Chapter 5 - MDHS-MathChapter 5 5.1 Monitoring Progress (pp. 236–238)4. a. x −3 −2 −1 01 y =...

Copyright © Big Ideas Learning, LLC Algebra 1 227All rights reserved. Worked-Out Solutions

Chapter 5

Chapter 5 Maintaining Mathematical Profi ciency (p. 233)

1. y + 4 = x

y + 4 − 4 = x − 4

y = x − 4

m = 1, b = −4

x

y2

−2

42−2−4

−6

2. 6x − y = −1

6x − 6x − y = −1 − 6x

−y = −6x − 1

−y — −1

= −6x − 1

— −1

y = 6x + 1

m = 6, b = 1

x

y

4

8

42−2−4

3. 4x + 5y = 20

4x − 4x + 5y = 20 − 4x

5y = −4x + 20

5y — 5 =

−4x + 20 —

5

y = − 4 —

5 x + 4

m = − 4 —

5 , b = 4

x

y

2

−4

−2

42−2

4. −2y + 12 = −3x

−2y + 12 − 12 = −3x − 12

−2y = −3x − 12

−2y — −2

= −3x − 12

— −2

y = 3 —

2 x + 6

m = 3 —

2 , b = 6

x

y

2

4

6

−2

2−2−6

5. m + 4 > 9

− 4 − 4 m > 5

The solution is m > 5.

0 2 4 6 8

5

6. 24 ≤ −6t

24

— −6

≥ −6t

— −6

−4 ≥ t

The solution is t ≤ −4.

0−2−4−6−8

7. 2a − 5 ≤ 13

+ 5 + 5 2a ≤ 18

2a — 2 ≤

18 —

2

a ≤ 9

The solution is a ≤ 9.

0 4 8 12 16

9

8. −5z + 1 < −14

− 1 − 1 −5z < −15

−5z — −5

> −15

— −5

z > 3

The solution is z > 3.

0 2 4 6 8

3

228 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.

Chapter 5

9. 4k − 16 < k + 2

− k − k 3k − 16 < 2

+ 16 + 16

3k < 18

3k — 3 <

18 —

3

k < 6

The solution is k < 6.

0 2 4 6 8

10. 7w + 12 ≥ 2w − 3

− 2w − 2w

5w + 12 ≥ −3

− 12 − 12

5w ≥ −15

5w — 5 ≥

−15 —

5

w ≥ −3

The solution is w ≥ −3.

0−2−4−6

−3

11. The lines intersect at the point (a, b).

Mathematical Practices (p. 234)

1.

6

−6

−6

2

IntersectionX=0 Y=-3

The point of intersection is (0, −3).

2.

6

−4

−6

4

IntersectionX=1.5 Y=-.5

The point of intersection is (1.5, −0.5).

3. 3x − 2y = 2 2x − y = 2

3x − 3x − 2y = 2 − 3x 2x − 2x − y = 2 − 2x

2y = −3x + 2 −y = −2x + 2

−2y — −2

= −3x + 2

— −2

−y

— −1

= −2x + 2

— −1

y = 3 —

2 x − 1 y = 2x − 2

6

−4

−6

4

IntersectionX=2 Y=2

The point of intersection is (2, 2).

Section 5.1

5.1 Explorations (p. 235)

1. a. An equation that represents the costs is C = 15x + 600.

b. An equation that represents the revenue is R = 75x.

c. A system of linear equations for this problem is

C = 15x + 600

R = 75x.

2. a. x (nights) 0 1 2 3 4 5 6

C (dollars) 600 615 630 645 660 675 690

R (dollars) 0 75 150 225 300 375 450

x (nights) 7 8 9 10 11

C (dollars) 705 720 735 750 765

R (dollars) 525 600 675 750 825

b. Your family must rent the bedroom for 10 nights before

breaking even.

c.

400

600

800

200

00

4 6 8 10 122

Days

Cost/Rev

enue

y = 15x + 600

y = 75x

x

y

d. The point of intersection is (10, 750). This point

represents where both functions have the same x- and

y-values. This is the same as the break-even point.

3. Graph both equations and fi nd the point of intersection.

To check your solution, substitute the x-coordinate for x

in both of the equations and verify that both results are the

y-coordinate of the point of intersection.


Chapter 5

4. a. x −3 −2 −1 0 1

y = −4.3x − 1.3 11.6 7.3 3 −1.3 −5.6

y = 1.7x + 4.7 −0.4 1.3 3 4.7 6.4

The solution is (−1, 3).

Sample answer: A table was chosen because the slopes

and intercepts are decimals and would be diffi cult to graph

accurately.

Check

6

−4

−6

4

IntersectionX=-1 Y=3

b.

x

y

4

6

8

2

4 62−2

y = x

y = −3x + 8

(2, 2)

The solution is (2, 2).

Sample answer: The graphing method was chosen

because both equations are in slope-intercept form with

slopes and y-intercepts that are whole numbers.

Check

6

−4

−6

4

IntersectionX=2 Y=2

c.

x

y

2

−2

2−4−6

y = −x − 1

y = 3x + 5

(−1.5, 0.5)

The solution is (−1.5, 0.5).

Sample answer: The graphing method was chosen

because the equations are in slope-intercept form and the

slopes and y-intercepts are whole numbers.

Check

6

−4

−6

4

IntersectionX=-1.5 Y=.5

5.1 Monitoring Progress (pp. 236–238)

1. Equation 1 Equation 2

2x + y = 0 −x + 2y = 5

2(1) + (−2) =?

0 −(1) + 2(−2) =?

5

2 − 2 =?

0 −1 − 4 =?

5

0 = 0 ✓ −5 ≠ 5 ✗

The ordered pair (1, −2) is a solution of the fi rst equation,

but it is not a solution of the second equation. So, (1, −2)

is not a solution of the linear system.


y = 3x + 1 y = −x + 5

4 =?

3(1) + 1 4 =?

−(1) + 5

4 =?

3 + 1 4 =?

−1 + 5

4 = 4 ✓ 4 = 4 ✓

Because the ordered pair (1, 4) is a solution of each equation,

it is a solution of the linear system.

3.

x

y

2

−4

42 6−2

y = −x + 4

y = x − 2

(3, 1)

Check Equation 1 Equation 2

y = x − 2 y = −x + 4

1 =?

3 − 2 1 =?

−3 + 4

1 = 1 ✓ 1 = 1 ✓


4.

x

y4

2

−2

−2−4−6

y = x + 312

y = − x − 532

(−4, 1)


y = 1 —

2 x + 3 y = −

3 — 2 x − 5

1 =?

1 —

2 (−4) + 3 1 =

? −

3 — 2 (−4) − 5

1 =?

−2 + 3 1 =?

