Ch 5 book - systems of linear equations

5 Solving Systems of
Linear Equations
5.1 Solving Systems of Linear Equations by Graphing
5.2 Solving Systems of Linear Equations by Substitution
5.3 Solving Systems of Linear Equations by Elimination
5.4 Solving Special Systems of Linear Equations
5.5 Solving Equations by Graphing
5.6 Graphing Linear Inequalities in Two Variables
5.7 Systems of Linear Inequalities
Fishing (p. 261)
Pets (p. 248)
Drama Club (p. 226)
Delivery Vans (p. 232)
Roofing Contractor (p. 220)
PPetts (((p. 2424 )8)8)
DDelilivery VVans ((p. 23232)2)
Drama Club (p 226)
Roofing Contractor (p. 220)
y q
FiFi hshiing ((p. 26261)1)
SEE the Big Idea
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215
Maintaining Mathematical ProficiencyMaintaining Mathematical Proficiency
Graphing Linear Functions
Example 1 Graph 3 + y =
1
—
2
x.
Step 1 Rewrite the equation in slope-intercept form.
y =
1
—
2
x − 3
Step 2 Find the slope and the y-intercept.
m =
1
—
2
and b = −3
Step 3 The y-intercept is −3. So, plot (0, −3).
Step 4 Use the slope to find another point on the line.
slope =
rise
—
run
=
1
—
2
Plot the point that is 2 units right and 1 unit up from (0, −3). Draw a line
through the two points.
Graph the equation.
1. y + 4 = x 2. 6x − y = −1 3. 4x + 5y = 20 4. −2y + 12 = −3x
Solving and Graphing Linear Inequalities
Example 2 Solve 2x − 17 ≤ 8x − 5. Graph the solution.
2x − 17 ≤ 8x − 5 Write the inequality.
+ 5 + 5 Add 5 to each side.
2x − 12 ≤ 8x Simplify.
− 2x − 2x Subtract 2x from each side.
−12 ≤ 6x Simplify.
−12
—
6
≤
6x
—
6
Divide each side by 6.
−2 ≤ x Simplify.
The solution is x ≥ −2.
0−5 −4 −3 −2 −1 1 32
x ≥ –2
Solve the inequality. Graph the solution.
5. m + 4 > 9 6. 24 ≤ −6t 7. 2a − 5 ≤ 13
8. −5z + 1 < −14 9. 4k − 16 < k + 2 10. 7w + 12 ≥ 2w − 3
11. ABSTRACT REASONING The graphs of the linear functions g and h have different slopes. The
value of both functions at x = a is b. When g and h are graphed in the same coordinate plane,
what happens at the point (a, b)?
x
y
2
−1
−4
42−2−4
1(0, −3)
2
Dynamic Solutions available at BigIdeasMath.com

216 Chapter 5 Solving Systems of Linear Equations
Using a Graphing Calculator
MathematicalMathematical
PracticesPractices
Monitoring ProgressMonitoring Progress
Use a graphing calculator to find the point of intersection of the graphs of the
two linear equations.
1. y = −2x − 3 2. y = −x + 1 3. 3x − 2y = 2
y = 1
—2
x − 3 y = x − 2 2x − y = 2
Mathematically proficient students use technological tools to explore concepts.
CoreCore ConceptConcept
Finding the Point of Intersection
You can use a graphing calculator to find the point of intersection, if it exists, of
the graphs of two linear equations.
1. Enter the equations into a graphing calculator.
2. Graph the equations in an appropriate viewing window, so that the point of
intersection is visible.
3. Use the intersect feature of the graphing calculator to find the point of
intersection.
Use a graphing calculator to find the point of intersection, if it exists, of the graphs of
the two linear equations.
y = −
1
—2x + 2 Equation 1
y = 3x − 5 Equation 2
SOLUTION
The slopes of the lines are not the same, so
you know that the lines intersect. Enter the
equations into a graphing calculator. Then
graph the equations in an appropriate
viewing window.
Use the intersect feature to find the point
of intersection of the lines.
The point of intersection is (2, 1).
−6
−4
4
66
y = − x + 2
1
2
y = 3x − 5
−6
−4
4
6
Intersection
X=2 Y=1

Section 5.1 Solving Systems of Linear Equations by Graphing 217
Writing a System of Linear Equations
Work with a partner. Your family opens a bed-and-breakfast. They spend $600
preparing a bedroom to rent. The cost to your family for food and utilities is
$15 per night. They charge $75 per night to rent the bedroom.
a. Write an equation that represents the costs.
Cost, C
(in dollars)
=
$15 per
night ⋅
Number of
nights, x
+ $600
b. Write an equation that represents the revenue (income).
Revenue, R
(in dollars)
=
$75 per
night ⋅
Number of
nights, x
c. A set of two (or more) linear equations is called a system of linear equations.
Write the system of linear equations for this problem.
Essential QuestionEssential Question How can you solve a system of linear
equations?
Using a Table or Graph to Solve a System
Work with a partner. Use the cost and revenue equations from Exploration 1 to
determine how many nights your family needs to rent the bedroom before recovering
the cost of preparing the bedroom. This is the break-even point.
a. Copy and complete the table.
b. How many nights does your family need to rent the bedroom before breaking even?
c. In the same coordinate plane, graph the cost equation and the revenue equation
from Exploration 1.
d. Find the point of intersection of the two graphs. What does this point represent?
How does this compare to the break-even point in part (b)? Explain.
Communicate Your AnswerCommunicate Your Answer
3. How can you solve a system of linear equations? How can you check your
solution?
4. Solve each system by using a table or sketching a graph. Explain why you chose
each method. Use a graphing calculator to check each solution.
a. y = −4.3x − 1.3 b. y = x c. y = −x − 1
y = 1.7x + 4.7 y = −3x + 8 y = 3x + 5
x (nights) 0 1 2 3 4 5 6 7 8 9 10 11
C (dollars)
R (dollars)
MODELING WITH
MATHEMATICS
To be proficient in math,
you need to identify
important quantities in
real-life problems and
map their relationships
using tools such as
diagrams, tables,
and graphs.
5.1 Solving Systems of Linear
Equations by Graphing
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5.1 Lesson What You Will LearnWhat You Will Learn
Check solutions of systems of linear equations.
Solve systems of linear equations by graphing.
Use systems of linear equations to solve real-life problems.
Systems of Linear Equationssystem of linear equations,
p. 218
solution of a system of linear
equations, p. 218
Previous
linear equation
ordered pair
Core VocabularyCore Vocabullarry
Checking Solutions
Tell whether the ordered pair is a solution of the system of linear equations.
a. (2, 5); x + y = 7 Equation 1
2x − 3y = −11 Equation 2
b. (−2, 0); y = −2x − 4 Equation 1
y = x + 4 Equation 2
SOLUTION
a. Substitute 2 for x and 5 for y in each equation.
Equation 1 Equation 2
x + y = 7 2x − 3y = −11
2 + 5 =
?
7 2(2) − 3(5) =
?
−11
7 = 7 ✓ −11 = −11 ✓
Because the ordered pair (2, 5) is a solution of each equation, it is a solution of
the linear system.
b. Substitute −2 for x and 0 for y in each equation.
y = −2x − 4 y = x + 4
0 =
?
−2(−2) − 4 0 =
?
−2 + 4
0 = 0 ✓ 0 ≠ 2 ✗
The ordered pair (−2, 0) is a solution of the first equation, but it is not a solution
of the second equation. So, (−2, 0) is not a solution of the linear system.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
Tell whether the ordered pair is a solution of the system of linear equations.
1. (1, −2);
2x + y = 0
−x + 2y = 5
2. (1, 4);
y = 3x + 1
y = −x + 5
READING
A system of linear
equations is also called
a linear system.
A system of linear equations is a set of two or more linear equations in the same
variables. An example is shown below.
x + y = 7 Equation 1
A solution of a system of linear equations in two variables is an ordered pair that is a
solution of each equation in the system.

Solving Systems of Linear Equations by Graphing
The solution of a system of linear equations is the point of intersection of the graphs of
the equations.
Solving a System of Linear Equations by Graphing
Step 1 Graph each equation in the same coordinate plane.
Step 2 Estimate the point of intersection.
Step 3 Check the point from Step 2 by substituting for x and y in each equation
of the original system.
Solving a System of Linear Equations by Graphing
Solve the system of linear equations by graphing.
y = −2x + 5 Equation 1
SOLUTION
Step 1 Graph each equation.
The graphs appear to intersect at (1, 3).
Step 3 Check your point from Step 2.
y = −2x + 5 y = 4x − 1
3 =
?
−2(1) + 5 3 =
?
4(1) − 1
3 = 3 ✓ 3 = 3 ✓
The solution is (1, 3).
3. y = x − 2 4. y = 1
—2
x + 3 5. 2x + y = 5
y = −x + 4 y = −
3
—2x − 5 3x − 2y = 4
REMEMBER
Note that the linear
equations are in
slope-intercept form. You
can use the method
presented in Section 3.5
to graph the equations.
x
y
2
42−2−4
(1, 3)
y = −2x + 5
1 3)
y = 4x − 1
−1
Check
Use the table or intersect feature of a graphing calculator to check your answer.
−6
−2
6
6
y = 4x − 1y = −2x + 5
Intersection
X=1 Y=3
X Y1
X=1
9
7
5
3
1
-1
-3
Y2
-9
-5
-1
3
7
11
15
-1
-2
0
1
2
3
4
When x = 1, the corresponding
y-values are equal.

Modeling with Mathematics
A roofing contractor buys 30 bundles of shingles and 4 rolls of roofing paper for
$1040. In a second purchase (at the same prices), the contractor buys 8 bundles of
shingles for $256. Find the price per bundle of shingles and the price per roll of
roofing paper.
SOLUTION
1. Understand the Problem You know the total price of each purchase and how
many of each item were purchased. You are asked to find the price of each item.
2. Make a Plan Use a verbal model to write a system of linear equations that
represents the problem. Then solve the system of linear equations.
3. Solve the Problem
Words 30 ⋅ Price per
bundle
+ 4 ⋅ Price
per roll
= 1040
8 ⋅
Price per
bundle
+ 0 ⋅ Price
per roll
= 256
Variables Let x be the price (in dollars) per bundle and let y be the
price (in dollars) per roll.
System 30x + 4y = 1040 Equation 1
8x = 256 Equation 2
Step 1 Graph each equation. Note that only
the first quadrant is shown because
x and y must be positive.
Step 2 Estimate the point of intersection. The
graphs appear to intersect at (32, 20).
30x + 4y = 1040 8x = 256
30(32) + 4(20) =
?
1040 8(32) =
?
256
1040 = 1040 ✓ 256 = 256 ✓
The solution is (32, 20). So, the price per bundle of shingles is $32, and the
price per roll of roofing paper is $20.
4. Look Back You can use estimation to check that your solution is reasonable.
A bundle of shingles costs about $30. So, 30 bundles of shingles and 4 rolls of
roofing paper (at $20 per roll) cost about 30(30) + 4(20) = $980, and 8 bundles
of shingles costs about 8(30) = $240. These prices are close to the given values,
so the solution seems reasonable.
6. You have a total of 18 math and science exercises for homework. You have
six more math exercises than science exercises. How many exercises do you
have in each subject?
Solving Real-Life Problems
8 16 240 32 x
80
160
240
320
0
y
(32, 20)
y = −7.5x + 260
x = 32

Exercises5.1 Dynamic Solutions available at BigIdeasMath.com
Monitoring Progress and Modeling with MathematicsMonitoring Progress and Modeling with Mathematics
In Exercises 3–8, tell whether the ordered pair is
a solution of the system of linear equations.
(See Example 1.)
3. (2, 6);
x + y = 8
3x − y = 0
4. (8, 2);
x − y = 6
2x − 10y = 4
5. (−1, 3);
y = −7x − 4
y = 8x + 5
6. (−4, −2);
y = 2x + 6
y = −3x − 14
7. (−2, 1);
6x + 5y = −7
2x − 4y = −8
8. (5, −6);
6x + 3y = 12
4x + y = 14
In Exercises 9–12, use the graph to solve the system of
linear equations. Check your solution.
9. x − y = 4 10. x + y = 5
4x + y = 1 y − 2x = −4
x
y
−2
42
x
y
2
4
41
11. 6y + 3x = 18 12. 2x − y = −2
−x + 4y = 24 2x + 4y = 8
x
y
2
4
−2−4−6
x
y
4
2−2
In Exercises 13–20, solve the system of linear equations
by graphing. (See Example 2.)
13. y = −x + 7 14. y = −x + 4
y = x + 1 y = 2x − 8
15. y = 1
—3
x + 2 16. y = 3
—4
x − 4
y = 2
—3
x + 5 y = −
1
—2x + 11
17. 9x + 3y = −3 18. 4x − 4y = 20
2x − y = −4 y = −5
19. x − 4y = −4 20. 3y + 4x = 3
−3x − 4y = 12 x + 3y = −6
ERROR ANALYSIS In Exercises 21 and 22, describe
and correct the error in solving the system of linear
equations.
21.
The solution of
the linear system
x − 3y = 6 and
2x − 3y = 3
is (3, −1).
✗
x
y
−1
2
2
22.
The solution of
the linear system
y = 2x − 1 and
y = x + 1
is x = 2.
✗
x
y
2
4
42
1. VOCABULARY Do the equations 5y − 2x = 18 and 6x = −4y − 10 form a system of linear
equations? Explain.
2. DIFFERENT WORDS, SAME QUESTION Consider the system of linear equations −4x + 2y = 4
and 4x − y = −6. Which is different? Find “both” answers.
Solve the system of linear equations. Solve each equation for y.
Find the point of intersection
of the graphs of the equations.
Find an ordered pair that is a solution
of each equation in the system.
Vocabulary and Core Concept CheckVocabulary and Core Concept Check

