2 or x ≥ 5
Algebra I
Page 62
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Solving Quadratic Functions Using Graphs
Back to Topic List
Example: Solve: 2x2 + 3x – 1 = 8



Construct the graph for f(x) = 2x2 + 3x - 1
Construct the graph for f(x) = 8
Determine the points where these two graphs intersect
The solutions for the
equation 2x2 + 3x – 1 = 8
are x = -3 and x = 1.5
Later in this course you will learn algebraic techniques to solve equations like this one.
To check the solutions, substitute each answer into the original equation.
x = -3
x = 1.5
2(-3)2 + 3(-3) – 1 = 8
2(1.5)2 + 3(1.5) – 1 = 8
2(9) + (-9) – 1 = 8
2(2.25) + (4.5) – 1 = 8
18 – 9 – 1 = 8
4.5 + 4.5 – 1 = 8
9–1=8
9–1=8
8=8
8=8
Algebra I
Page 63
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Solving Quadratic Inequalities Using Graphs
Back to Topic List
Example: Solve: 2x2 + 3x – 1 < 8



Construct the graph for f(x) = 2x2 + 3x - 1
Construct the graph for f(x) = 8
Determine the points where these two graphs intersect
The solutions for the
equation 2x2 + 3x – 1 = 8
are x = -3 and x = 1.5
But we want to know what x values will
make the inequality true.
Notice in this example the inequality does not include the equals.
For this reason, the points will be open circles.
Think about the values of x that would make the inequality true.
Try -1
2(-1)2 + 3(-1) – 1 < 8
2(1) – 3 – 1 < 8
2–3–1<8
-1 – 1 < 8
-2 < 8
-2 is less than 8, so -1 is a solution.
So we want all the x values between -3 and 1.5.
Solve: 2x2 + 3x – 1 < 8
Answer: -3 < x < 1.5
Algebra I
Page 64
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Solving Exponential Functions Using Graphs
Back to Topic List
Example: Solve: 4x – 7 = 9



Construct the graph for f(x) = 4x – 7
Construct the graph for f(x) = 9
Determine the points where these two graphs intersect
The solution for the
equation 4x – 7 = 9
is x = 2
Later in this course you will learn algebraic techniques to solve equations like this one.
To check the solutions, substitute the answer into the original equation.
4x – 7 = 9
42 – 7 = 9
16 – 7 = 9
9=9
Algebra I
Page 65
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities
Solving Exponential Inequalities Using Graphs
Back to Topic List
Example: Solve: 4x – 7 ≥ 9



Construct the graph for f(x) = 4x – 7
Construct the graph for f(x) = 9
Determine the points where these two graphs intersect
The solution for the
equation 4x – 7 = 9
is x = 2
But we want to know what x values
will make the inequality true.
Think about the values of x that would make the inequality true.
Try 1
Try 3
41 – 7 ≥ 9
4–7≥9
-3 ≥ 9
-3 is not greater than 9, so 1 is not a solution
43 – 7 ≥ 9
64 – 7 ≥ 9
57 ≥ 9
57 is greater than 9, so 3 is a solution
So we want all the x values greater than or equal to 2.
Solve: 4x – 7 ≥ 9
Answer: x ≥ 2
Algebra I
Page 66
Unit 1B. Graphs, Equations, Linear Functions, and Inequalities