6 − 5

1 = 1 ✓ 1 = 1 ✓



Chapter 5

5. 2x + y = 5 3x − 2y = 4

2x − 2x + y = 5 − 2x 3x − 3x − 2y = 4 − 3x

y = −2x + 5 −2y = −3x + 4

−2y

— − 2

= −3x + 4

— −2

x

y

2

−4

−2

4 62−2

y = x − 232

y = −2x + 5

(2, 1) y =

3 —

2 x − 2


2x + y = 5 3x − 2y = 4

2(2) + 1 =?

5 3(2) − 2(1) =?

4

4 + 1 =?

5 6 − 2 =?

4

5 = 5 ✓ 4 = 4 ✓


6. Words Number

of math

exercises

+Number

of science

exercises

=

18

Number

of math

exercises

=

6 +Number of

science

exercises

Variables Let x be the number of science exercises, and let y

be the number of math exercises.

System y + x = 18

x

y

8

12

16

4

08 12 1640

y = x + 6

y = −x + 18

(6, 12)

y = 6 + x

y + x = 18

y + x − x = 18 − x

y = −x + 18


y + x = 18 y = 6 + x

12 + 6 =?

18 12 =?

6 + 6

18 = 18 ✓ 12 = 12 ✓

The solution is (6, 12). So, you have 6 exercises in science

and 12 exercises in math.

5.1 Exercises (pp. 239 –240)

Vocabulary and Core Concept Check

1. yes; Because 5y − 2x = 18 and 6x = −4y − 10 are both

linear equations in the same variables, they form a system of

linear equations.

2. The one that is different is “Solve each equation for y.” When

you solve both equations for y, you get y = 2x + 2 and

y = 4x + 6. The other three ask for the solution of the

system, which is (−2, −2).

−4x + 2y = 4 4x − y = −6

−4x + 4x + 2y = 4 + 4x 4x − 4x − y = −6 − 4x

2y = 4x + 4 −y = −4x − 6

2y — 2 =

4x + 4 —

2

−y —

−1 =

−4x − 6 —

−1

y = 2x + 2 y = 4x + 6

x

y

2

−4

−2

2−2−4−6

y = 2x + 2

y = 4x + 6

(−2, −2)


−4x + 2y = 4 4x − y = −6

−4(−2) + 2(−2) =?

4 4(−2) − (−2) =?

−6

8 − 4 =?

4 −8 + 2 =?

−6

4 = 4 ✓ −6 = −6 ✓

Monitoring Progress and Modeling with Mathematics


x + y = 8 3x − y = 0

2 + 6 =?

8 3(2) − 6 =?

0

8 = 8 ✓ 6 − 6 =?

0

0 = 0 ✓

Because the ordered pair (2, 6) is a solution of each equation,

it is a solution of the linear system.


x − y = 6 2x − 10y = 4

8 − 2 =?

6 2(8) − 10(2) =?

4

6 = 6 ✓ 16 − 20 =?

4

−4 ≠ 4 ✗

The ordered pair (8, 2) is a solution of the fi rst equation, but

it is not a solution of the second equation. So, (8, 2) is not a

solution of the linear system.


Chapter 5


y = −7x − 4 y = 8x + 5

3 =?

−7(−1) − 4 3 =?

8(−1) + 5

3 =?

7 − 4 3 =?

−8 + 5

3 = 3 ✓ 3 ≠ −3 ✗

The ordered pair (−1, 3) is a solution of the fi rst equation,

but it is not a solution of the second equation. So, (−1, 3) is

not a solution of the linear system.


y = 2x + 6 y = −3x − 14

−2 =?

2(−4) + 6 −2 =?

−3(−4) − 14

−2 =?

−8 + 6 −2 =?

12 − 14

−2 = −2 ✓ −2 = −2 ✓

Because the ordered pair (−4, −2) is a solution of each

equation, it is a solution of the linear system.


6x + 5y = −7 2x − 4y = −8

6(−2) + 5(1) =?

−7 2(−2) − 4(1) =?

−8

−12 + 5 =?

−7 −4 − 4 = −8 ✓

−7 = −7 ✓

Because the ordered pair (−2, 1) is a solution of each



6x + 3y = 12 4x + y = 14

6(5) + 3(−6) =?

12 4(5) + (−6) =?

14

30 − 18 =?

12 20 − 6 =?

14 ✓

12 = 12 ✓ 14 = 14 ✓

Because the ordered pair (5, −6) is a solution of each


9. The lines appear to intersect at (1, −3).


x − y = 4 4x + y = 1

1 − (−3) =?

4 4(1) + (−3) =?

1

1 + 3 =?

4 4 − 3 =?

1

4 = 4 ✓ 1 = 1 ✓

The solution is (1, −3).

10. The lines appear to intersect at (3, 2).


x + y = 5 y − 2x = −4

3 + 2 =?

5 2 − 2(3) =?

−4

5 = 5 ✓ 2 − 6 =?

−4

−4 = −4 ✓


11. The lines appear to intersect at (−4, 5).


6y + 3x = 18 −x + 4y = 24

6(5) + 3(−4) =?

18 −(−4) + 4(5) =?

24

30 − 12 =?

18 4 + 20 =?

24

18 = 18 ✓ 24 = 24 ✓




2x − y = −2 2x + 4y = 8

2(0) − 2 =?

−2 2(0) + 4(2) =?

8

0 − 2 =?

−2 0 + 8 =?

8

−2 = −2 ✓ 8 = 8 ✓


13.

x

y

4

6

2

4 62

y = −x + 7

y = x + 1

(3, 4)


y = −x + 7 y = x + 1

4 =?

−3 + 7 4 =?

3 + 1

4 = 4 ✓ 4 = 4 ✓



Chapter 5

14.

x

y8

−4

8 12−4

y = −x + 4

y = 2x − 8

(4, 0)


y = −x + 4 y = 2x − 8

0 =?

−4 + 4 0 =?

2(4) − 8

0 = 0 ✓ 0 =?

8 − 8

0 = 0 ✓


15.

x

y

−4

−4−12

y = x + 523

y = x + 213

(−9, −1)


y = 1 —

3 x + 2 y =

2 —

3 x + 5

−1 =?

1 —

3 (−9) + 2 −1 =

? 2 —

3 (−9) + 5

−1 =?

−3 + 2 −1 =?

−6 + 5

−1 = −1 ✓ −1 = −1 ✓

The solution is (−9, −1).

16.

x

y

8

4

8 12

y = x − 434

y = − x + 1112

(12, 5)


y = 3 —

4 x − 4 y = −

1 — 2 x + 11

5 =?

3 —

4 (12) − 4 5 =

? −

1 — 2 (12) + 11

5 =?

9 − 4 5 =?