Solve the literal equation for y. (Section 1.5)
34. 10x + 5y = 5x + 20 35. 9x + 18 = 6y − 3x 36. 3
—4
x + 1
—4
y = 5
Reviewing what you learned in previous grades and lessons
USING TOOLS In Exercises 23–26, use a graphing
calculator to solve the system of linear equations.
23. 0.2x + 0.4y = 4 24. −1.6x − 3.2y = −24
−0.6x + 0.6y = −3 2.6x + 2.6y = 26
25. −7x + 6y = 0 26. 4x − y = 1.5
0.5x + y = 2 2x + y = 1.5
27. MODELING WITH MATHEMATICS You have
40 minutes to exercise at the gym, and you want to
burn 300 calories total using both machines. How
much time should you spend on each machine?
(See Example 3.)
Elliptical Trainer
8 calories
per minute
Stationary Bike
6 calories
per minute
28. MODELING WITH MATHEMATICS
You sell small and large candles
at a craft fair. You collect $144
selling a total of 28 candles.
How many of each type of candle
did you sell?
29. MATHEMATICAL CONNECTIONS Write a linear
equation that represents the area and a linear equation
that represents the perimeter of the rectangle. Solve
the system of linear equations by graphing. Interpret
your solution.
6 cm
(3x − 3) cm
30. THOUGHT PROVOKING Your friend’s bank account
balance (in dollars) is represented by the equation
y = 25x + 250, where x is the number of months.
Graph this equation. After 6 months, you want to
have the same account balance as your friend. Write a
linear equation that represents your account balance.
Interpret the slope and y-intercept of the line that
represents your account balance.
31. COMPARING METHODS Consider the equation
x + 2 = 3x − 4.
a. Solve the equation using algebra.
b. Solve the system of linear equations y = x + 2
and y = 3x − 4 by graphing.
c. How is the linear system and the solution in part
(b) related to the original equation and the solution
in part (a)?
32. HOW DO YOU SEE IT? A teacher is purchasing
binders for students. The graph shows the total costs
of ordering x binders from three different companies.
150
0
50
75
100
125
150
20 25 30 35
Cost(dollars)
40 45 50 x
y
Number of binders
Buying Binders
Company A
Company B
Company C
a. For what numbers of binders are the costs the
same at two different companies? Explain.
b. How do your answers in part (a) relate to systems
of linear equations?
33. MAKING AN ARGUMENT You and a friend are going
hiking but start at different locations. You start at the
trailhead and walk 5 miles per hour. Your friend starts
3 miles from the trailhead and walks 3 miles per hour.
you
your friend
a. Write and graph a system of linear equations that
represents this situation.
b. Your friend says that after an hour of hiking you
will both be at the same location on the trail. Is
your friend correct? Use the graph from part (a) to
explain your answer.
$6
each
$4
each

Section 5.2 Solving Systems of Linear Equations by Substitution 223
Using Substitution to Solve Systems
Work with a partner. Solve each system of linear equations using two methods.
Method 1 Solve for x first.
Solve for x in one of the equations. Substitute the expression for x into the other
equation to find y. Then substitute the value of y into one of the original equations
to find x.
Method 2 Solve for y first.
Solve for y in one of the equations. Substitute the expression for y into the other
equation to find x. Then substitute the value of x into one of the original equations
to find y.
Is the solution the same using both methods? Explain which method you would
prefer to use for each system.
a. x + y = −7 b. x − 6y = −11 c. 4x + y = −1
−5x + y = 5 3x + 2y = 7 3x − 5y = −18
Essential QuestionEssential Question How can you use substitution to solve a
system of linear equations?
Writing and Solving a System of Equations
Work with a partner.
a. Write a random ordered pair with integer
coordinates. One way to do this is to use
a graphing calculator. The ordered pair
generated at the right is (−2, −3).
b. Write a system of linear equations that has
your ordered pair as its solution.
c. Exchange systems with your partner and use
one of the methods from Exploration 1 to
solve the system. Explain your choice
of method.
3. How can you use substitution to solve a system of linear equations?
4. Use one of the methods from Exploration 1 to solve each system of linear
equations. Explain your choice of method. Check your solutions.
a. x + 2y = −7 b. x − 2y = −6 c. −3x + 2y = −10
2x − y = −9 2x + y = −2 −2x + y = −6
d. 3x + 2y = 13 e. 3x − 2y = 9 f. 3x − y = −6
x − 3y = −3 −x − 3y = 8 4x + 5y = 11
ATTENDING TO
PRECISION
you need to communicate
precisely with others.
randInt(-5‚5‚2)
{-2 -3}
Choose two
random integers
between −5 and 5.
Equations by Substitution

Solve systems of linear equations by substitution.
Solving Linear Systems by Substitution
Previous
system of linear equations
solution of a system of
linear equations
Solving a System of Linear Equations by Substitution
Step 1 Solve one of the equations for one of the variables.
Step 2 Substitute the expression from Step 1 into the other equation and
solve for the other variable.
Step 3 Substitute the value from Step 2 into one of the original equations
and solve.
Solving a System of Linear Equations
by Substitution
Solve the system of linear equations by substitution.
y = −2x − 9 Equation 1
SOLUTION
Step 1 Equation 1 is already solved for y.
Step 2 Substitute −2x − 9 for y in Equation 2 and solve for x.
6x − 5(−2x − 9) = −19 Substitute −2x − 9 for y.
6x + 10x + 45 = −19 Distributive Property
16x + 45 = −19 Combine like terms.
16x = −64 Subtract 45 from each side.
x = −4 Divide each side by 16.
Step 3 Substitute −4 for x in Equation 1 and solve for y.
= −2(−4) − 9 Substitute −4 for x.
= 8 − 9 Multiply.
= −1 Subtract.
The solution is (−4, −1).
Solve the system of linear equations by substitution. Check your solution.
1. y = 3x + 14 2. 3x + 2y = 0 3. x = 6y − 7
y = −4x y = 1
—2
x − 1 4x + y = −3
Another way to solve a system of linear equations is to use substitution.
Check
Equation 1
y = −2x − 9
−1 =
?
−2(−4) − 9
−1 = −1 ✓
Equation 2
6x − 5y = −19
6(−4) − 5(−1) =
?
−19
−19 = −19 ✓

by Substitution
Solve the system of linear equations by substitution.
−x + y = 3 Equation 1
3x + y = −1 Equation 2
SOLUTION
Step 1 Solve for y in Equation 1.
y = x + 3 Revised Equation 1
Step 2 Substitute x + 3 for y in Equation 2 and solve for x.
3x + y = −1 Equation 2
3x + (x + 3) = −1 Substitute x + 3 for y.
4x + 3 = −1 Combine like terms.
4x = −4 Subtract 3 from each side.
Step 3 Substitute −1 for x in Equation 1 and solve for y.
−x + y = 3 Equation 1
−(−1) + y = 3 Substitute −1 for x.
y = 2 Subtract 1 from each side.
The solution is (−1, 2).
Algebraic Check
Equation 1
−x + y = 3
−(−1) + 2 =
?
3
3 = 3 ✓
Equation 2
3x + y = −1
3(−1) + 2 =
?
−1
−1 = −1 ✓
Graphical Check
−5
−2
4
4
y = x + 3
4
y = −3x − 1
Intersection
X=-1 Y=2
4. x + y = −2 5. −x + y = −4
−3x + y = 6 4x − y = 10
6. 2x − y = −5 7. x − 2y = 7
3x − y = 1 3x − 2y = 3
ANOTHER WAY
You could also begin by
solving for x in Equation 1,
solving for y in Equation 2,
or solving for x in
Equation 2.

A drama club earns $1040 from a production. A total of 64 adult tickets and
132 student tickets are sold. An adult ticket costs twice as much as a student ticket.
Write a system of linear equations that represents this situation. What is the price
of each type of ticket?
SOLUTION
1. Understand the Problem You know the amount earned, the total numbers of adult
and student tickets sold, and the relationship between the price of an adult ticket
and the price of a student ticket. You are asked to write a system of linear equations
that represents the situation and find the price of each type of ticket.
Words 64 ⋅ Adult ticket
price
+ 132 ⋅ Student
ticket price
= 1040
Adult ticket
price
= 2 ⋅ Student
ticket price
Variables Let x be the price (in dollars) of an adult ticket and let y be the
price (in dollars) of a student ticket.
x = 2y Equation 2
Step 1 Equation 2 is already solved for x.
Step 2 Substitute 2y for x in Equation 1 and solve for y.
64x + 132y = 1040 Equation 1
64(2y) + 132y = 1040 Substitute 2y for x.
260y = 1040 Simplify.
y = 4 Simplify.
Step 3 Substitute 4 for y in Equation 2 and solve for x.
x = 2y Equation 2
x = 2(4) Substitute 4 for y.
x = 8 Simplify.
The solution is (8, 4). So, an adult ticket costs $8 and a student ticket costs $4.
4. Look Back To check that your solution is correct, substitute the values of x and
y into both of the original equations and simplify.
64(8) + 132(4) = 1040 8 = 2(4)
1040 = 1040 ✓ 8 = 8 ✓
8. There are a total of 64 students in a drama club and a yearbook club. The drama
club has 10 more students than the yearbook club. Write a system of linear
equations that represents this situation. How many students are in each club?
STUDY TIP
You can use either of
the original equations
to solve for x. However,
using Equation 2 requires
fewer calculations.
A
W
o
S
1
2
3

In Exercises 3−8, tell which equation you would choose
to solve for one of the variables. Explain.
3. x + 4y = 30 4. 3x − y = 0
x − 2y = 0 2x + y = −10
5. 5x + 3y = 11 6. 3x − 2y = 19
5x − y = 5 x + y = 8
7. x − y = −3 8. 3x + 5y = 25
4x + 3y = −5 x − 2y = −6
In Exercises 9–16, solve the sytem of linear equations
by substitution. Check your solution. (See Examples 1
and 2.)
9. x = 17 − 4y 10. 6x − 9 = y
y = x − 2 y = −3x
11. x = 16 − 4y 12. −5x + 3y = 51
3x + 4y = 8 y = 10x − 8
13. 2x = 12 14. 2x − y = 23
x − 5y = −29 x − 9 = −1
15. 5x + 2y = 9 16. 11x − 7y = −14
x + y = −3 x − 2y = −4
17. ERROR ANALYSIS Describe and correct the error in
solving for one of the variables in the linear system
8x + 2y = −12 and 5x − y = 4.
Step 1 5x − y = 4
−y = −5x + 4
y = 5x − 4
Step 2 5x − (5x − 4) = 4
5x − 5x + 4 = 4
4 = 4
✗
4x + 2y = 6 and 3x + y = 9.
Step 1 3x + y = 9
y = 9 − 3x
Step 2 4x + 2(9 − 3x) = 6
4x + 18 − 6x = 6
−2x = −12
x = 6
Step 3 3x + y = 9
3x + 6 = 9
3x = 3
x = 1
✗
19. MODELING WITH MATHEMATICS A farmer plants
corn and wheat on a 180-acre farm. The farmer wants
to plant three times as many acres of corn as wheat.
Write a system of linear equations that represents this
situation. How many acres of each crop should the
farmer plant? (See Example 3.)
20. MODELING WITH MATHEMATICS A company that
offers tubing trips down a river rents tubes for a
person to use and “cooler” tubes to carry food and
water. A group spends $270 to rent a total of 15 tubes.
Write a system of linear equations that represents this
situation. How many of each type of tube does the
group rent?
1. WRITING Describe how to solve a system of linear equations by substitution.
2. NUMBER SENSE When solving a system of linear equations by substitution, how do you decide
which variable to solve for in Step 1?

Find the sum or difference. (Skills Review Handbook)
36. (x − 4) + (2x − 7) 37. (5y − 12) + (−5y − 1)
38. (t − 8) − (t + 15) 39. (6d + 2) − (3d − 3)
40. 4(m + 2) + 3(6m − 4) 41. 2(5v + 6) − 6(−9v + 2)
In Exercises 21–24, write a system of linear equations
that has the ordered pair as its solution.
21. (3, 5) 22. (−2, 8)
23. (−4, −12) 24. (15, −25)
25. PROBLEM SOLVING A math test is worth 100 points
and has 38 problems. Each problem is worth either
5 points or 2 points. How many problems of each
point value are on the test?
26. PROBLEM SOLVING An investor owns shares of
Stock A and Stock B. The investor owns a total of
200 shares with a total value of $4000. How many
shares of each stock does the investor own?
Stock Price
A $9.50
B $27.00
MATHEMATICAL CONNECTIONS In Exercises 27 and 28,
(a) write an equation that represents the sum of the
angle measures of the triangle and (b) use your equation
and the equation shown to find the values of x and y.
27.
x + 2 = 3y
x°
y°
28.
x°
y°(y − 18)°
3x − 5y = −22
29. REASONING Find the values of a and b so that the
solution of the linear system is (−9, 1).
ax + by = −31 Equation 1
ax − by = −41 Equation 2
30. MAKING AN ARGUMENT Your friend says that given
a linear system with an equation of a horizontal line
and an equation of a vertical line, you cannot solve
the system by substitution. Is your friend correct?
Explain.
31. OPEN-ENDED Write a system of linear equations in
which (3, −5) is a solution of Equation 1 but not a
solution of Equation 2, and (−1, 7) is a solution of
the system.
32. HOW DO YOU SEE IT? The graphs of two linear
equations are shown.
2 4 6 x
2
4
6
y
y = x + 1
y = 6 − x
1
4
a. At what point do the lines appear to intersect?
b. Could you solve a system of linear equations by
substitution to check your answer in part (a)?
Explain.
33. REPEATED REASONING A radio station plays a total
of 272 pop, rock, and hip-hop songs during a day. The
number of pop songs is 3 times the number of rock
songs. The number of hip-hop songs is 32 more than
the number of rock songs. How many of each type of
song does the radio station play?
34. THOUGHT PROVOKING You have $2.65 in coins.
Write a system of equations that represents this
situation. Use variables to represent the number of
each type of coin.
35. NUMBER SENSE The sum of the digits of a
two-digit number is 11. When the digits are reversed,
the number increases by 27. Find the original number.

