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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. The 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.
VIDEO LINK: Khan Academy Solving Linear Systems by Graphing
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
19. Given the graph of the line to the right represents
the first equation is a system.
10
a. Graph the line y = x - 10
b. Find the solution to the system of equations.
Explain how you identified the solution.
-10
c. If the equation for the given line is
1
y = - x + 2, prove you found the
2
correct solution using substitution.
10
-10
10
20. Given the graph of the line to the right.
Which of the graphs below shows the point
of intersection of the given line with y = -x - 2.
a)
-10
-10
b)
10
10
-10
10
-10
10
10
-10
-10
c)
d)
10
10
-10
10
-10
-10
10
-10
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
19. Given the graph of the line to the right represents
the first equation is a system.
a. Graph the line y =
10
1
x-4
2
-10
10
b. Find the solution to the system of equations.
Explain how you identified the solution.
-10
c. If the equation for the given line is
y = 3x + 6, prove you found the
correct solution using substitution.
10
20. Given the graph of the line to the right.
Which of the graphs below shows the point
of intersection of the given line with y = .5x + 11.
a)
-10
10
-10
b)
10
10
-10
-10
10
10
-10
-10
c)
d)
10
-10
10
10
-10
-10
10
-10
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
Examples B
Given the equations
1
y= x–1
2
Given the equations
3
y= x+4
5
5
y=- x-2
3
y = -2x + 4
The slopes are opposite
in sign. The slope of
m=
The slopes are opposite
in sign. The slope of
1
2
m=
has a reciprocal of
3
5
has a reciprocal of
2
m=
=2
1
m=
Since the slopes are
opposite reciprocals
the lines are
.
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 =
Example B (continued)
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.
VIDEO LINK: Khan Academy Equations of Parallel and Perpendicular Lines
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)
10. a perpendicular line through (3, 0)
11. a  line through (1, 2)
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.
VIDEO LINK: Khan Academy Solving Linear Systems by Substitution
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
34. Given the system of linear equations
y = 4x + 5
3x – 5y = 9
a. Solve the system of equations using substitution. Write your answer as an
ordered pair. Show your work algebraically and explain how you got the
solution.
b. Verify that your solution is correct. Show your work algebraically.
35. Given the system of linear equations
y =x-2
2y – 3x = -1
a. Solve the system of equations using substitution. Write your answer as an
ordered pair. Show your work algebraically and explain how you got the
solution.
b. Verify that your solution is correct. Show your work algebraically.
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
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
1
32. Find the equation of the line perpendicular to y = - x + 4 passing through (-1, 4)
3
33. Given the system of linear equations
2y - 4x = 14
-2y + 5x = -12
a. Solve the system of equations using substitution. Write your answer as an
ordered pair. Show your work and explain how you got the solution.
b. Verify that your solution is correct. Show your work algebraically.
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
VIDEO LINK: Khan Academy Solving Systems of Equations by Elimination
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%
34%
8%
20%
Winter Sports
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?
16. Given the system of linear equations
2y - 5x = 13
2y + 7x = 1
a. Solve the system of equations using elimination. Write your answer as an
ordered pair. Show your work and explain how you got the solution.
b. Verify that your solution is correct. Show your work algebraically.
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.
x + 9y = 38
2x – 5y = 7
4x – 7y = -6
-2x + 4y = 3
5. 2x + 3y = 20
8x – 6y = 80
6.
7.
8.
9.
9x + 2y = -4
-4x + 3y = 29
x - 2y = 11
3x - 6y = 2
11. -3x - 2y = -31
3x - y = 25
4x – 2y = -4
8x – 4y = -8
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
x
16. a + 0 = a
18. 1
.
19. -w + w = 0
20. b
.
21. (b + c) + t = b + (c + t)
22. x(y – z) = xy – xz
17. x
.
.
y = y
.
x
z=z
1
=1
b
23. Jake’s algebraic work and solution for the system of equations is shown below.
3x + 4y = -1
y + 2x = 6
Step 1
Step 2
Step 3
Step 4
Step 5
Step 6
Step 7
Step 8
y + 2x = 6
3x + 4y = -1
3x + 8x + 6 = -1
10x + 6 = -1
10x = 5
x = .5
y = 2(.5) + 6
(.5, 7)
y = 2x + 6
3x + 4(2x + 6) = -1
y=7
a. Explain what Jake did wrong in at least three of the steps.
b. Check Jake’s solution algebraically and explain what the result tells us about
Jake’s solution.
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.
VIDEO LINK: Khan Academy Graphing Linear Inequalities in Two Variables
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.
-3y > 2x – 6
-3
-3 -3
2
y<- x+2
3
2y + 4 < 6x + 4
–4
–4
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)
(-1, 0)
(6, 0)
(0, -3)
Step 2. Picking a point from
each region. We
couldn’t pick (0,0)
since it is on one of
the lines.
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.
VIDEO LINK: Khan Academy Graphing Systems of Inequalities in 2 Variables
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
21. Mark’s algebraic work and solution for the system of equations is shown below.
5x + 2y = -2
-2x + 3y = 14
Step 1
Step 2
Step 3
Step 4
Step 5
Step 6
Step 7
Step 8
5x + 2y = -2
3(5x + 2y = -2)
-2x + 3y = 14
-2(-2x + 3y = 14)
15x + 6y = -6
-4x - 6y = -28
12x = -36
x = -3
-2x + 3y = 14
-2x + 3(-3) = 14
-2x - 6 = 14
-2x = 20
x = -10
(-3, -10)
15x + 6y = -6
-4x - 6y = -28
a. Explain what Mark did wrong in at least three of the steps.
b. Check Mark’s solution algebraically and explain what the result tells us about
Mark’s solution.
Translate each sentence into an inequality.
22. Twice a number ‘x’ increased by six is at least ninety.
23. Half a number ‘a’ decreased by seven is at most thirteen.
24. Jane has sold more than four times as many raffle tickets as Sally.
25.
Alex’s high score in the video game “Angry Kangaroos” is less than one-third
of martin’s score.
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
So there are 42 nickels and 29 dimes.
The number of dimes
Now we find the number of nickels.
the number of nickels
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?