−6 + 11

5 = 5 ✓ 5 = 5 ✓


17. 9x + 3y = −3 2x − y = −4

9x − 9x + 3y = −3 − 9x 2x − 2x − y = −4 − 2x

3y = −9x − 3 −y = −2x − 4

3y — 3 =

−9x − 3 —

3

−y —

−1 =

−2x − 4 —

−1

y = −3x − 1 y = 2x + 4

x

y

4

2

−2

42−4

y = −3x − 1

y = 2x + 4(−1, 2)


9x + 3y = −3 2x − y = −4

9(−1) + 3(2) =?

−3 2(−1) − 2 =?

−4

−9 + 6 =?

−3 −2 − 2 =?

−4

−3 = −3 ✓ −4 = −4 ✓


18. 4x − 4y = 20

4x − 4x − 4y = 20 − 4x

−4y = −4x + 20

−4y — −4

= −4x + 20

— −4

y = x − 5

x

y

−4

−8

−2

2−2

y = −5

y = x − 5

(0, −5)


4x − 4y = 20 y = −5

4(0) − 4(−5) =?

20 −5 = −5 ✓

0 + 20 =?

20

20 = 20 ✓



Chapter 5

19. x − 4y = −4 −3x − 4y = 12

x − x − 4y = −4 − x −3x + 3x − 4y = 12 + 3x

−4y = −x − 4 −4y = 3x + 12

−4y

— −4

= −x − 4

— −4

−4y

— −4

= 3x + 12

— −4

y = 1 —

4 x + 1 y = −

3 —

4 x − 3

x

y4

2

−4

−2

y = x + 114

y = − x − 334

(−4, 0)


x − 4y = −4 −3x − 4y = 12

−4 − 4(0) =?

−4 −3(−4) − 4(0) =?

12

−4 − 0 =?

−4 12 − 0 =?

12

−4 = −4 ✓ 12 = 12 ✓


20. 3y + 4x = 3 x + 3y = −6

3y + 4x − 4x = 3 − 4x x − x + 3y = −6 − x

3y = −4x + 3 3y = −x − 6

3y

— 3 =

−4x + 3 —

3

3y —

3 =

−x − 6 —

3

y = − 4 —

3 x + 1 y = −

1 —

3 x − 2

x

y

−4

−6

4 62

y = − x + 143

y = − x − 213

(3, −3)


3y + 4x = 3 x + 3y = −6

3(−3) + 4(3) =?

3 3 + 3(−3) =?

−6

−9 + 12 =?

3 3 − 9 =?

−6

3 = 3 ✓ −6 = −6 ✓


21. The graph of 2x − 3y = 3 should have a y-intercept of −1,

not −3.

x − 3y = 6 2x − 3y = 3

x − x − 3y = 6 − x 2x − 2x − 3y = 3 − 2x

−3y = −x + 6 −3y = −2x + 3

−3y — −3

= −x + 6

— −3

−3y

— −3

= −2x + 3

— −3

y = 1 —

3 x − 2 y =

2 —

3 x − 1

x

y

2

−4

2−2−4−6

y = x − 213

y = x − 123


x − 3y = 6 2x − 3y = 3

−3 − 3(−3) =?

6 2(−3) − 3(−3) =?

3

−3 + 9 =?

6 −6 + 9 =?

3

6 = 6 ✓ 3 = 3 ✓

The solution of the linear system x − 3y = 6 and 2x − 3y = 3

is (−3, −3).

22. The solution of the system should be the ordered pair for

the point of intersection, not just the x-value where the

lines intersect.

x

y

4

6

2

4 62

y = x + 1

y = 2x − 1

(2, 3)


y = 2x − 1 y = x + 1

3 =?

2(2) − 1 3 =?

2 + 1

3 =?

4 − 1 3 = 3 ✓

3 = 3 ✓

The solution of the linear system y = 2x − 1 and y = x + 1

is (2, 3).


Chapter 5

23. 0.2x + 0.4y = 4

0.2x − 0.2x + 0.4y = 4 − 0.2x

0.4y = −0.2x + 4

0.4y — 0.4

= −0.2x + 4

— 0.4

y = −0.5x + 10

−0.6x + 0.6y = −3

−0.6x + 0.6x + 0.6y = −3 + 0.6x

0.6y = 0.6x − 3

0.6y

— 0.6

= 0.6x − 3

— 0.6

y = x − 5

120

0

8

IntersectionX=10 Y=5


24. −1.6x − 3.2y = −24

−1.6x + 1.6x − 3.2y = −24 + 1.6x

−3.2y = 1.6x − 24

−3.2y — −3.2

= 1.6x − 24

— −3.2

y = −0.5x + 7.5

2.6x + 2.6y = 26

2.6x − 2.6x + 2.6y = 26 − 2.6x

2.6y = −2.6x + 26

2.6y

— 2.6

= −2.6x + 26

— 2.6

y = −x + 10

120

0

8

IntersectionX=5 Y=5


25. −7x + 6y = 0 0.5x + y = 2

−7x + 7x + 6y = 0 + 7x 0.5x − 0.5x + y = 2 − 0.5x

6y = 7x y = −0.5x + 2

6y — 6 =

7x —

6

6

−4

−6

4

IntersectionX=1.2 Y=1.4

y = 7 —

6 x

The solution is (1.2, 1.4).

26. 4x − y = 1.5 2x + y = 1.5

4x − 4x − y = 1.5 − 4x 2x − 2x + y = 1.5 − 2x

−y = −4x + 1.5 y = −2x + 1.5

−y — −1

= −4x + 1.5

— −1

y = 4x − 1.5

6

−4

−6

4

IntersectionX=.5 Y=.5

The solution is (0.5, 0.5).

27. Words Time on

elliptical +Time on

bike = 40

8 ⋅ Time on

elliptical + 6 ⋅ Time on

bike = 300

Variables Let x be how much time (in minutes) you

spend on the elliptical trainer, and let y be

how much time (in minutes) you spend on the

stationary bike.

System x + y = 40

8x + 6y = 300

x + y = 40 8x + 6y = 300

x − x + y = 40 − x 8x − 8x + 6y = 300 − 8x

y = −x + 40 6y = −8x + 300

20 40 600 x

y

40

20

0

Time (minutes)

Exercise with an EllipticalTrainer and Stationary Bike

Cal

ori

es b

urn

ed

(30, 10)

y = −x + 40

y = − x + 5043

6y — 6 =

−8x + 300 —

6

y = − 4 —

3 x + 50


x + y = 40 8x + 6y = 300

30 + 10 =?

40 8(30) + 6(10) =?

300

40 = 40 ✓ 240 + 60 =?

300

300 = 300 ✓

The solution is (30, 10). So, you should spend 30 minutes on

the elliptical trainer and 10 minutes on the stationary bike.


Chapter 5

28. Words Number of

small candles +Number of

large candles = 28

4 ⋅ Number of

small candles + 6 ⋅ Number of

large candles = 144

Variables Let x be how many small candles you sell, and

let y be how many large candles you sell.