Section 5.3 Solving Systems of Linear Equations by Elimination 229
Equations by Elimination
Writing and Solving a System of Equations
Work with a partner. You purchase a drink and a sandwich for $4.50. Your friend
purchases a drink and five sandwiches for $16.50. You want to determine the price of
a drink and the price of a sandwich.
a. Let x represent the price (in dollars) of one drink. Let y represent the price
(in dollars) of one sandwich. Write a system of equations for the situation. Use
the following verbal model.
Number
of drinks ⋅ Price
per drink
+
Number of
sandwiches ⋅ Price per
sandwich
=
Total
price
Label one of the equations Equation 1 and the other equation Equation 2.
b. Subtract Equation 1 from Equation 2. Explain how you can use the result to solve
the system of equations. Then find and interpret the solution.
Essential QuestionEssential Question How can you use elimination to solve a system
of linear equations?
Using Elimination to Solve a System
2x + y = 7 Equation 1
x + 5y = 17 Equation 2
a. Can you eliminate a variable by adding or subtracting the equations as they are?
If not, what do you need to do to one or both equations so that you can?
b. Solve the system individually. Then exchange solutions with your partner and
compare and check the solutions.
4. How can you use elimination to solve a system of linear equations?
5. When can you add or subtract the equations in a system to solve the system?
When do you have to multiply first? Justify your answers with examples.
6. In Exploration 3, why can you multiply an equation in the system by a constant
and not change the solution of the system? Explain your reasoning.
Using Elimination to Solve Systems
Work with a partner. Solve each system of linear equations using two methods.
Method 1 Subtract. Subtract Equation 2 from Equation 1. Then use the result to
solve the system.
Method 2 Add. Add the two equations. Then use the result to solve the system.
Is the solution the same using both methods? Which method do you prefer?
a. 3x − y = 6 b. 2x + y = 6 c. x − 2y = −7
3x + y = 0 2x − y = 2 x + 2y = 5
CHANGING COURSE
you need to monitor and
evaluate your progress
and change course using
a different solution
method, if necessary.

Solve systems of linear equations by elimination.
Solving Linear Systems by Elimination
Previous
coefficient
Solving a System of Linear Equations by Elimination
Step 1 Multiply, if necessary, one or both equations by a constant so at least one
pair of like terms has the same or opposite coefficients.
Step 2 Add or subtract the equations to eliminate one of the variables.
Step 3 Solve the resulting equation.
Step 4 Substitute the value from Step 3 into one of the original equations and
solve for the other variable.
by Elimination
Solve the system of linear equations by elimination.
3x + 2y = 4 Equation 1
SOLUTION
Step 1 Because the coefficients of the y-terms are opposites, you do not need to
multiply either equation by a constant.
Step 2 Add the equations.
6x = 0 Add the equations.
Step 3 Solve for x.
6x = 0 Resulting equation from Step 2
x = 0 Divide each side by 6.
Step 4 Substitute 0 for x in one of the original equations and solve for y.
3(0) + 2y = 4 Substitute 0 for x.
y = 2 Solve for y.
The solution is (0, 2).
Check
Equation 1
3x + 2y = 4
3(0) + 2(2) =
?
4
4 = 4 ✓
Equation 2
3x − 2y = −4
3(0) − 2(2) =
?
−4
−4 = −4 ✓
You can use elimination to solve a system of equations because replacing one
equation in the system with the sum of that equation and a multiple of the other
produces a system that has the same solution. Here is why.
Consider System 1. In this system, a and c are algebraic expressions, and b and d are
constants. Begin by multiplying each side of Equation 2 by a constant k. By the
Multiplication Property of Equality, kc = kd. You can rewrite Equation 1 as
Equation 3 by adding kc on the left and kd on the right. You can rewrite Equation 3 as
Equation 1 by subtracting kc on the left and kd on the right. Because you can rewrite
either system as the other, System 1 and System 2 have the same solution.
System 1
a = b Equation 1
c = d Equation 2
System 2
a + kc = b + kd Equation 3
c = d Equation 2

by Elimination
Solve the system of linear equations by elimination.
−10x + 3y = 1 Equation 1
−5x − 6y = 23 Equation 2
SOLUTION
Step 1 Multiply Equation 2 by −2 so that the coefficients of the x-terms
are opposites.
−10x + 3y = 1 −10x + 3y = 1 Equation 1
−5x − 6y = 23 Multiply by −2. 10x + 12y = −46 Revised Equation 2
10x + 12y = −46 Revised Equation 2
15y = −45 Add the equations.
Step 3 Solve for y.
15y = −45 Resulting equation from Step 2
y = −3 Divide each side by 15.
Step 4 Substitute −3 for y in one of the original equations and solve for x.
−5x − 6y = 23 Equation 2
−5x − 6(−3) = 23 Substitute −3 for y.
−5x + 18 = 23 Multiply.
−5x = 5 Subtract 18 from each side.
x = −1 Divide each side by −5.
The solution is (−1, −3).
Monitoring Progress Help in English and Spanish at BigIdeasMath.com
Solve the system of linear equations by elimination. Check your solution.
1. 3x + 2y = 7 2. x − 3y = 24 3. x + 4y = 22
−3x + 4y = 5 3x + y = 12 4x + y = 13
ANOTHER WAY
To use subtraction to
eliminate one of the
variables, multiply
Equation 2 by 2 and then
subtract the equations.
−10x + 3y = 1
−(−10x − 12y = 46)
15y = −45
Methods for Solving Systems of Linear Equations
Concept SummaryConcept Summary
Method When to Use
Graphing (Lesson 5.1) To estimate solutions
Substitution
(Lesson 5.2)
When one of the variables in one of the
equations has a coefficient of 1 or −1
Elimination
(Lesson 5.3)
When at least one pair of like terms has the
same or opposite coefficients
Elimination (Multiply First)
(Lesson 5.3)
When one of the variables cannot be eliminated
by adding or subtracting the equations
Check
10
−10
−10
10
Intersection
X=-1 Y=-3
10
Equation 1
10
3
Equation 2

A business with two locations buys seven large delivery vans and five small delivery
vans. Location A receives five large vans and two small vans for a total cost of
$235,000. Location B receives two large vans and three small vans for a total cost of
$160,000. What is the cost of each type of van?
SOLUTION
1. Understand the Problem You know how many of each type of van each location
receives. You also know the total cost of the vans for each location. You are asked
to find the cost of each type of van.
Words 5 ⋅ Cost of
large van
+ 2 ⋅ Cost of
small van
= 235,000
2 ⋅ Cost of
large van
+ 3 ⋅ Cost of
small van
= 160,000
Variables Let x be the cost (in dollars) of a large van and let y be the
cost (in dollars) of a small van.
System 5x + 2y = 235,000 Equation 1
2x + 3y = 160,000 Equation 2
Step 1 Multiply Equation 1 by −3. Multiply Equation 2 by 2.
5x + 2y = 235,000 Multiply by −3. −15x − 6y = −705,000 Revised Equation 1
2x + 3y = 160,000 Multiply by 2. 4x + 6y = 320,000 Revised Equation 2
−15x − 6y = −705,000 Revised Equation 1
4x + 6y = 320,000 Revised Equation 2
−11x = −385,000 Add the equations.
Step 3 Solving the equation −11x = −385,000 gives x = 35,000.
Step 4 Substitute 35,000 for x in one of the original equations and solve for y.
5x + 2y = 235,000 Equation 1
5(35,000) + 2y = 235,000 Substitute 35,000 for x.
y = 30,000 Solve for y.
The solution is (35,000, 30,000). So, a large van costs $35,000 and a small van
costs $30,000.
4. Look Back Check to make sure your solution makes sense with the given
information. For Location A, the total cost is 5(35,000) + 2(30,000) = $235,000.
For Location B, the total cost is 2(35,000) + 3(30,000) = $160,000. So, the
solution makes sense.
4. Solve the system in Example 3 by eliminating x.
STUDY TIP
In Example 3, both
equations are multiplied
by a constant so that the
coefficients of the y-terms
are opposites.

In Exercises 3−10, solve the system of linear equations
by elimination. Check your solution. (See Example 1.)
3. x + 2y = 13 4. 9x + y = 2
−x + y = 5 −4x − y = −17
5. 5x + 6y = 50 6. −x + y = 4
x − 6y = −26 x + 3y = 4
7. −3x − 5y = −7 8. 4x − 9y = −21
−4x + 5y = 14 −4x − 3y = 9
9. −y − 10 = 6x 10. 3x − 30 = y
5x + y = −10 7y − 6 = 3x
by elimination. Check your solution. (See Examples 2
and 3.)
11. x + y = 2 12. 8x − 5y = 11
2x + 7y = 9 4x − 3y = 5
13. 11x − 20y = 28 14. 10x − 9y = 46
3x + 4y = 36 −2x + 3y = 10
15. 4x − 3y = 8 16. −2x − 5y = 9
5x − 2y = −11 3x + 11y = 4
17. 9x + 2y = 39 18. 12x − 7y = −2
6x + 13y = −9 8x + 11y = 30
5x − 7y = 16 and x + 7y = 8.
5x − 7y = 16
x + 7y = 8
4x = 24
x = 6
✗
4x + 3y = 8 and x − 2y = −13.
21. MODELING WITH MATHEMATICS A service center
charges a fee of x dollars for an oil change plus
y dollars per quart of oil used. A sample of its sales
record is shown. Write a system of linear equations
that represents this situation. Find the fee and cost per
quart of oil.
A B
2
1
3
4
Customer
Oil Tank Size
(quarts)
Total
Cost
A 5
B 7
C
$22.45
$25.45
22. MODELING WITH MATHEMATICS A music website
charges x dollars for individual songs and y dollars
for entire albums. Person A pays $25.92
to download 6 individual songs and
2 albums. Person B pays $33.93 to
download 4 individual songs and
3 albums. Write a system of linear
equations that represents this
situation. How much does the
website charge to download
a song? an entire album?
1. OPEN-ENDED Give an example of a system of linear equations that can be solved by first adding the
equations to eliminate one variable.
2. WRITING Explain how to solve the system of linear equations 2x − 3y = −4 Equation 1
−5x + 9y = 7 Equation 2by elimination.
4x + 3y = 8 4x + 3y = 8
x − 2y = −13 Multiply by −4. −4x + 8y = −13
11y = −5
y =
−5
—
11
✗

Solve the equation. Determine whether the equation has one solution, no solution, or
infinitely many solutions. (Section 1.3)
36. 5d − 8 = 1 + 5d 37. 9 + 4t = 12 − 4t
38. 3n + 2 = 2(n − 3) 39. −3(4 − 2v) = 6v − 12
Write an equation of the line that passes through the given point and is parallel to the
given line. (Section 4.3)
40. (4, −1); y = −2x + 7 41. (0, 6); y = 5x − 3 42. (−5, −2); y = 2
—3
x + 1
using any method. Explain why you chose the method.
23. 3x + 2y = 4 24. −6y + 2 = −4x
2y = 8 − 5x y − 2 = x
25. y − x = 2 26. 3x + y = 1
—3
y = −
1
—4 x + 7 2x − 3y = 8
—3
27. WRITING For what values of a can you solve the
linear system ax + 3y = 2 and 4x + 5y = 6 by
elimination without multiplying first? Explain.
28. HOW DO YOU SEE IT? The circle graph shows the
results of a survey in which 50 students were asked
about their favorite meal.
Favorite Meal
Dinner
25
Breakfast
Lunch
a. Estimate the numbers of students who chose
breakfast and lunch.
b. The number of students who chose lunch was
5 more than the number of students who chose
breakfast. Write a system of linear equations that
represents the numbers of students who chose
breakfast and lunch.
c. Explain how you can solve the linear system in
part (b) to check your answers in part (a).
29. MAKING AN ARGUMENT Your friend says that any
system of equations that can be solved by elimination
can be solved by substitution in an equal or fewer
number of steps. Is your friend correct? Explain.
30. THOUGHT PROVOKING Write a system of linear
equations that can be added to eliminate a variable
or subtracted to eliminate a variable.
31. MATHEMATICAL CONNECTIONS A rectangle has a
perimeter of 18 inches. A new rectangle is formed
by doubling the width w and tripling the lengthℓ,
as shown. The new rectangle has a perimeter P
of 46 inches.
P = 46 in. 2w
3
a. Write and solve a system of linear equations to
find the length and width of the original rectangle.
b. Find the length and width of the new rectangle.
32. CRITICAL THINKING Refer to the discussion of
System 1 and System 2 on page 230. Without solving,
explain why the two systems shown have the
same solution.
System 1 System 2
3x − 2y = 8 Equation 1 5x = 20 Equation 3
x + y = 6 Equation 2 x + y = 6 Equation 2
33. PROBLEM SOLVING You are making 6 quarts of
fruit punch for a party. You have bottles of 100% fruit
juice and 20% fruit juice. How many quarts of each
type of juice should you mix to make 6 quarts of 80%
fruit juice?
34. PROBLEM SOLVING A motorboat takes 40 minutes
to travel 20 miles downstream. The return trip takes
60 minutes. What is the speed of the current?
35. CRITICAL THINKING Solve for x, y, and z in the
system of equations. Explain your steps.
x + 7y + 3z = 29 Equation 1
3z + x − 2y = −7 Equation 2
5y = 10 − 2x Equation 3