System x + y = 28

4x + 6y = 144

x + y = 28 4x + 6y = 144

x − x + y = 28 − x 4x − 4x + 6y = 144 − 4x

y = −x + 28 6y = −4x + 144

8 16 24 320 x

y

16

24

32

8

0

Number of candles

Selling Candles at a Craft Fair

Mo

ney

ear

ned

(d

olla

rs)

y = −x + 28

y = − x + 2423

(12, 16)

6y — 6 =

−4x + 144 —

6

y = − 2 —

3 x + 24


x + y = 28 4x + 6y = 144

12 + 16 =?

28 4(12) + 6(16) =?

144

28 = 28 ✓ 48 + 96 =?

144

144 = 144 ✓

The solution is (12, 16). So, you sell 12 small candles and

16 large candles.

29. A =ℓ⋅ w P = 2ℓ+ 2w

y = 6 (3x − 3) y = 2(6) + 2 (3x − 3)

y = 6 (3x) − 6(3) y = 12 + 2 (3x) − 2(3)

y = 18x − 18 y = 12 + 6x − 6

y = 6x + 6

x

y

12

18

6

02 310

y = 6x + 6

y = 18x − 18

(2, 18)


y = 18x − 18 y = 6x + 6

18 =?

18(2) − 18 18 =?

6(2) + 6

18 =?

36 − 18 18 =?

12 + 6

18 = 18 ✓ 18 = 18 ✓

The solution is (2, 18). This means that when x = 2, the

area of the rectangle will be 18 square centimeters, and

the perimeter will be 18 centimeters.

30.

200

300

400

100

00 4 6 82

Time (months)

Bal

ance

(d

olla

rs)

y = 25x + 250

x

y

Sample answers: A linear equation that could represent my

account balance is y = 12.5x + 325. The slope is 12.5, and

the y-intercept is 325. So, you must deposit $325 initially

and then deposit an additional $12.50 each month for 6

months.

31. a. x + 2 = 3x − 4

−x −x

2 = 2x − 4

+4 +4

6 = 2x

6 —

2 =

2x —

2

3 = x

The solution is x = 3.

b.

x

y

4

6

−2

4 62

y = x + 2

y = 3x − 4

(3, 5)


y = x + 2 y = 3x − 4

5 =?

3 + 2 5 =?

3(3) − 4

5 = 5 ✓ 5 =?

9 − 4

5 = 5 ✓



Chapter 5

c. The x-value of the solution in part (b) is the solution of the

equation in part (a). The expressions with x on the right

sides of the equations in part (b) make up each side of

the equation in part (a). So, when x = 3, each side of the

equation in part (a) is equal to the y-value of the solution

in part (b).

32. a. If the teacher buys 20 binders, companies B and C charge

the same. For 35 binders, companies A and B charge

the same. For 50 binders, companies A and C charge the

same. These are the x-values of the points of intersection

in the graph.

b. Each of the answers in part (a) is the x-value of a solution

of a system of two linear equations formed by each pair of

equations.

33. a. Let x be the time (in hours) you and your friend have

been hiking, and let y be the distance (in miles) from the

trailhead. A system of linear equations that represents this

situation is

y = 5x

y = 3x + 3.

8

12

16

4

00 2 3 41

Time (hours)

Dis

tan

ce (

mile

s)

y = 3x + 3

y = 5x

x

y

b. no; According to the graph, after 1 hour of hiking, you

will be 5 miles from the trailhead, and your friend will

be 6 miles from the trailhead. So, you will still be 1 mile

apart. The graphs intersect at (1.5, 7.5). So, after hiking

for 1.5 hours, you will meet at a location that is 7.5 miles

from the trailhead.

Maintaining Mathematical Profi ciency

34. 10x + 5y = 5x + 20

10x − 10x + 5y = 5x − 10x + 20

5y = −5x + 20

5y

— 5 =

−5x + 20 —

5

y = −x + 4

The rewritten literal equation is y = −x + 4.

35. 9x + 18 = 6y − 3x

9x + 3x + 18 = 6y − 3x + 3x

12x + 18 = 6y

12x + 18

— 6 =

6y —

6

2x + 3 = y

The rewritten literal equation is y = 2x + 3.

36. 3 — 4 x +

1 —

4 y = 5

3 — 4 x −

3 —

4 x +

1 —

4 y = 5 −

3 —

4 x

1 — 4 y = −

3 —

4 x + 5

4 ⋅ 1 —

4 y = 4 ⋅ ( − 3 — 4 x + 5 )

y = 4 ( − 3 — 4 x ) + 4(5)

y = −3x + 20

The rewritten literal equation is y = −3x + 20.

Section 5.2


1. a. Method 1 Method 2

Equation 1 Equation 1

x + y = −7 x + y = −7

x + y − y = −7 − y x − x + y = −7 − x

x = −y − 7 y = −x − 7


−5x + y = 5 −5x + y = 5

−5(−y − 7) + y = 5 −5x + (−x − 7) = 5

−5(−y) − 5(−7) + y = 5 −5x − x − 7 = 5

5y + 35 + y = 5 −6x − 7 = 5 6y + 35 = 5 +7 +7

−35 −35 −6x = 12

6y = −30 −6x

— −6

= 12

— −6

6y — 6 = −30

— 6 x = −2

y = −5


x + y = −7 x + y = −7

x + (−5) = −7 −2 + y = −7

+5 +5 +2 +2

x = −2 y = −5

The solution (−2, −5) is the same using both methods.

For this problem, Method 2 is preferred because both

equations can be solved for y easily.


Chapter 5

b. Method 1 Method 2


x − 6y = −11 x − 6y = −11

x − 6y + 6y = −11 + 6y x − x − 6y = −11 − x

x = 6y − 11 −6y = −x − 11

−6y

— −6

= −x − 11

— −6

y = 1 —

6 x +

11 —

6


3x + 2y = 7 3x + 2y = 7

3(6y − 11) + 2y = 7 3x + 2 ( 1 — 6 x +

11 —

6 ) = 7

3(6y) − 3(11) + 2y = 7 3x + 2 ( 1 — 6 x ) + 2 ( 11

— 6 ) = 7

18y − 33 + 2y = 7 3x + 1 —

3 x +

11 —

3 = 7

20y − 33 = 7 10

— 3 x +

11 —

3 = 7

+33 +33 − 11

— 3 −

11 —

3

20y = 40

20y

— 20

= 40

— 20

10

— 3 x = 10

— 3

y = 2 3 —

10 ∙ 10

— 3 x = 3 —

10 ∙ 10

— 3

x = 1


x − 6y = −11 x − 6y = −11

x − 6(2) = −11 1 − 6y = −11

x − 12 = −11 −1 −1

+12 +12 −6y = −12

x = 1 −6y

— −6

= −12

— −6

y = 2

The solution (1, 2) is the same using both methods.