Section 5.4 Solving Special Systems of Linear Equations 235
5.4 Solving Special Systems
of Linear Equations
Using a Table to Solve a System
Work with a partner. You invest $450 for equipment to make skateboards. The
materials for each skateboard cost $20. You sell each skateboard for $20.
a. Write the cost and revenue equations. Then copy and complete the table for your
cost C and your revenue R.
b. When will your company break even? What is wrong?
Essential QuestionEssential Question Can a system of linear equations have no
solution or infinitely many solutions?
Writing and Analyzing a System
Work with a partner. A necklace and matching bracelet have two types of beads.
The necklace has 40 small beads and 6 large beads and weighs 10 grams. The bracelet
has 20 small beads and 3 large beads and weighs 5 grams. The threads holding the
beads have no significant weight.
a. Write a system of linear equations that represents the situation. Let x be the weight
(in grams) of a small bead and let y be the weight (in grams) of a large bead.
b. Graph the system in the coordinate plane shown.
What do you notice about the two lines?
c. Can you find the weight of each type of bead?
Explain your reasoning.
3. Can a system of linear equations have no solution or infinitely many solutions?
Give examples to support your answers.
4. Does the system of linear equations represented by each graph have no solution,
one solution, or infinitely many solutions? Explain.
a.
x
y
4
1
42−1
y = x + 2
x + y = 2
b.
x
y
3
6
42
3
y = x + 2
−x + y = 1
c.
x
y
3
1
6
42
3
y = x + 2
−2x + 2y = 4
x (skateboards) 0 1 2 3 4 5 6 7 8 9 10
C (dollars)
R (dollars)
MODELING WITH
MATHEMATICS
you need to interpret
mathematical results in
real-life contexts.
x
y
1
1.5
2
0.5
0
0.2 0.3 0.40.10

Determine the numbers of solutions of linear systems.
Use linear systems to solve real-life problems.
The Numbers of Solutions of Linear Systems
Previous
parallel
Solutions of Systems of Linear Equations
A system of linear equations can have one solution, no solution, or infinitely
many solutions.
One solution No solution Infinitely many solutions
x
y
x
y
x
y
The lines intersect. The lines are parallel. The lines are the same.
Solving a System: No Solution
Solve the system of linear equations.
y = 2x + 1 Equation 1
SOLUTION
Method 1 Solve by graphing.
Graph each equation.
The lines have the same slope and different
y-intercepts. So, the lines are parallel.
Because parallel lines do not intersect,
there is no point that is a solution
of both equations.
So, the system of linear equations
has no solution.
Method 2 Solve by substitution.
Substitute 2x − 5 for y in Equation 1.
y = 2x + 1 Equation 1
2x − 5 = 2x + 1 Substitute 2x − 5 for y.
−5 = 1 ✗ Subtract 2x from each side.
The equation −5 = 1 is never true. So, the system of linear equations
has no solution.
ANOTHER WAY
You can solve some linear
systems by inspection. In
Example 1, notice you can
rewrite the system as
−2x + y = 1
−2x + y = −5.
This system has no
solution because −2x + y
cannot be equal to both
1 and −5.
STUDY TIP
A linear system with
no solution is called an
inconsistent system.
x
y
2
−2
−4
41−2
1
2
1
2
2
−22
y = 2x + 1
x44
y = 2x − 5

Solving a System: Infinitely Many Solutions
−2x + y = 3 Equation 1
SOLUTION
The lines have the same slope and the same
y-intercept. So, the lines are the same. Because
the lines are the same, all points on the line
are solutions of both equations.
So, the system of linear equations has
infinitely many solutions.
Method 2 Solve by elimination.
Step 1 Multiply Equation 1 by −2.
−2x + y = 3 Multiply by −2. 4x − 2y = −6 Revised Equation 1
−4x + 2y = 6 −4x + 2y = 6 Equation 2
4x − 2y = −6 Revised Equation 1
0 = 0 Add the equations.
The equation 0 = 0 is always true. So, the solutions are all the points on the line
−2x + y = 3. The system of linear equations has infinitely many solutions.
Check
Use the table feature of
a graphing calculator to
check your answer. You
can see that for any x-value,
the corresponding y-values
are equal.
1. x + y = 3 2. y = −x + 3
2x + 2y = 6 2x + 2y = 4
3. x + y = 3 4. y = −10x + 2
x + 2y = 4 10x + y = 10
STUDY TIP
A linear system with
infinitely many solutions
is called a consistent
dependent system.
X Y1
X=0
-3
-1
1
3
5
7
9
Y2
-3
-1
1
3
5
7
9
-2
-3
-1
0
1
2
3
x
y
4
6
1
−2
42−4
−2x + y = 3
44
−4x + 2y = 6

The perimeter of the trapezoidal piece of land is 48 kilometers. The perimeter of
the rectangular piece of land is 144 kilometers. Write and solve a system of linear
equations to find the values of x and y.
SOLUTION
1. Understand the Problem You know the perimeter of each piece of land and the
side lengths in terms of x or y. You are asked to write and solve a system of linear
equations to find the values of x and y.
2. Make a Plan Use the figures and the definition of perimeter to write a
system of linear equations that represents the problem. Then solve the system
of linear equations.
Perimeter of trapezoid Perimeter of rectangle
2x + 4x + 6y + 6y = 48 9x + 9x + 18y + 18y = 144
6x + 12y = 48 Equation 1 18x + 36y = 144 Equation 2
18x + 36y = 144 Equation 2
The lines have the same slope and the same
y-intercept. So, the lines are the same.
In this context, x and y must be positive.
Because the lines are the same, all the points
on the line in Quadrant I are solutions of
both equations.
So, the system of linear equations has infinitely many solutions.
Method 2 Solve by elimination.
Multiply Equation 1 by −3 and add the equations.
6x + 12y = 48 Multiply by −3. −18x − 36y = −144 Revised Equation 1
18x + 36y = 144 18x + 36y = 144 Equation 2
0 = 0 Add the equations.
The equation 0 = 0 is always true. In this context, x and y must be positive.
So, the solutions are all the points on the line 6x + 12y = 48 in Quadrant I.
The system of linear equations has infinitely many solutions.
4. Look Back Choose a few of the ordered pairs (x, y) that are solutions of Equation 1.
You should find that no matter which ordered pairs you choose, they will also be
solutions of Equation 2. So, infinitely many solutions seems reasonable.
5. WHAT IF? What happens to the solution in Example 3 when the perimeter of the
trapezoidal piece of land is 96 kilometers? Explain.
2 4 60 x
2
4
6
0
y
6x + 12y = 486x + 12y =
18x + 36y = 144
4x
2x
6y 6y
18y
18y
9x9x
CONNECTIONS TO
GEOMETRY
When two lines have the
same slope and the same
y-intercept, the lines
coincide. You will learn
more about coincident lines
in Chapter 10.

In Exercises 3−8, match the system of linear equations
with its graph. Then determine whether the system has
one solution, no solution, or infinitely many solutions.
3. −x + y = 1 4. 2x − 2y = 4
x − y = 1 −x + y = −2
5. 2x + y = 4 6. x − y = 0
−4x − 2y = −8 5x − 2y = 6
7. −2x + 4y = 1 8. 5x + 3y = 17
3x − 6y = 9 x − 3y = −2
A.
x
y
2
4
2−1
B.
x
y
2
4
6
1 4−2
C.
x
y
2
−2
2 4−2
D.
x
y
2
−3
1 4
E.
x
y
2
−1
2 4
F.
x
y
2
−3
2−3
In Exercises 9–16, solve the system of linear equations.
(See Examples 1 and 2.)
9. y = −2x − 4 10. y = −6x − 8
y = 2x − 4 y = −6x + 8
11. 3x − y = 6 12. −x + 2y = 7
−3x + y = −6 x − 2y = 7
13. 4x + 4y = −8 14. 15x − 5y = −20
−2x − 2y = 4 −3x + y = 4
15. 9x − 15y = 24 16. 3x − 2y = −5
6x − 10y = −16 4x + 5y = 47
In Exercises 17–22, use only the slopes and y-intercepts
of the graphs of the equations to determine whether the
system of linear equations has one solution, no solution,
or infinitely many solutions. Explain.
17. y = 7x + 13 18. y = −6x − 2
−21x + 3y = 39 12x + 2y = −6
19. 4x + 3y = 27 20. −7x + 7y = 1
4x − 3y = −27 2x − 2y = −18
21. −18x + 6y = 24 22. 2x − 2y = 16
3x − y = −2 3x − 6y = 30
and correct the error in solving the system of linear
equations.
23.
−4x + y = 4
4x + y = 12
The lines do not intersect. So, the system
has no solution.
✗
x
y
1
−3
2−2
24.
y = 3x − 8
y = 3x − 12
The lines have the same slope. So, the system
has infinitely many solutions.
✗
1. REASONING Is it possible for a system of linear equations to have exactly two solutions? Explain.
2. WRITING Compare the graph of a system of linear equations that has infinitely many solutions and
the graph of a system of linear equations that has no solution.

Solve the equation. Check your solutions. (Section 1.4)
33. ∣2x + 6∣ = ∣x∣ 34. ∣3x − 45∣ = ∣12x∣
35. ∣x − 7∣ = ∣2x − 8∣ 36. ∣2x + 1∣ = ∣3x − 11∣
25. MODELING WITH MATHEMATICS A small bag of
trail mix contains 3 cups of dried fruit and 4 cups of
almonds. A large bag contains 41
—2
cups of dried fruit
and 6 cups of almonds. Write and solve a system of
linear equations to find the price of 1 cup of dried fruit
and 1 cup of almonds. (See Example 3.)
$9 $6
26. MODELING WITH MATHEMATICS In a canoe race,
Team A is traveling 6 miles per hour and is 2 miles
ahead of Team B. Team B is also traveling 6 miles
per hour. The teams continue traveling at their current
rates for the remainder of the race. Write a system
of linear equations that represents this situation. Will
Team B catch up to Team A? Explain.
27. PROBLEM SOLVING A train travels from New York
City to Washington, D.C., and then back to New York
City. The table shows the number of tickets purchased
for each leg of the trip. The cost per ticket is the same
for each leg of the trip. Is there enough information to
determine the cost of one coach ticket? Explain.
Destination
Coach
tickets
Business
class
tickets
Money
collected
(dollars)
Washington, D.C. 150 80 22,860
New York City 170 100 27,280
28. THOUGHT PROVOKING Write a system of three
linear equations in two variables so that any two of
the equations have exactly one solution, but the entire
system of equations has no solution.
29. REASONING In a system of linear equations, one
equation has a slope of 2 and the other equation has
a slope of −
1
—3. How many solutions does the system
have? Explain.
30. HOW DO YOU SEE IT? The graph shows information
about the last leg of a 4 × 200-meter relay for
three relay teams. Team A’s runner ran about
7.8 meters per second, Team B’s runner ran about
7.8 meters per second, and Team C’s runner ran about
8.8 meters per second.
40
0
50
100
150
8 12 16 20 24 28
Distance(meters)
x
y
Time (seconds)
Last Leg of 4 × 200-Meter Relay
Team C
Team B
Team A
a. Estimate the distance at which Team C’s runner
passed Team B’s runner.
b. If the race was longer, could Team C’s runner have
passed Team A’s runner? Explain.
c. If the race was longer, could Team B’s runner have
passed Team A’s runner? Explain.
31. ABSTRACT REASONING Consider the system of
linear equations y = ax + 4 and y = bx − 2, where
a and b are real numbers. Determine whether each
statement is always, sometimes, or never true. Explain
your reasoning.
a. The system has infinitely many solutions.
b. The system has no solution.
c. When a < b, the system has one solution.
32. MAKING AN ARGUMENT One admission to an ice
skating rink costs x dollars, and renting a pair of
ice skates costs y dollars. Your friend says she can
determine the exact cost of one admission and one
skate rental. Is your friend correct? Explain.
Total
2
Admissions3
38.00
Skate Rentals
$ Total
15 Admissions
10
190.00
Skate Rentals
$

241
5.1–5.4 What Did You Learn?
Core VocabularyCore Vocabulary
system of linear equations, p. 218 solution of a system of linear equations, p. 218
Core ConceptsCore Concepts
Section 5.1
Solving a System of Linear Equations by Graphing, p. 219
Section 5.2
Solving a System of Linear Equations by Substitution, p. 224
Section 5.3
Solving a System of Linear Equations by Elimination, p. 230
Section 5.4
Solutions of Systems of Linear Equations, p. 236
Mathematical PracticesMathematical Practices
1. Describe the given information in Exercise 33 on page 228 and your plan for finding the solution.
2. Describe another real-life situation similar to Exercise 22 on page 233 and the mathematics that you
can apply to solve the problem.
3. What question(s) can you ask your friend to help her understand the error in the statement she made
in Exercise 32 on page 240?
Analyzing Your Errors
2414111111
Study Errors
What Happens: You do not study the right material
or you do not learn it well enough to remember
it on a test without resources such as notes.
How to Avoid This Error: Take a practice test.
Work with a study group. Discuss the topics on
the test with your teacher. Do not try to learn a
whole chapter’s worth of material in one night.
int_math1_pe_05mc.indd 241int_math1_pe_05mc.indd 241 1/29/15 2:39 PM1/29/15 2:39 PM

5.1–5.4 Quiz
Use the graph to solve the system of linear equations. Check your solution. (Section 5.1)
1. y = −
1
—3 x + 2 2. y = 1
—2
x − 1 3. y = 1
y = x − 2 y = 4x + 6 y = 2x + 1
x
y
3
1
−1
42−2
x
y
2
−2
−4
−2−4
x
y
2
−2
2−2
Solve the system of linear equations by substitution. Check your solution. (Section 5.2)
4. y = x − 4 5. 2y + x = −4 6. 3x − 5y = 13
−2x + y = 18 y − x = −5 x + 4y = 10
Solve the system of linear equations by elimination. Check your solution. (Section 5.3)
7. x + y = 4 8. x + 3y = 1 9. 2x − 3y = −5
−3x − y = −8 5x + 6y = 14 5x + 2y = 16
Solve the system of linear equations. (Section 5.4)
10. x − y = 1 11. 6x + 2y = 16 12. 3x − 3y = −2
x − y = 6 2x − y = 2 −6x + 6y = 4
13. You plant a spruce tree that grows 4 inches per year and a hemlock tree
that grows 6 inches per year. The initial heights are shown. (Section 5.1)
a. Write a system of linear equations that represents this situation.
b. Solve the system by graphing. Interpret your solution.
14. It takes you 3 hours to drive to a concert 135 miles away. You drive
55 miles per hour on highways and 40 miles per hour on the rest of
the roads. (Section 5.1, Section 5.2, and Section 5.3)
a. How much time do you spend driving at each speed?
b. How many miles do you drive on highways? the rest of the roads?
15. In a football game, all of the home team’s points are from 7-point
touchdowns and 3-point field goals. The team scores six times.
Write and solve a system of linear equations to find the numbers
of touchdowns and field goals that the home team scores.
(Section 5.1, Section 5.2, and Section 5.3)
spruce
tree
hemlock
tree
14 in.
8 in.
int_math1_pe_05mc.indd 242int_math1_pe_05mc.indd 242 1/29/15 2:39 PM1/29/15 2:39 PM

Section 5.5 Solving Equations by Graphing 243
Solving Equations by Graphing5.5
Solving an Equation by Graphing
Work with a partner. Solve 2x − 1 = −
1
—2x + 4 by graphing.
a. Use the left side to write a linear equation. Then use the right side to write
another linear equation.
b. Graph the two linear equations from
part (a). Find the x-value of the point of
intersection. Check that the x-value is the
solution of
2x − 1 = −
1
—2
x + 4.
c. Explain why this “graphical method” works.
Essential QuestionEssential Question How can you use a system of linear
equations to solve an equation with variables on both sides?
Previously, you learned how to use algebra to solve equations with variables
on both sides. Another way is to use a system of linear equations.
Solving Equations Algebraically
and Graphically
Work with a partner. Solve each equation using two methods.
Method 1 Use an algebraic method.
Method 2 Use a graphical method.
Is the solution the same using both methods?
a. 1
—2
x + 4 = −
1
—4 x + 1 b. 2
—3
x + 4 = 1
—3
x + 3
c. −
2
—3 x − 1 = 1
—3
x − 4 d. 4
—5
x + 7
—5
= 3x − 3
e. −x + 2.5 = 2x − 0.5 f. −3x + 1.5 = x + 1.5
3. How can you use a system of linear equations to solve an equation with
variables on both sides?
4. Compare the algebraic method and the graphical method for solving a
linear equation with variables on both sides. Describe the advantages and
disadvantages of each method.
USING TOOLS
STRATEGICALLY
you need to consider the
available tools, which
may include pencil and
paper or a graphing
calculator, when solving
a mathematical problem.
x
y
2
4
6
2 4 6−2
−2

Solve linear equations by graphing.
Solve absolute value equations by graphing.
Use linear equations to solve real-life problems.
Solving Linear Equations by Graphing
You can use a system of linear equations to solve an equation with variables
on both sides.
Previous
absolute value equation
Solving Linear Equations by Graphing
Step 1 To solve the equation ax + b = cx + d, write two linear equations.
ax + b = cx + d
and
Step 2 Graph the system of linear equations. The x-value of the solution
of the system of linear equations is the solution of the equation
ax + b = cx + d.
y = cx + dy = ax + b
Solving an Equation by Graphing
Solve −x + 1 = 2x − 5 by graphing. Check your solution.
SOLUTION
Step 1 Write a system of linear equations using each side of the original equation.
−x + 1 = 2x − 5
Step 2 Graph the system.
y = −x + 1 Equation 1
The graphs intersect at (2, −1).
So, the solution of the equation is x = 2.
Solve the equation by graphing. Check your solution.
1. 1
—2
x − 3 = 2x 2. −4 + 9x = −3x + 2
Check
−x + 1 = 2x − 5
−(2) + 1 =
?
2(2) − 5
−1 = −1 ✓
y = 2x − 5y = −x + 1
x
y
1
1
y = −x + 1
y = 2x − 5
−1
−2
−4
(2, −1)

Solving Absolute Value Equations by Graphing
Solving an Absolute Value Equation by Graphing
Solve ∣x + 1∣ = ∣2x − 4∣ by graphing. Check your solutions.
SOLUTION
Recall that an absolute value equation of the form ∣ax + b∣ = ∣cx + d∣ has
two related equations.
ax + b = cx + d Equation 1
ax + b = −(cx + d) Equation 2
So, the related equations of ∣x + 1∣ = ∣2x − 4∣ are as follows.
x + 1 = 2x − 4 Equation 1
x + 1 = −(2x − 4) Equation 2
Apply the steps for solving an equation by graphing to each of the related equations.
Step 1 Write a system of linear equations for each related equation.
x + 1 = 2x − 4 x + 1 = −(2x − 4)
x + 1 = −2x + 4
System 1
System 2
Step 2 Graph each system.
System 1 System 2
y = x + 1 y = x + 1
y = 2x − 4 y = −2x + 4
x
y
2
4
6
1 64
y = x + 1
y = 2x − 4
(5, 6)
x
y
2
4
6
1 64
y = x + 1
y = −2x + 4
(1, 2)
The graphs intersect at (5, 6). The graphs intersect at (1, 2).
So, the solutions of the equation are x = 5 and x = 1.
Solve the equation by graphing. Check your solutions.
3. ∣2x + 2∣ = ∣x − 2∣
4. ∣x − 6∣ = ∣−x + 4∣
y = 2x − 4y = x + 1
y = −2x + 4y = x + 1
Check
∣x + 1∣ = ∣2x − 4∣
∣5 + 1∣ =
?
∣2(5) − 4∣
∣6∣ =
?
∣6∣
6 = 6 ✓
∣x + 1∣ = ∣2x − 4∣
∣1 + 1∣ =
?
∣2(1) − 4∣
∣2∣ =
?
∣−2∣
2 = 2 ✓

You are studying two glaciers. In 2000, Glacier A had an area of about 40 square miles
and Glacier B had an area of about 32 square miles. You estimate that Glacier A will
melt at a rate of 2 square miles per decade and Glacier B will melt at a rate of
0.25 square mile per decade. In what year will the areas of the glaciers be the same?
SOLUTION
Step 1 Use a verbal model to write an equation that represents the problem. Let x
be the number of decades after 2000. Then write a system of linear equations
using each side of the equation.
Glacier A Glacier B
Area of
Glacier A
in 2000
−
Area lost
per decade
after 2000
⋅
Number
of decades
after 2000
=
Area of
Glacier B
in 2000
−
Area lost
per decade
after 2000
⋅
Number
of decades
after 2000
40 − 2x = 32 − 0.25x
Step 2 Graph the system. The graphs intersect between x = 4 and x = 5. Make a table
using x-values between 4 and 5. Use an increment of 0.1.
Notice when x = 4.5, the area of Glacier A is greater than the area of Glacier
B. But when x = 4.6, the area of Glacier A is less than the area of Glacier B.
So, the solution must be between x = 4.5 and x = 4.6. Make another table
using x-values between 4.5 and 4.6. Use an increment of 0.01.
When x = 4.57, the corresponding y-values are about the same. So, the graphs
intersect at about (4.57, 30.86).
So, the areas of the glaciers will be the same after about 4.57 decades, or around
the year 2046.
5. WHAT IF? In 2000, Glacier C had an area of about 30 square miles. You estimate
that it will melt at a rate of 0.45 square mile per decade. In what year will the
areas of Glacier A and Glacier C be the same?
y = 40 − 2x y = 32 − 0.25x
x 4.1 4.2 4.3 4.4 4.5 4.6 4.7
y = 40 − 2x 31.8 31.6 31.4 31.2 31 30.8 30.6
y = 32 − 0.25x 30.98 30.95 30.93 30.9 30.88 30.85 30.83
x 4.51 4.52 4.53 4.54 4.55 4.56 4.57 4.58
y = 40 − 2x 30.98 30.96 30.94 30.92 30.9 30.88 30.86 30.84
y = 32 − 0.25x 30.87 30.87 30.87 30.87 30.86 30.86 30.86 30.86
Check
0
0
45
10
Intersection
X=4.5714286 Y=30.857143
y = 32 − 0.25x
y = 40 − 2x
x
y
20
10
0
40
30
420 86
y = 32 − 0.25x
y = 40 − 2x

In Exercises 3–6, use the graph to solve the equation.
Check your solution.
3. −2x + 3 = x 4. −3 = 4x + 1
x
y
1
3
1 3−3
x
y
1
2−2
5. −x − 1 = 1
—3
x + 3 6. −
3
—2 x − 2 = −4x + 3
x
y
2
4
1−2−4
x
y
2 4
−2
−4
−6
In Exercises 7−14, solve the equation by graphing.
Check your solution. (See Example 1.)
7. x + 4 = −x 8. 4x = x + 3
9. x + 5 = −2x − 4 10. −2x + 6 = 5x − 1
11. 1
—2
x − 2 = 9 − 5x 12. −5 + 1
—4
x = 3x + 6
13. 5x − 7 = 2(x + 1) 14. −6(x + 4) = −3x − 6
Determine whether the equation has one solution,
no solution, or infinitely many solutions.
15. 3x − 1 = −x + 7 16. 5x − 4 = 5x + 1
17. −4(2 − x) = 4x − 8
18. −2x − 3 = 2(x − 2)
19. −x − 5 = −
1
—3(3x + 5)
20. 1
—2
(8x + 3) = 4x + 3
—2
In Exercises 21 and 22, use the graphs to solve the
equation. Check your solutions.
21. ∣x − 4∣ = ∣3x∣
xy
2−2
−6
x
y
2−2
−4
−2
22. ∣2x + 4∣ = ∣x − 1∣
xy
−4−6
−6
−4
x
y
3−1
4
−3
Check your solutions. (See Example 2.)
23. ∣2x∣ = ∣x + 3∣ 24. ∣2x − 6∣ = ∣x∣
25. ∣−x + 4∣ = ∣2x − 2∣
26. ∣x + 2∣ = ∣−3x + 6∣
27. ∣x + 1∣ = ∣x − 5∣
28. ∣2x + 5∣ = ∣−2x + 1∣
29. ∣x − 3∣ = 2∣x∣ 30. 4∣x + 2∣ = ∣2x + 7∣
1. REASONING The graphs of the equations y = 3x − 20 and y = −2x + 10 intersect at the
point (6, −2). Without solving, find the solution of the equation 3x − 20 = −2x + 10.
2. WRITING Explain how to rewrite the absolute value equation ∣2x − 4∣ = ∣−5x + 1∣ as two systems
of linear equations.