For this system, Method 1 is preferred because the fi rst

equation can be solved for x easily.

c. Method 1 Method 2


3x − 5y = −18 4x + y = −1

3x − 5y + 5y = −18 + 5y 4x − 4x + y = −1 − 4x

3x = 5y − 18 y = −4x − 1

3x

— 3 =

5y − 18 —

3

x = 5 —

3 y − 6


4x + y = −1 3x − 5y = −18

4 ( 5 — 3 y − 6 ) + y = −1 3x − 5(−4x − 1) = −18

4 ( 5 — 3 y ) − 4(6) + y = −1 3x − 5(−4x) − 5(−1) = −18

20

— 3 y − 24 + y = −1 3x + 20x + 5 = −18

23

— 3 y − 24 = −1 23x + 5 = −18

+24 +24 −5 −5

23

— 3 y = 23 23x = −23

3 —

23 ⋅

23 —

3 y =

3 —

23 ⋅ 23

23x —

23 =

−23 —

23

y = 3 x = −1


4x + y = −1 4x + y = −1

4x + 3 = −1 4(−1) + y = −1

−3 −3 −4 + y = −1

4x = −4 +4 +4

4x

— 4 =

−4 —

4 y = 3

x = −1

The solution (−1, 3) is the same using both methods.

For this system, Method 2 is preferred because the fi rst

equation can be solved for y easily.

2. a. Sample answer: randInt (−5, 5, 2)

{−1 3}

b. Sample answer: 2x − y = 2(−1) − 3

= −2 − 3

= −5

One linear equation that has (−1, 3) as a solution is

2x − y = −5.

x + 5y = −1 + 5(3)

= −1 + 15

= 14

Another linear equation that has (−1, 3) as a solution is

x + 5y = 14. So, a system of linear equations that has

(−1, 3) as its solution is

2x − y = −5.x + 5y = 14


Chapter 5

c. Sample answer: I would solve my system by Method 1.

First, I would solve Equation 2 for x, because x has a

coeffi cient of 1.


x + 5y = 14 2x − y = −5

x + 5y − 5y = 14 − 5y 2(−5y + 14) − y = −5

x = − 5y + 14 2(−5y) + 2(14) − y = −5

Equation 2 −10y + 28 − y = −5

x + 5y = 14 −11y + 28 = −5

x + 5(3) = 14 −28 −28

x + 15 = 14 −11y = −33

−15 −15 −11y

— −11

= −33

— −11

x = −1 y = 3

The solution of the system is (−1, 3).

3. To solve a system of linear equations, fi rst solve one of the

equations for one of the variables. Substitute the expression

for this variable into the other equation to fi nd the value of

the other variable. Then substitute this value into one of the

original equations to fi nd the value of the other variable.

4. a. Sample answer: Start by solving Equation 1 for x, because

x has a coeffi cient of 1. So, it will take fewer steps to

isolate x and I will get a relatively simple expression with

no fractions or decimals.


x + 2y = −7 2x − y = −9

x + 2y −2y = −7 − 2y 2(−2y − 7) − y = −9

x = −2y − 7 2(−2y) − 2(7) − y = −9

Equation 1 −4y − 14 − y = −9

x + 2y = −7 −5y − 14 = −9

x + 2(−1) = −7 +14 +14

x − 2 = −7 −5y = 5

+2 +2 −5y

— −5

= 5 —

−5

x = −5 y = −1


x + 2y = −7 2x − y = −9

−5 + 2(−1) =?

−7 2(−5) − (−1) =?

−9

−5 − 2 =?

−7 −10 + 1 =?

−9

−7 = −7 ✓ −9 = −9 ✓

The solution of the system is (−5, −1).

b. Sample answer: Start by solving Equation 2 for y, because

y has a coeffi cient of 1. So, it will take fewer steps to

isolate y and I will get a relatively simple expression with



2x + y = −2 x − 2y = −6

2x − 2x + y = −2 − 2x x − 2(−2x − 2) = −6

y = −2x − 2 x − 2(−2x) − 2(−2) = −6

Equation 2 x + 4x + 4 = −6

2x + y = −2 5x + 4 = −6

2(−2) + y = −2 −4 −4

−4 + y = −2 5x = −10

+4 +4 5x

— 5 =

−10 —

5

y = 2 x = −2


x − 2y = −6 2x + y = −2

−2 − 2(2) =?

−6 2(−2) + 2 =?

−2

−2 − 4 =?

−6 −4 + 2 =?

−2

−6 = −6 ✓ −2 = −2 ✓

The solution of the system is (−2, 2).

c. Sample answer: Start by solving Equation 2 for y, because

y has a coeffi cient of 1. So, it will take fewer steps to

isolate y and I will get a relatively simple expression with



−2x + y = −6 −3x + 2y = −10

−2x + 2x + y = −6 + 2x −3x + 2(2x − 6) = −10

y = 2x − 6 −3x + 2(2x) − 2(6) = −10

Equation 2 −3x + 4x − 12 = −10

−2x + y = −6 x − 12 = −10

−2(2) + y = −6 +12 +12

−4 + y = −6 x = 2

+4 +4

y = −2


−3x + 2y = −10 −2x + y = −6

−3(2) + 2(−2) =?

−10 −2(2) − 2 =?

−6

−6 − 4 =?

−10 −4 − 2 =?

−6

−10 = −10 ✓ −6 = −6 ✓

The solution of the system is (2, –2).


Chapter 5

d. Sample answer: Start by solving Equation 2 for x, because

x has a coeffi cient of 1. So, it will take fewer steps to

isolate x and I will get a relatively simple expression with



x – 3y = –3 3x + 2y = 13

x – 3y + 3y = –3 + 3y 3(3y – 3) + 2y = 13

x = 3y – 3 3(3y) – 3(3) + 2y = 13

Equation 2 9y – 9 + 2y = 13

x – 3y = –3 11y – 9 = 13

x – 3(2) = –3 +9 +9

x – 6 = –3 11y = 22

+6 +6 11y

— 11

= 22

— 11

x = 3 y = 2


3x + 2y = 13 x – 3y = –3

3(3) + 2(2) =?

13 3 – 3(2) =?

–3

9 + 4 =?

13 3 – 6 =?

–3

13 = 13 ✓ –3 = –3 ✓

The solution of the system is (3, 2).

e. Sample answer: Start by solving Equation 2 for x, because

x has a coeffi cient of −1. So, I will get a relatively simple

expression with no fractions or decimals.


−x − 3y = 8 3x − 2y = 9

−x − 3y + 3y = 8 + 3y 3(−3y − 8) − 2y = 9

−x = 3y + 8 3(−3y) − 3(8) − 2y = 9

−x

— −1

= 3y + 8

— −1

−9y − 24 − 2y = 9

x = −3y − 8 −11y − 24 = 9

Equation 1 +24 +24

3x − 2y = 9 −11y = 33

3x − 2(−3) = 9 −11y

— −11

= 33

— −11

3x + 6 = 9 y = −3

−6 −6

3x = 3

3x

— 3 =

3 —

3

x = 1


3x − 2y = 9 −x − 3y = 8

3(1) − 2(−3) =?