Graph the inequality. (Section 2.1)
42. y > 5 43. x ≤ −2 44. n ≥ 9 45. c < −6
Use the graphs of f and g to describe the transformation from the graph of f to the
graph of g. (Section 3.6)
46. f(x) = x − 5; g(x) = f(x + 2) 47. f(x) = 6x; g(x) = −f(x)
48. f(x) = −2x + 1; g(x) = f(4x) 49. f(x) = 1
—2
x − 2; g(x) = f(x − 1)
USING TOOLS In Exercises 31 and 32, use a graphing
calculator to solve the equation.
31. 0.7x + 0.5 = −0.2x − 1.3
32. 2.1x + 0.6 = −1.4x + 6.9
33. MODELING WITH MATHEMATICS There are about
34 million gallons of water in Reservoir A and about
38 million gallons in Reservoir B. During a drought,
Reservoir A loses about 0.8 million gallons per month
and Reservoir B loses about 1.1 million gallons per
month. After how many months will the reservoirs
contain the same amount of water? (See Example 3.)
34. MODELING WITH MATHEMATICS Your dog is
16 years old in dog years. Your cat is 28 years old in
cat years. For every human year, your dog ages by
7 dog years and your cat ages by 4 cat years. In how
many human years will both pets be the same age in
their respective types of years?
35. MODELING WITH MATHEMATICS You and a friend
race across a field to a fence and back. Your friend has
a 50-meter head start. The equations shown represent
you and your friend’s distances d (in meters) from the
fence t seconds after the race begins. Find the time at
which you catch up to your friend.
You: d = ∣−5t + 100∣
Your friend: d = ∣−31
—3
t + 50∣
36. MAKING AN ARGUMENT The graphs of y = −x + 4
and y = 2x − 8 intersect at the point (4, 0). So,
your friend says the solution of the equation
−x + 4 = 2x − 8 is (4, 0). Is your friend correct?
Explain.
37. OPEN-ENDED Find values for m and b so that the
solution of the equation mx + b = −2x − 1 is
x = −3.
38. HOW DO YOU SEE IT? The graph shows the total
revenue and expenses of a company x years after it
opens for business.
20
0
2
4
6
4 6 8 10
Millionsofdollars
x
y
Year
Revenue and Expenses
revenue
expenses
a. Estimate the point of intersection of the graphs.
b. Interpret your answer in part (a).
39. MATHEMATICAL CONNECTIONS The value of the
perimeter of the triangle (in feet) is equal to the value
of the area of the triangle (in square feet). Use a graph
to find x.
6 ft
x ft
(x − 2) ft
40. THOUGHT PROVOKING A car has an initial value
of $20,000 and decreases in value at a rate of
$1500 per year. Describe a different car that will be
worth the same amount as this car in exactly 5 years.
Specify the initial value and the rate at which the
value decreases.
41. ABSTRACT REASONING Use a graph to determine the
sign of the solution of the equation ax + b = cx + d
in each situation.
a. 0 < b < d and a < c b. d < b < 0 and a < c

Section 5.6 Graphing Linear Inequalities in Two Variables 249
Essential QuestionEssential Question How can you graph a linear inequality in
two variables?
A solution of a linear inequality in two variables is an ordered pair (x, y) that makes
the inequality true. The graph of a linear inequality in two variables shows all the
solutions of the inequality in a coordinate plane.
Writing a Linear Inequality in Two Variables
a. Write an equation represented by
the dashed line.
b. The solutions of an inequality are
represented by the shaded region. In words,
describe the solutions of the inequality.
c. Write an inequality represented by the graph.
Which inequality symbol did you use?
Explain your reasoning.
Graphing Linear Inequalities in Two Variables
Work with a partner. Graph each linear inequality in two variables. Explain your
steps. Use a graphing calculator to check your graphs.
a. y > x + 5 b. y ≤ −
1
—2 x + 1 c. y ≥ −x − 5
4. How can you graph a linear inequality in two variables?
5. Give an example of a real-life situation that can be modeled using a linear
inequality in two variables.
USING TOOLS
STRATEGICALLY
To be proficient in
math, you need to use
technological tools to
explore and deepen
your understanding
of concepts.
x
y
2
4
2 4−2−4
−2
Work with a partner. Use a graphing calculator to graph y ≥ 1
—4
x − 3.
a. Enter the equation y = 1
—4
x − 3 into your calculator.
b. The inequality has the symbol ≥. So, the
region to be shaded is above the graph of
y = 1
—4
x − 3, as shown. Verify this by testing
a point in this region, such as (0, 0), to make
sure it is a solution of the inequality.
Because the inequality symbol is greater than or equal to, the line is solid and not
dashed. Some graphing calculators always use a solid line when graphing inequalities.
In this case, you have to determine whether the line should be solid or dashed, based
on the inequality symbol used in the original inequality.
10
−10
−10
10
y ≥ x − 3
1
4
5.6 Graphing Linear Inequalities
in Two Variables

Check solutions of linear inequalities.
Graph linear inequalities in two variables.
Use linear inequalities to solve real-life problems.
Linear Inequalities
A linear inequality in two variables, x and y, can be written as
ax + by < c ax + by ≤ c ax + by > c ax + by ≥ c
where a, b, and c are real numbers. A solution of a linear inequality in two variables
is an ordered pair (x, y) that makes the inequality true.
linear inequality in
two variables, p. 250
solution of a linear inequality
in two variables, p. 250
graph of a linear inequality,
p. 250
half-planes, p. 250
Previous
ordered pair
Checking Solutions
Tell whether the ordered pair is a solution of the inequality.
a. 2x + y < −3; (−1, 9) b. x − 3y ≥ 8; (2, −2)
SOLUTION
a. 2x + y < −3 Write the inequality.
2(−1) + 9 <
?
−3 Substitute −1 for x and 9 for y.
7 < −3 ✗ Simplify. 7 is not less than −3.
So, (−1, 9) is not a solution of the inequality.
b. x − 3y ≥ 8 Write the inequality.
2 − 3(−2) ≥
?
8 Substitute 2 for x and −2 for y.
8 ≥ 8 ✓ Simplify. 8 is equal to 8.
So, (2, −2) is a solution of the inequality.
Tell whether the ordered pair is a solution of the inequality.
1. x + y > 0; (−2, 2) 2. 4x − y ≥ 5; (0, 0)
3. 5x − 2y ≤ −1; (−4, −1) 4. −2x − 3y < 15; (5, −7)
Graphing Linear Inequalities in Two Variables
The graph of a linear inequality in two variables shows all the solutions of the
inequality in a coordinate plane.
x
y
4
2
2−2
All solutions of < 2
lie on one side of the
= 2 .
The boundary line divides
the coordinate plane into two
half-planes.The shaded
half-plane is the graph of < 2 .
y
y
x
x
y x
boundary line
READING
A dashed boundary line
means that points on the
line are not solutions. A
solid boundary line means
that points on the line
are solutions.

Graphing a Linear Inequality in One Variable
Graph y ≤ 2 in a coordinate plane.
SOLUTION
Step 1 Graph y = 2. Use a solid line because the
x
y
1
3
2 4−1
(0, 0)
inequality symbol is ≤.
Step 2 Test (0, 0).
y ≤ 2 Write the inequality.
0 ≤ 2 ✓ Substitute.
Step 3 Because (0, 0) is a solution, shade the
half-plane that contains (0, 0).
Check
3
−1
−2
5
Graphing a Linear Inequality in Two Variables
Step 1 Graph the boundary line for the inequality. Use a dashed line for < or >.
Use a solid line for ≤ or ≥.
Step 2 Test a point that is not on the boundary line to determine whether it is a
solution of the inequality.
Step 3 When the test point is a solution, shade the half-plane that contains the
point. When the test point is not a solution, shade the half-plane that
does not contain the point.
Graphing a Linear Inequality in Two Variables
Graph −x + 2y > 2 in a coordinate plane.
SOLUTION
Step 1 Graph −x + 2y = 2, or y = 1
—2
x + 1. Use a
x
y
2
4
2−2
(0, 0)
dashed line because the inequality symbol is >.
Step 2 Test (0, 0).
−x + 2y > 2 Write the inequality.
−(0) + 2(0) >
?
2 Substitute.
0 > 2 ✗ Simplify.
Step 3 Because (0, 0) is not a solution, shade the
half-plane that does not contain (0, 0).
Graph the inequality in a coordinate plane.
5. y > −1 6. x ≤ −4
7. x + y ≤ −4 8. x − 2y < 0
STUDY TIP
It is often convenient to
use the origin as a test
point. However, you must
choose a different test
point when the origin is
on the boundary line.

You can spend at most $10 on grapes and apples for a fruit salad. Grapes cost
$2.50 per pound, and apples cost $1 per pound. Write and graph an inequality that
represents the amounts of grapes and apples you can buy. Identify and interpret
two solutions of the inequality.
SOLUTION
1. Understand the Problem You know the most that you can spend and the prices
per pound for grapes and apples. You are asked to write and graph an inequality
and then identify and interpret two solutions.
2. Make a Plan Use a verbal model to write an inequality that represents the
problem. Then graph the inequality. Use the graph to identify two solutions.
Then interpret the solutions.
Words
Cost per
pound of
grapes
⋅
Pounds
of grapes
+
Cost per
pound of
apples
⋅
Pounds
of apples
≤
Amount
you can
spend
Variables Let x be pounds of grapes and y be pounds of apples.
Inequality 2.50 ⋅x + 1 ⋅y ≤ 10
Step 1 Graph 2.5x + y = 10, or y = −2.5x + 10. Use a solid line because the
inequality symbol is ≤. Restrict the graph to positive values of x and y
because negative values do not make sense in this real-life context.
Step 2 Test (0, 0).
2.5x + y ≤ 10 Write the inequality.
2.5(0) + 0 ≤
?
10 Substitute.
0 ≤ 10 ✓ Simplify.
Step 3 Because (0, 0) is a solution, shade the half-plane that contains (0, 0).
One possible solution is (1, 6) because it lies in the shaded half-plane. Another
possible solution is (2, 5) because it lies on the solid line. So, you can buy
1 pound of grapes and 6 pounds of apples, or 2 pounds of grapes and
5 pounds of apples.
4. Look Back Check your solutions by substituting them into the original inequality,
as shown.
9. You can spend at most $12 on red peppers and tomatoes for salsa. Red peppers
cost $4 per pound, and tomatoes cost $3 per pound. Write and graph an inequality
that represents the amounts of red peppers and tomatoes you can buy. Identify and
interpret two solutions of the inequality.
Check
2.5x + y ≤ 10
2.5(1) + 6 ≤
?
10
8.5 ≤ 10 ✓
2.5x + y ≤ 10
2.5(2) + 5 ≤
?
10
10 ≤ 10 ✓
Fruit Salad
2 4 60 3 51 x
1
2
3
4
5
6
7
8
9
10
0
y
(1, 6)
(2, 5)
Poundsofapples
Pounds of grapes

In Exercises 3–10, tell whether the ordered pair is a
solution of the inequality. (See Example 1.)
3. x + y < 7; (2, 3) 4. x − y ≤ 0; (5, 2)
5. x + 3y ≥ −2; (−9, 2) 6. 8x + y > −6; (−1, 2)
7. −6x + 4y ≤ 6; (−3, −3)
8. 3x − 5y ≥ 2; (−1, −1) 9. −x − 6y > 12; (−8, 2)
10. −4x − 8y < 15; (−6, 3)
In Exercises 11−16, tell whether the ordered pair is a
solution of the inequality whose graph is shown.
11. (0, −1) 12. (−1, 3)
x
y
2
4
2−2
−2
13. (1, 4) 14. (0, 0)
15. (3, 3) 16. (2, 1)
17. MODELING WITH MATHEMATICS A carpenter has
at most $250 to spend on lumber. The inequality
8x + 12y ≤ 250 represents the numbers x of 2-by-8
boards and the numbers y of 4-by-4 boards the
carpenter can buy. Can the carpenter buy twelve
2-by-8 boards and fourteen 4-by-4 boards? Explain.
2 in. x 8 in. x 8 ft
$8 each
4 in. x 4 in. x 8 ft
$12 each
18. MODELING WITH MATHEMATICS The inequality
3x + 2y ≥ 93 represents the numbers x of multiple-
choice questions and the numbers y of matching
questions you can answer correctly to receive an A on
a test. You answer 20 multiple-choice questions and
18 matching questions correctly. Do you receive an A
on the test? Explain.
In Exercises 19–24, graph the inequality in a coordinate
plane. (See Example 2.)
19. y ≤ 5 20. y > 6
21. x < 2 22. x ≥ −3
23. y > −7 24. x < 9
In Exercises 25−30, graph the inequality in a coordinate
plane. (See Example 3.)
25. y > −2x − 4 26. y ≤ 3x − 1
27. −4x + y < −7 28. 3x − y ≥ 5
29. 5x − 2y ≤ 6 30. −x + 4y > −12
ERROR ANALYSIS In Exercises 31 and 32, describe and
correct the error in graphing the inequality.
31. y < −x + 1
x
y
3
3−2
−2
✗
32. y ≤ 3x − 2
x
y
4
2
2−2
−1
✗
1. VOCABULARY How can you tell whether an ordered pair is a solution of a linear inequality?
2. WRITING Compare the graph of a linear inequality in two variables with the graph of a linear
equation in two variables.

Write the next three terms of the arithmetic sequence. (Section 4.6)
46. 0, 8, 16, 24, 32, . . . 47. −5, −8, −11, −14, −17, . . . 48. −
3
—2, −1
—2
, 1
—2
, 3
—2
, 5
—2
, . . .
33. MODELING WITH MATHEMATICS You have at
most $20 to spend at an arcade. Arcade games cost
$0.75 each, and snacks cost $2.25 each. Write and
graph an inequality that represents the numbers of
games you can play and snacks you can buy. Identify
and interpret two solutions of the inequality.
(See Example 4.)
34. MODELING WITH MATHEMATICS A drama club
must sell at least $1500 worth of tickets to cover the
expenses of producing a play. Write and graph an
inequality that represents
how many adult and
student tickets the club
must sell. Identify and
interpret two solutions
of the inequality.
In Exercises 35–38, write an inequality that represents
the graph.
35.
x
y
2
4
2−2
36.
x
y
1
4
2−2
37.
x
y
1
2−1
−3
−5
38.
x
y
1
2−2
−2
39. PROBLEM SOLVING Large boxes weigh 75 pounds,
and small boxes weigh 40 pounds.
a. Write and graph
an inequality that
represents the numbers
of large and small boxes
a 200-pound delivery
person can take on
the elevator.
b. Explain why some
solutions of the
inequality might not
be practical in real life.
40. HOW DO YOU SEE IT? Match each inequality with
its graph.
a. 3x − 2y ≤ 6 b. 3x − 2y < 6
c. 3x − 2y > 6 d. 3x − 2y ≥ 6
A.
x
y
1
1 3−2
−2
B.
x
y
1
1 3−2
−2
C.
x
y
1
1 3−2
−2
D.
x
y
1
1 3−2
−2
41. REASONING When graphing a linear inequality in
two variables, why must you choose a test point that
is not on the boundary line?
42. THOUGHT PROVOKING Write a linear inequality in
two variables that has the following two properties.
• (0, 0), (0, −1), and (0, 1) are not solutions.
• (1, 1), (3, −1), and (−1, 3) are solutions.
43. WRITING Can you always use (0, 0) as a test point
when graphing an inequality? Explain.
CRITICAL THINKING In Exercises 44 and 45, write and
graph an inequality whose graph is described by the
given information.
44. The points (2, 5) and (−3, −5) lie on the boundary
line. The points (6, 5) and (−2, −3) are solutions of
the inequality.
45. The points (−7, −16) and (1, 8) lie on the boundary
line. The points (−7, 0) and (3, 14) are not solutions
of the inequality.
Weight limit:
2000 lb

Section 5.7 Systems of Linear Inequalities 255
Systems of Linear Inequalities5.7
Essential QuestionEssential Question How can you graph a system of
linear inequalities?
Graphing Linear Inequalities
Work with a partner. Match each linear inequality with its graph. Explain
your reasoning.
2x + y ≤ 4 Inequality 1
2x − y ≤ 0 Inequality 2
A.
x
y
4
2
−2
−4
1 4−2−4
B.
x
y
4
2
−4
2 4−2−4
Graphing a System of Linear Inequalities
Work with a partner. Consider the linear inequalities given in Exploration 1.
2x + y ≤ 4 Inequality 1
2x − y ≤ 0 Inequality 2
a. Use two different colors to graph the inequalities in the same coordinate plane.
What is the result?
b. Describe each of the shaded regions of the graph. What does the unshaded
region represent?
3. How can you graph a system of linear inequalities?
4. When graphing a system of linear inequalities, which region represents the
solution of the system?
5. Do you think all systems of linear inequalities
have a solution? Explain your reasoning.
6. Write a system of linear inequalities
represented by the graph.
x
y
4
6
2
−4
−2
1 4−2−4
MAKING SENSE
OF PROBLEMS
you need to explain to
yourself the meaning
of a problem.