9 −1 − 3(−3) =?

8

3 + 6 =?

9 −1 + 9 =?

8

9 = 9 ✓ 8 = 8 ✓

The solution of the system is (1, –3).

f. Sample answer: Start by solving Equation 1 for y, because

y has a coeffi cient of −1. So, I will get a relatively simple

expression with no fractions or decimals.


3x − y = −6 4x + 5y = 11

3x − 3x − y = −6 − 3x 4x + 5(3x + 6) = 11

−y = −3x − 6 4x + 5(3x) + 5(6) = 11

−y

— −1

= −3x − 6

— −1

4x + 15x + 30 = 11

y = 3x + 6 19x + 30 = 11

Equation 2 −30 −30

4x + 5y = 11 19x = −19

4(−1) + 5y = 11 19x

— 19

= −19

— 19

−4 + 5y = 11 x = −1

+4 +4

5y = 15

5y

— 5 =

15 —

5

y = 3


3x − y = −6 4x + 5y = 11

3(−1) − 3 =?

−6 4(−1) + 5(3) =?

11

−3 − 3 =?

−6 −4 + 15 =?

11

−6 = −6 ✓ 11 = 11 ✓

The solution of the system is (–1, 3).


1. Substitute −4x for y in Equation 1 and solve for x.

y = 3x + 14

−4x = 3x + 14

−3x −3x

−7x = 14

−7x — −7

= 14

— −7

x = −2

Substitute −2 for x in Equation 2 and solve for y.

y = −4x

y = −4(−2)

y = 8

Check y = 3x + 14 y = −4x

8 =?

3(−2) + 14 8 =?

−4(−2)

8 =?

−6 + 14 8 = 8 ✓

8 = 8 ✓



Chapter 5

2. Substitute 1 —

2 x − 1 for y in Equation 1 and solve for x.

3x + 2y = 0

3x + 2 ( 1 — 2 x − 1 ) = 0

3x + 2 ( 1 — 2 x ) − 2(1) = 0

3x + x −2 = 0

4x −2 = 0

+2 +2

4x = 2

4x — 4 =

2 —

4

x = 1 —

2

Substitute 1 —

2 for x in Equation 2 and solve for y.

y = 1 —

2 x − 1

y = 1 —

2 ( 1 — 2 ) − 1

y = 1 —

4 − 1

y = − 3 —

4

Check 3x + 2y = 0 y = 1 —

2 x − 1

3 ( 1 — 2 ) + 2 ( −

3 —

4 ) =? 0 −

3 —

4 =

? 1 —

2 ( 1 — 2 ) − 1

3 —

2 −

3 —

2 =

? 0 −

3 —

4 =

? 1 —

4 −1

0 = 0 ✓ − 3 —

4 = −

3 —

4 ✓

The solution is ( 1 — 2 , −

3 —

4 ) .

3. Substitute 6y − 7 for x in Equation 2 and solve for y.

4x + y = −3

4(6y − 7) + y = −3

4(6y) − 4(7) + y = −3

24y − 28 + y = −3

25y − 28 = −3

+28 +28

25y = 25

25y — 25

= 25

— 25

y = 1

Substitute 1 for y in Equation 1 and solve for x.

x = 6y − 7

x = 6(1) − 7

x = 6 − 7

x = −1

Check x = 6y − 7 4x + y = −3

−1 =?

6(1) − 7 4(−1) + 1 =?

−3

−1 =?

6 − 7 −4 + 1 =?

−3

−1 = −1 ✓ −3 = −3 ✓


4. Step 1 x + y = −2

x + y − y = −2 − y

x = −y − 2

Step 2 −3x + y = 6

−3(−y − 2) + y = 6

−3(−y) − 3(−2) + y = 6

3y + 6 + y = 6

4y + 6 = 6

−6 −6

4y = 0

4y

— 4 =

0 —

4

y = 0

Step 3 x + y = −2

x + 0 = −2

x = −2

Check x + y = −2 −3x + y = 6

−2 + 0 =?

−2 −3(−2) + 0 =?

6

−2 = −2 ✓ 6 + 0 =?

6

6 = 6 ✓


5. Step 1 −x + y = −4

−x + x + y = −4 + x

y = x − 4

Step 2 4x − y = 10

4x − (x − 4) = 10

4x − x + 4 = 10

3x + 4 = 10

−4 −4

3x = 6

3x

— 3 =

6 —

3

x = 2

Step 3 −x + y = −4

−2 + y = −4

+2 +2

y = −2


Chapter 5

Check −x + y = −4 4x − y = 10

−2 − 2 =?

−4 4(2) − (−2) =?

10

−4 = −4 ✓ 8 + 2 =?

10

10 = 10 ✓


6. Step 1 2x − y = −5

2x − 2x − y = −5 − 2x

−y = −2x − 5

−y

— −1

= −2x − 5

— −1

y = 2x + 5

Step 2 3x − y = 1

3x − (2x + 5) = 1

3x −2x − 5 = 1

x − 5 = 1

+5 +5

x = 6

Step 3 3x − y = 1

3(6) − y = 1

18 − y = 1

−18 −18

−y = −17

−y

— −1

= −17

— −1

y = 17

Check 2x − y = −5 3x − y = 1

2(6) − 17 =?

−5 3(6) − 17 =?

1

12 − 17 =?

−5 18 − 17 =?

1

−5 = −5 ✓ 1 = 1 ✓


7. Step 1 x − 2y = 7

x − 2y + 2y = 7 + 2y

x = 2y + 7

Step 2 3x − 2y = 3

3(2y + 7) − 2y = 3

3(2y) + 3(7) − 2y = 3

6y + 21 − 2y = 3

4y + 21 = 3

−21 −21

4y = −18

4y

— 4 = −

18 —

4

y = − 9 —

2

Step 3 x −2y = 7

x − 2 ( − 9 —

2 ) = 7

x + 9 = 7

−9 −9

x = −2

Check x − 2y = 7 3x − 2y = 3

−2 − 2 ( − 9 —

2 ) =? 7 3(−2) − 2 ( −

9 —

2 ) =? 3

−2 + 9 =?

7 −6 + 9 =?

3

7 = 7 ✓ 3 = 3 ✓

The solution is ( −2, − 9 —

2 ) .

8. Words Number of

students in

drama club+

Number of

students in

yearbook club= 64

Number of

students in

drama club= 10 +

Number of

students in

yearbook club

Variables Let x be the number of students in the drama

club, and let y be the number of students in the

yearbook club.

System x + y = 64 Equation 1

x = 10 + y Equation 2

Substitute 10 + y for x in Equation 1 and solve for y.

x + y = 64

(10 + y) + y = 64

10 + 2y = 64

−10 −10

2y = 54

2y — 2 =

54 —

2

y = 27


x = 10 + y

x = 10 + 27

x = 37

The solution is (37, 27). So, there are 37 students in the

drama club and 27 students in the yearbook club.