Check solutions of systems of linear inequalities.
Graph systems of linear inequalities.
Write systems of linear inequalities.
Use systems of linear inequalities to solve real-life problems.
Systems of Linear Inequalities
A system of linear inequalities is a set of two or more linear inequalities in the same
variables. An example is shown below.
y < x + 2 Inequality 1
y ≥ 2x − 1 Inequality 2
A solution of a system of linear inequalities in two variables is an ordered pair that
is a solution of each inequality in the system.
system of linear inequalities,
p. 256
solution of a system of linear
inequalities, p. 256
graph of a system of linear
inequalities, p. 257
Previous
linear inequality in
two variables
Checking Solutions
Tell whether each ordered pair is a solution of the system of linear inequalities.
y < 2x Inequality 1
y ≥ x + 1 Inequality 2
a. (3, 5) b. (−2, 0)
SOLUTION
a. Substitute 3 for x and 5 for y in each inequality.
Inequality 1 Inequality 2
y < 2x y ≥ x + 1
5 <
?
2(3) 5 ≥
?
3 + 1
5 < 6 ✓ 5 ≥ 4 ✓
Because the ordered pair (3, 5) is a solution of each inequality, it is a solution
of the system.
b. Substitute −2 for x and 0 for y in each inequality.
Inequality 1 Inequality 2
y < 2x y ≥ x + 1
0 <
?
2(−2) 0 ≥
?
−2 + 1
0 < −4 ✗ 0 ≥ −1 ✓
Because (−2, 0) is not a solution of each inequality, it is not a solution of
the system.
Tell whether the ordered pair is a solution of the system of linear inequalities.
1. (−1, 5);
y < 5
y > x − 4
2. (1, 4);
y ≥ 3x + 1
y > x − 1

Graph the system of linear inequalities.
y ≤ 3 Inequality 1
y > x + 2 Inequality 2
SOLUTION
Step 1 Graph each inequality.
Step 2 Find the intersection of
the half-planes. One
solution is (−3, 1).
Step 1 Graph each inequality in the same
coordinate plane.
Step 2 Find the intersection of the half-planes
that are solutions of the inequalities. This
intersection is the graph of the system.
Graphing a System of Linear Inequalities:
No Solution
2x + y < −1 Inequality 1
2x + y > 3 Inequality 2
SOLUTION
Step 2 Find the intersection of the half-planes. Notice that the lines are parallel,
and the half-planes do not intersect.
So, the system has no solution.
3. y ≥ −x + 4 4. y > 2x − 3 5. −2x + y < 4
x + y ≤ 0 y ≥ 1
—2
x + 1 2x + y > 4
Graphing Systems of Linear Inequalities
The graph of a system of linear inequalities is the graph of all the solutions of
the system.
x
y
4
6
2 4−1
y < x + 2
y ≥ 2x − 1
x
y
4
1
−2
2−1−4
(−3, 1)
The solution is the
purple-shaded region.
Check
Verify that (−3, 1) is a
solution of each inequality.
Inequality 1
y ≤ 3
1 ≤ 3 ✓
Inequality 2
y > x + 2
1 >
?
−3 + 2
1 > −1 ✓
x
y
2
−2
−4
1 3−2−4

Writing Systems of Linear Inequalities
Writing a System of Linear Inequalities
Write a system of linear inequalities represented
by the graph.
SOLUTION
Inequality 1 The horizontal boundary line passes
through (0, −2). So, an equation of the
line is y = −2. The shaded region is
above the solid boundary line, so the
inequality is y ≥ −2.
Inequality 2 The slope of the other boundary line is 1, and the y-intercept
is 0. So, an equation of the line is y = x. The shaded region
is below the dashed boundary line, so the inequality is y < x.
The system of linear inequalities represented by the graph is
y ≥ −2 Inequality 1
y < x. Inequality 2
x
y
2
−4
2 4−2−4
Writing a System of Linear Inequalities
Write a system of linear inequalities represented
by the graph.
SOLUTION
Inequality 1 The vertical boundary line passes
through (3, 0). So, an equation of the
line is x = 3. The shaded region is to
the left of the solid boundary line,
so the inequality is x ≤ 3.
Inequality 2 The slope of the other boundary line is 2
—3
, and the y-intercept is −1.
So, an equation of the line is y = 2
—3
x − 1. The shaded region is above
the dashed boundary line, so the inequality is y > 2
—3
x − 1.
The system of linear inequalities represented by the graph is
x ≤ 3 Inequality 1
y > 2
—3
x − 1. Inequality 2
Write a system of linear inequalities represented by the graph.
6.
x
y
1
2 4
−2
7.
x
y
2
4
2 4
−2
x
y
2
4
−4
2 4 6

You have at most 8 hours to spend at the mall
and at the beach. You want to spend at least
2 hours at the mall and more than 4 hours
at the beach. Write and graph a system that
represents the situation. How much time can
you spend at each location?
SOLUTION
1. Understand the Problem You know the total amount of time you can spend
at the mall and at the beach. You also know how much time you want to spend
at each location. You are asked to write and graph a system that represents the
situation and determine how much time you can spend at each location.
2. Make a Plan Use the given information to write a system of linear inequalities.
Then graph the system and identify an ordered pair in the solution region.
3. Solve the Problem Let x be the number of hours at the mall and let y be the
number of hours at the beach.
x + y ≤ 8 at most 8 hours at the mall and at the beach
x ≥ 2 at least 2 hours at the mall
y > 4 more than 4 hours at the beach
Graph the system.
Time at the Mall and at the Beach
2 4 6 7 8 90 3 51 x
1
2
3
4
5
6
7
8
0
y
Hoursatthebeach
Hours at the mall
One ordered pair in the solution region is (2.5, 5).
So, you can spend 2.5 hours at the mall and 5 hours at the beach.
4. Look Back Check your solution by substituting it into the inequalities in the
system, as shown.
8. Name another solution of Example 6.
9. WHAT IF? You want to spend at least 3 hours at the mall. How does this change
the system? Is (2.5, 5) still a solution? Explain.
Check
x + y ≤ 8
2.5 + 5 ≤
?
8
7.5 ≤ 8 ✓
x ≥ 2
2.5 ≥ 2 ✓
y > 4
5 > 4 ✓

Exercises5.7
In Exercises 3−6, tell whether the ordered pair is a
solution of the system of linear inequalities.
3. (−4, 3)
4. (−3, −1)
5. (−2, 5)
6. (1, 1)
In Exercises 7−10, tell whether the ordered pair
is a solution of the system of linear inequalities.
(See Example 1.)
7. (−5, 2);
y < 4
y > x + 3
8. (1, −1);
y > −2
y > x − 5
9. (0, 0); y ≤ x + 7
y ≥ 2x + 3
10. (4, −3);
y ≤ −x + 1
y ≤ 5x − 2
In Exercises 11−20, graph the system of linear
inequalities. (See Examples 2 and 3.)
11. y > −3 12. y < −1
y ≥ 5x x > 4
13. y < −2 14. y < x − 1
y > 2 y ≥ x + 1
15. y ≥ −5 16. x + y > 4
y − 1 < 3x y ≥ 3
—2
x − 9
17. x + y > 1 18. 2x + y ≤ 5
−x − y < −3 y + 2 ≥ −2x
19. x < 4 20. x + y ≤ 10
y > 1 x − y ≥ 2
y ≥ −x + 1 y > 2
In Exercises 21−26, write a system of linear inequalities
represented by the graph. (See Examples 4 and 5.)
21.
x
y
2
4
2−2
22.
x
y
2
1 53−1
−2
23.
x
y
2
4−2
−1
24.
x
y
1
2−2
25.
x
y
2
−3
2−2−4
26.
x
y
2
−3
2−3
1. VOCABULARY How can you verify that an ordered pair is a solution
of a system of linear inequalities?
2. WHICH ONE DOESN’T BELONG? Use the graph shown. Which of
the ordered pairs does not belong with the other three? Explain
your reasoning.
(1, −2) (0, −4) (−1, −6) (2, −4)
x
y
−2
−4
−6
52
x
y
1−3−5
2
4

and correct the error in graphing the system of
linear inequalities.
27.
y ≤≤ x − 1
x
y
1
−3
2−2
y ≥≥ x + 3
✗
28.
y ≤≤ 3x + 4
x
y
1
4
2−2−4
y > 1
—2
x + 2
✗
29. MODELING WITH MATHEMATICS You can spend at
most $21 on fruit. Blueberries cost $4 per pound, and
strawberries cost $3 per pound. You need at least
3 pounds of fruit to make muffins. (See Example 6.)
a. Write and graph a system of linear inequalities
that represents the situation.
b. Identify and interpret a solution of the system.
c. Use the graph to
determine whether
you can buy
4 pounds of
blueberries
and 1 pound
of strawberries.
30. MODELING WITH MATHEMATICS You earn
$10 per hour working as a manager at a grocery store.
You are required to work at the grocery store at least
8 hours per week. You also teach music lessons for
$15 per hour. You need to earn at least $120 per week,
but you do not want to work more than 20 hours
per week.
b. Identify and interpret a solution of the system.
c. Use the graph to determine whether you can work
8 hours at the grocery store and teach 1 hour of
music lessons.
31. MODELING WITH MATHEMATICS You are fishing
for surfperch and rockfish, which are species of
bottomfish. Gaming laws allow you to catch no more
than 15 surfperch per day, no more than 10 rockfish
per day, and no more than 20 total bottomfish per day.
b. Use the graph to determine whether you can catch
11 surfperch and 9 rockfish in 1 day.
surfperch rockfish
32. REASONING Describe the intersection of the
half-planes of the system shown.
x − y ≤ 4
x − y ≥ 4
33. MATHEMATICAL CONNECTIONS The following points
are the vertices of a shaded rectangle.
(−1, 1), (6, 1), (6, −3), (−1, −3)
a. Write a system of linear inequalities represented
by the shaded rectangle.
b. Find the area of the rectangle.
34. MATHEMATICAL CONNECTIONS The following points
are the vertices of a shaded triangle.
(2, 5), (6, −3), (−2, −3)
a. Write a system of linear inequalities represented
by the shaded triangle.
b. Find the area of the triangle.
35. PROBLEM SOLVING You plan to spend less than
half of your monthly $2000 paycheck on housing
and savings. You want to spend at least 10% of
your paycheck on savings and at most 30% of it on
housing. How much money can you spend on savings
and housing?
36. PROBLEM SOLVING On a road trip with a friend, you
drive about 70 miles per hour, and your friend drives
about 60 miles per hour. The plan is to drive less than
15 hours and at least 600 miles each day. Your friend
will drive more hours than you. How many hours can
you and your friend each drive in 1 day?