Chapter 5

5.2 Exercises (pp. 245–246)


1. To solve a system of linear equations by substitution, solve

one of the equations for one of the variables. Then substitute

the expression for this variable into the other equation to fi nd

the value of the other variable. Finally, substitute this value

into one of the original equations to fi nd the value of the

other variable.

2. Sample answer: If one of the variables has a coeffi cient of

1 or −1, I solve for that variable in Step 1.


3. Sample answer: Solve Equation 2 for x because x has a

coeffi cient of 1.

4. Sample answer: Solve Equation 2 for y because y has a

coeffi cient of 1.

5. Sample answer: Solve Equation 2 for y because y has a

coeffi cient of −1.

6. Sample answer: Solve Equation 2 for either x or y because

x and y each have a coeffi cient of 1.


coeffi cient of 1.


coeffi cient of 1.

9. Substitute 17 − 4y for x in Equation 2 and solve for y.

y = x − 2

y = (17 − 4y) − 2

y = 17 − 4y −2

y = −4y + 15

+4y +4y

5y = 15

5y

— 5 =

15 —

5

y = 3


x = 17 − 4y

x = 17 − 4(3)

x = 17 − 12

x = 5

Check x = 17 − 4y y = x − 2

5 =?

17 − 4(3) 3 =?

5 − 2

5 =?

17 − 12 3 = 3 ✓

5 = 5 ✓


10. Substitute −3x for y in Equation 1 and solve for x.

6x − 9 = y

6x − 9 = −3x

−6x −6x

−9 = −9x

−9

— −9

= −9x

— −9

1 = x

Substitute 1 for x in Equation 2 and solve for y.

y = −3x

y = −3(1)

y = −3

Check 6x − 9 = y y = −3x

6(1) − 9 =?

−3 −3 =?

−3(1)

6 − 9 =?

−3 −3 = −3 ✓

−3 = −3 ✓


11. Substitute 16 − 4y for x in Equation 2 and solve for y.

3x + 4y = 8

3(16 − 4y) + 4y = 8

3(16) − 3(4y) + 4y = 8

48 − 12y + 4y = 8

48 − 8y = 8

−48 −48

−8y = −40

−8y

— −8

= −40

— −8

y = 5


x = 16 − 4y

x = 16 − 4(5)

x = 16 − 20

x = −4

Check x = 16 − 4y 3x + 4y = 8

−4 =?

16 − 4(5) 3(−4) + 4(5) =?

8

−4 =?

16 − 20 −12 + 20 =?

8

−4 = −4 ✓ 8 = 8 ✓



Chapter 5

12. Substitute 10x − 8 for y in Equation 1 and solve for x.

−5x + 3y = 51

−5x + 3(10x − 8) = 51

−5x + 3(10x) −3(8) = 51

−5x + 30x − 24 = 51

25x − 24 = 51

+24 +24

25x = 75

25x — 25

= 75

— 25

x = 3


y = 10x − 8

y = 10(3) − 8

y = 30 − 8

y = 22

Check −5x + 3y = 51 y = 10x − 8

−5(3) + 3(22) =?

51 22 =?

10(3) − 8

−15 + 66 =?

51 22 =?

30 − 8

51 = 51 ✓ 22 = 22 ✓


13. Solve Equation 1 for x.

2x = 12

2x — 2 =

12 —

2

x = 6


x − 5y = −29

6 − 5y = −29

−6 −6

−5y = −35

−5y — −5

= −35

— −5

y = 7

Check 2x = 12 x − 5y = −29

2(6) =?

12 6 − 5(7) =?

−29

12 = 12 ✓ 6 − 35 =?

−29

−29 = −29 ✓


14. Solve Equation 2 for x.

x − 9 = −1

+9 +9

x = 8


2x − y = 23

2(8) − y = 23

16 − y = 23

−16 −16

−y = 7

−y — −1

= 7 —

−1

y = −7

Check 2x − y = 23 x − 9 = −1

2(8) − (−7) =?

23 8 − 9 =?

−1

16 + 7 =?

23 −1 = −1 ✓

23 = 23 ✓


15. Step 1 x + y = −3

x − x + y = −3 − x

y = −x − 3

Step 2 5x + 2y = 9

5x + 2(−x − 3) = 9

5x + 2(−x) − 2(3) = 9

5x − 2x − 6 = 9

3x − 6 = 9

+6 +6

3x = 15

3x

— 3 =

15 —

3

x = 5

Step 3 x + y = −3

5 + y = −3

−5 −5

y = −8

Check: 5x + 2y = 9 x + y = −3

5(5) + 2(−8) =?

9 5 − 8 =?

−3

25 − 16 =?

9 −3 = −3 ✓

9 = 9 ✓



Chapter 5

16. Step 1 x − 2y = −4

x − 2y + 2y = −4 + 2y

x = 2y − 4

Step 2 11x − 7y = −14

11(2y − 4) − 7y = −14

11(2y) − 11(4) − 7y = −14

22y − 44 − 7y = −14

15y − 44 = −14

+44 +44

15y = 30

15y

— 15

= 30

— 15

y = 2

Step 3 x − 2y = −4

x − 2(2) = −4

x − 4 = −4

+4 +4

x = 0

Check 11x − 7y = −14 x − 2y = − 4

11(0) − 7(2) =?

−14 0 − 2(2) =?

−4

0 − 14 =?

−14 0 − 4 =?

−4

−14 = −14 ✓ −4 = −4 ✓


17. In Step 2, the expression for y should be substituted in the

other equation.

Step 1 5x − y = 4

5x − 5x − y = 4 − 5x

−y = −5x + 4

−y

— −1 =

−5x + 4 —

−1

y = 5x − 4

Step 2 8x + 2y = −12

8x + 2(5x − 4) = −12

8x + 2(5x) − 2(4) = −12

8x + 10x − 8 = −12

18x − 8 = −12

+8 +8

18x = −4

18x

— 18

= −4

— 18

x = − 2 —

9

18. In Step 3, 6 should be substituted for x, not for y.

Step 3 3x + y = 9

3(6) + y = 9

18 + y = 9

−18 −18

y = −9

19. WordsAcres of

corn + Acres of

wheat = 180

Acres of

corn = 3 ⋅ Acres of

wheat

Variables Let x be the number of acres of corn the farmer

should plant, and let y be the number of acres of

wheat the farmer should plant.


x = 3y Equation 2

Substitute 3y for x in Equation 1 and solve for y.

x + y = 180

3y + y = 180

4y = 180

4y

— 4 =

180 —

4

y = 45


x = 3y

x = 3(45)

x = 135

The solution is (135, 45). So, the farmer should plant

135 acres of corn and 45 acres of wheat.