Write the product using exponents. (Skills Review Handbook)
49. 4 ⋅4 ⋅4 ⋅4 ⋅4 50. (−13) ⋅(−13) ⋅(−13) 51. x ⋅x ⋅x ⋅x ⋅x ⋅x
Write an equation of the line with the given slope and y-intercept. (Section 4.1)
52. slope: 1 53. slope: −3 54. slope: −
1
—4 55. slope: 4
—3
y-intercept: −6 y-intercept: 5 y-intercept: −1 y-intercept: 0
37. WRITING How are solving systems of linear
inequalities and solving systems of linear equations
similar? How are they different?
38. HOW DO YOU SEE IT? The graphs of two linear
equations are shown.
x
y
2
−2
−4
1 3−2−4
y = −3x + 4
A B
C
Dy = 2x + 1
Replace the equal signs with inequality symbols to
create a system of linear inequalities that has point C
as a solution, but not points A, B, and D. Explain
your reasoning.
y −3x + 4
y 2x + 1
39. USING STRUCTURE Write a system of linear
inequalities that is equivalent to ∣y∣ < x, where x > 0.
Graph the system.
40. MAKING AN ARGUMENT Your friend says that a
system of linear inequalities in which the boundary
lines are parallel must have no solution. Is your friend
correct? Explain.
41. CRITICAL THINKING Is it possible for the solution
set of a system of linear inequalities to be all real
numbers? Explain your reasoning.
OPEN-ENDED In Exercises 42−44, write a system of
linear inequalities with the given characteristic.
42. All solutions are in Quadrant I.
43. All solutions have one positive coordinate and one
negative coordinate.
44. There are no solutions.
45. OPEN-ENDED One inequality in a system is
−4x + 2y > 6. Write another inequality so the system
has (a) no solution and (b) infinitely many solutions.
46. THOUGHT PROVOKING You receive a gift certificate
for a clothing store and plan to use it to buy T-shirts
and sweatshirts. Describe a situation in which you can
buy 9 T-shirts and 1 sweatshirt, but you cannot buy
3 T-shirts and 8 sweatshirts. Write and graph a system
of linear inequalities that represents the situation.
47. CRITICAL THINKING Write a system of linear
inequalities that has exactly one solution.
48. MODELING WITH MATHEMATICS You make
necklaces and key chains to sell at a craft fair. The
table shows the amounts of time and money it takes to
make a necklace and a key chain, and the amounts of
time and money you have available for making them.
Necklace Key chain Available
Time to make
(hours)
0.5 0.25 20
Cost to make
(dollars)
2 3 120
a. Write and graph a system of four linear
inequalities that represents the number x of
necklaces and the number y of key chains that
you can make.
b. Find the vertices (corner points) of the graph of
the system.
c. You sell each necklace for $10 and each key chain
for $8. The revenue R is given by the equation
R = 10x + 8y. Find the revenue corresponding to
each ordered pair in part (b). Which vertex results
in the maximum revenue?

263
5.5–5.7 What Did You Learn?
Core VocabularyCore Vocabulary
linear inequality in two variables, p. 250
solution of a linear inequality in two variables,
p. 250
graph of a linear inequality, p. 250
half-planes, p. 250
system of linear inequalities, p. 256
solution of a system of linear inequalities, p. 256
graph of a system of linear inequalities, p. 257
Core ConceptsCore Concepts
Section 5.5
Solving Linear Equations by Graphing, p. 244
Solving Absolute Value Equations by Graphing, p. 245
Section 5.6
Graphing a Linear Inequality in Two Variables, p. 251
Section 5.7
Graphing a System of Linear Inequalities, p. 257
Writing a System of Linear Inequalities, p. 258
Mathematical PracticesMathematical Practices
1. Why do the equations in Exercise 35 on page 248 contain absolute value expressions?
2. Why is it important to be precise when answering part (a) of Exercise 39 on page 254?
3. Describe the overall step-by-step process you used to solve Exercise 35 on page 261.
Performance Task:
Fishing Limits
Do oceans support unlimited numbers of fish? Can you
use mathematics to set fishing limits so that this valuable
food resource is not endangered?
To explore the answers to these questions and
more, check out the Performance Task and
Real-Life STEM video at BigIdeasMath.com.
int_math1_pe_05ec.indd 263int_math1_pe_05ec.indd 263 1/29/15 2:38 PM1/29/15 2:38 PM

55 Chapter Review
Solving Systems of Linear Equations by Graphing (pp. 217–222)5.1
Solve the system by graphing. y = x − 2 Equation 1
Step 1 Graph each equation.
The graphs appear to intersect at (1, −1).
y = x − 2 y = −3x + 2
−1 =
?
1 − 2 −1 =
?
−3(1) + 2
−1 = −1 ✓ −1 = −1 ✓
The solution is (1, −1).
1. y = −3x + 1 2. y = −4x + 3 3. 5x + 5y = 15
y = x − 7 4x − 2y = 6 2x − 2y = 10
Solving Systems of Linear Equations by Substitution (pp. 223–228)5.2
Solve the system by substitution. −2x + y = −8 Equation 1
Step 1 Solve for y in Equation 1.
y = 2x − 8 Revised Equation 1
Step 2 Substitute 2x − 8 for y in Equation 2 and solve for x.
7x + (2x − 8) = 10 Substitute 2x − 8 for y.
9x − 8 = 10 Combine like terms.
9x = 18 Add 8 to each side.
x = 2 Divide each side by 9.
Step 3 Substituting 2 for x in Equation 1 and solving for y gives y = −4.
The solution is (2, −4).
4. 3x + y = −9 5. x + 4y = 6 6. 2x + 3y = 4
y = 5x + 7 x − y = 1 y + 3x = 6
7. You spend $20 total on tubes of paint and disposable brushes for an art project. Tubes of paint
cost $4.00 each and paintbrushes cost $0.50 each. You purchase twice as many brushes as tubes
of paint. How many brushes and tubes of paint do you purchase?
x
y
2
42
(1, −1)
y = −3x + 2
y y = x − 2
−1

Chapter 5 Chapter Review 265
Solving Systems of Linear Equations by Elimination (pp. 229–234)5.3
Solve the system by elimination. 4x + 6y = −8 Equation 1
x − 2y = −2 Equation 2
Step 1 Multiply Equation 2 by 3 so that the coefficients of the y-terms are opposites.
4x + 6y = −8 4x + 6y = −8 Equation 1
x − 2y = −2 3x − 6y = −6 Revised Equation 2
4x + 6y = −8 Equation 1
3x − 6y = −6 Revised Equation 2
7x = −14 Add the equations.
Step 3 Solve for x.
7x = −14 Resulting equation from Step 2
Step 4 Substitute −2 for x in one of the original equations
and solve for y.
4(−2) + 6y = −8 Substitute −2 for x.
−8 + 6y = −8 Multiply.
y = 0 Solve for y.
The solution is (−2, 0).
Solve the system of linear equations by elimination. Check your solution.
8. 9x − 2y = 34 9. x + 6y = 28 10. 8x − 7y = −3
5x + 2y = −6 2x − 3y = −19 6x − 5y = −1
Solving Special Systems of Linear Equations (pp. 235–240)5.4
Solve the system. 4x + 2y = −14 Equation 1
Solve by substitution. Substitute −2x − 6 for y in Equation 1.
4x + 2(−2x − 6) = −14 Substitute −2x − 6 for y.
4x − 4x − 12 = −14 Distributive Property
−12 = −14 ✗ Combine like terms.
The equation −12 = −14 is never true. So, the system has no solution.
11. x = y + 2 12. 3x − 6y = −9 13. −4x + 4y = 32
−3x + 3y = 6 −5x + 10y = 10 3x + 24 = 3y
Check
Equation 1
4x + 6y = −8
4(−2) + 6(0) =
?
−8
−8 = −8 ✓
Equation 2
x − 2y = −2
(−2) − 2(0) =
?
−2
−2 = −2 ✓
Multiply by 3.

Solving Equations by Graphing (pp. 243–248)5.5
Solve 3x − 1 = −2x + 4 by graphing. Check your solution.
Step 1 Write a system of linear equations using each side of the original equation.
3x − 1 = −2x + 4
Step 2 Graph the system.
x
y
2
4
1 3 5
y = 3x − 1
y = −2x + 4
−1
(1, 2)
The graphs intersect at (1, 2).
So, the solution of the equation is x = 1.
Solve the equation by graphing. Check your solution(s).
14. 1
—3
x + 5 = −2x − 2 15. ∣x + 1∣ = ∣−x − 9∣ 16. ∣2x − 8∣ = ∣x + 5∣
Check
3x − 1 = −2x + 4
3(1) − 1 =
?
−2(1) + 4
2 = 2 ✓
y = −2x + 4
h th t
y = 3x − 1
Graphing Linear Inequalities in Two Variables (pp. 249–254)5.6
Graph 4x + 2y ≥≥ −6 in a coordinate plane.
Step 1 Graph 4x + 2y = −6, or y = −2x − 3. Use a solid line
x
y
1
−3
2−2
because the inequality symbol is ≥.
Step 2 Test (0, 0).
4x + 2y ≥ −6 Write the inequality.
4(0) + 2(0) ≥
?
−6 Substitute.
0 ≥ −6 ✓ Simplify.
Step 3 Because (0, 0) is a solution, shade the half-plane that contains (0, 0).
Graph the inequality in a coordinate plane.
17. y > −4 18. −9x + 3y ≥ 3 19. 5x + 10y < 40
Systems of Linear Inequalities (pp. 255–262)5.7
Graph the system. y < x − 2 Inequality 1
y ≥ 2x − 4 Inequality 2
Step 2 Find the intersection of the
half-planes. One solution is (0, −3).
20. y ≤ x − 3 21. y > −2x + 3 22. x + 3y > 6
y ≥ x + 1 y ≥ 1
—4
x − 1 2x + y < 7
x
y
1
−4
3−2
(0, −3)
The solution is the
purple-shaded region.

Chapter 5 Chapter Test 267
Chapter Test
55Solve the system of linear equations using any method. Explain why you chose
the method.
1. 8x + 3y = −9 2. 1
—2
x + y = −6 3. y = 4x + 4
−8x + y = 29 y = 3
—5
x + 5 −8x + 2y = 8
4. x = y − 11 5. 6x − 4y = 9 6. y = 5x − 7
x − 3y = 1 9x − 6y = 15 −4x + y = −1
7. Write a system of linear inequalities so the points (1, 2) and (4, −3) are solutions of the
system, but the point (−2, 8) is not a solution of the system.
8. How is solving the equation ∣2x + 1∣ = ∣x − 7∣ by graphing similar to solving the
equation 4x + 3 = −2x + 9 by graphing? How is it different?
9. y > 1
—2
x + 4 10. x + y < 1 11. y ≥ −
2
—3 x + 1
2y ≤ x + 4 5x + y > 4 −3x + y > −2
12. You pay $45.50 for 10 gallons of gasoline and 2 quarts of oil at a gas station.
Your friend pays $22.75 for 5 gallons of the same gasoline and 1 quart of the
same oil.
a. Is there enough information to determine the cost of 1 gallon of gasoline
and 1 quart of oil? Explain.
b. The receipt shown is for buying the same gasoline and same oil. Is there
now enough information to determine the cost of 1 gallon of gasoline and
1 quart of oil? Explain.
c. Determine the cost of 1 gallon of gasoline and 1 quart of oil.
13. Describe the advantages and disadvantages of solving a system of linear
equations by graphing.
14. You have at most $60 to spend on trophies and medals to give as
prizes for a contest.
a. Write and graph an inequality that represents the numbers of
trophies and medals you can buy. Identify and interpret a solution
of the inequality.
b. You want to purchase at least 6 items. Write and graph a system
that represents the situation. How many of each item can you buy?
15. Compare the slopes and y-intercepts of the graphs of the equations
in the linear system 8x + 4y = 12 and 3y = −6x − 15 to determine
whether the system has one solution, no solution, or infinitely many
solutions. Explain.
Trophies
$12 each
Medals
$3 each

55 Cumulative Assessment
1. The graph of which equation is shown?
x
y
2
−4
−8
84
(2, 0)
(0, −9)
○A 9x − 2y = −18
○B −9x − 2y = 18
○C 9x + 2y = 18
○D −9x + 2y = −18
2. A van rental company rents out 6-, 8-, 12-, and 16-passenger vans. The function
C(x) = 100 + 5x represents the cost C (in dollars) of renting an x-passenger van for a
day. Choose the numbers that are in the range of the function.
130 200190180170160150140
3. Fill in the system of linear inequalities with <, ≤, >, or ≥ so that the graph represents
the system.
y 3x − 2
x
y
4
2
42
y −x + 5
4. Your friend claims to be able to fill in each box with a constant so that when you set
each side of the equation equal to y and graph the resulting equations, the lines will
intersect exactly once. Do you support your friend’s claim? Explain.
4x + = 4x +
5. The tables represent the numbers of items sold at a concession stand on days with
different average temperatures. Determine whether the data represented by each table
show a positive, a negative, or no correlation.
Temperature (°F), x 14 27 32 41 48 62 73
Cups of hot chocolate, y 35 28 22 9 4 2 1
Temperature (°F), x 14 27 32 41 48 62 73
Bottles of sports drink, y 8 12 13 16 19 27 29

Chapter 5 Cumulative Assessment 269
6. Which two equations form a system of linear equations that has no solution?
y = 3x + 2 y =
1
—3x + 2 y = 2x + 3 y = 3x +
1
—2
7. Fill in a value for a so that each statement is true for the equation ax − 8 = 4 − x.
a. When a = , the solution is x = −2.
b. When a = , the solution is x = 12.
c. When a = , the solution is x = 3.
8. Which ordered pair is a solution of the linear inequality whose graph is shown?
○A (1, 1)
x
y
2
−2
2−2
○B (−1, 1)
○C (−1, −1)
○D (1, −1)
9. Which of the systems of linear equations are equivalent?
4x − 5y = 3
2x + 15y = −1
4x − 5y = 3
−4x − 30y = 2
4x − 5y = 3
4x + 30y = −1
12x − 15y = 9
2x + 15y = −1
10. The graph shows the amounts y (in dollars) that a referee earns
for refereeing x high school volleyball games.
a. Does the graph represent a linear or nonlinear function?
Explain.
b. Describe the domain of the function. Is the domain discrete
or continuous?
c. Write a function that models the data.
d. Can the referee earn exactly $500? Explain.
Amountearned(dollars)
Games
Volleyball Referee
0 2 4 6 x
y
0
60
120
180
240

Ch 5 book - systems of linear equations

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