20. WordsNumber of

1 person tubes + Number of

“cooler” tubes = 15

20 ⋅ Number of

1 person tubes + 12.5 ⋅ Number of

“cooler” tubes = 270

Variables Let x be the number of 1 person tubes the group

rents, and let y be the number of “cooler” tubes

the group rents.


20x + 12.5y = 270 Equation 2

Step 1 x + y = 15

x + y − y = 15 − y

x = 15 − y


Chapter 5

Step 2 20x + 12.5y = 270

20(15 − y) + 12.5y = 270

20(15) − 20(y) + 12.5y = 270

300 − 20y + 12.5y = 270

300 − 7.5y = 270

−300 −300

−7.5y = −30

−7.5y

— −7.5

= −30

— −7.5

y = 4

Step 3 x + y = 15

x + 4 = 15

−4 = −4

x = 11

The solution is (11, 4). So, the group rents 11 tubes for

people to use and 4 “cooler” tubes.

21. Sample answer: 4x − 5y = 4(3) − 5(5)

= 12 − 25

= −13

One linear equation that has (3, 5) as a solution is

4x − 5y = −13.

−2x − y = −2(3) − 5

= −6 − 5

= −11

Another linear equation that has (3, 5) as a solution is

−2x − y = −11. So, a system of linear equations that has

(3, 5) as its solution is

4x − 5y = −13

−2x − y = −11.

22. Sample answer:

4x + y = 4(−2) + 8

= −8 + 8

= 0

One linear equation that has (−2, 8) as a solution is

4x + y = 0.

−3x + 7y = −3(−2) + 7(8)

= 6 + 56

= 62

Another linear equation that has (−2, 8) as a solution is

−3x + 7y = 62. So, a system of linear equations that has

(−2, 8) as its solution is

4x + y = 62

−3x + 7y = 62 .

23. Sample answer:

x − y = −4 − (−12)

= −4 + 12

= 8

One linear equation that has (−4, −12) as a solution is

x − y = 8.

5x − 2y = 5(−4) − 2(−12)

= −20 + 24

= 4

Another linear equation that has (−4, −12) as a solution is

5x − 2y = 4. So, a system of linear equations that has

(−4, −12) as its solution is

x − y = 8.5x − 2y = 4

24. Sample answer:

3x + y = 3(15) + (−25)

= 45 − 25

= 20

One linear equation that has (15, −25) as a solution is

3x + y = 20.

−2x − y = −2(15) − (−25)

= −30 + 25

= −5

Another linear equation that has (15, −25) as a solution is

−2x − y = −5. So, a system of linear equations that has

(15, −25) as its solution is

3x + y = 20.−2x − y = −5

25. WordsNumber

of 5-point

problems+

Number

of 2-point

problems= 38

5 ⋅

Number

of 5-point

problems+ 2 ⋅

Number

of 2-point

problems= 100

Variables Let x be the number of 5-point problems on

the test, and let y be the number of 2-point

problems on the test.


5x + 2y = 100 Equation 2

Step 1 x + y = 38

x − x + y = 38 − x y = 38 − x


Chapter 5

Step 2 5x + 2y = 100

5x + 2(38 − x) = 100

5x + 2(38) − 2(x) = 100

5x + 76 − 2x = 100

3x + 76 = 100

−76 −76

3x = 24

3x

— 3 =

24 —

3

x = 8

Step 3 x + y = 38

8 + y = 38

−8 −8

y = 30

The solution is (8, 30). So, 8 of the questions on the math

test are worth 5 points, and 30 questions are worth 2 points.

26. WordsShares of

Stock A +Shares of

Stock B = 200

9.5 ⋅ Shares of

Stock A + 27 ⋅ Shares of

Stock B = 4000

Variables Let x be the number of shares of Stock A the

investor owns, and let y be the number of shares

of Stock B the investor owns.


9.5x + 27y = 4000 Equation 2

Step 1 x + y = 200

x − x + y = 200 − x

y = 200 − x

Step 2 9.5x + 27y = 4000

9.5x + 27(200 − x) = 4000

9.5x + 27(200) − 27(x) = 4000

9.5x + 5400 − 27x = 4000

−17.5x + 5400 = 4000

−5400 −5400

−17.5x = −1400

−17.5x

— −17.5

= −1400

— −17.5

x = 80

Step 3 x + y = 200

80 + y = 200

−80 −80

y = 120

The solution is (80, 120). So, the investor owns 80 shares of

Stock A and 120 shares of Stock B.

27. a. x + y + 90 = 180

b. x + y + 90 = 180 Equation 1

x + 2 = 3y Equation 2

Step 1 x + 2 = 3y x + 2 − 2 = 3y − 2

x = 3y − 2

Step 2 x + y + 90 = 180

3y − 2 + y + 90 = 180

4y + 88 = 180

−88 −88

4y = 92

4y

— 4 =

92 —

4

y = 23

Step 3 x + 2 = 3y

x + 2 = 3(23)

x + 2 = 69

−2 −2

x = 67

So, x = 67 and y = 23.

28. a. x + y + (y − 18) = 180

b. x + 2y − 18 = 180 Equation 1

3x − 5y = −22 Equation 2

Step 1 x + 2y − 18 = 180

x + 2y − 18 + 18 = 180 + 18

x + 2y = 198

x + 2y − 2y = 198 − 2y

x = 198 − 2y

Step 2 3x − 5y = −22

3(198 − 2y) − 5y = −22

3(198) −3 (2y) − 5y = −22

594 − 6y − 5y = −22

594 − 11y = −22

−594 −594

−11y = −616

−11y — −11

= −616

— −11

y = 56

Step 3 3x − 5y = −22

3x − 5(56) = −22

3x − 280 = −22

+280 +280

3x = 258

3x — 3 =

258 —

3

x = 86

So, x = 86 and y = 56.


Chapter 5

29. ax + by = −31 ⇒ a(−9) + b(1) = −31 ⇒ −9a + b = −31

ax − by = −41 ⇒ a(−9) − b(1) = −41 ⇒ −9a − b = −41

Step 1 −9a + b = −31

−9a + 9a + b = −31 + 9a

b = 9a − 31

Step 2 −9a − b = −41

−9a − (9a − 31) = −41

−9a − 9a + 31 = −41

−18a + 31 = −41

−31 −31

−18a = −72

−18a — −18

= −72

— −18

a = 4

Step 3 –9a + b = –31

–9(4) + b = –31

–36 + b = –31

+36 +36

b = 5

When a = 4 and b = 5, the solution of the linear system is

(−9, 1).

30. yes; The equation of a horizontal line is of the form x = a,

and the equation of a vertical line is of the form y = b.

Each equation has either x or y but not both. So, Step 2 is

impossible because an expression for a variable from one

equation cannot be substituted for that same variable in the

other equation. Therefore, the system cannot be solved by

substitution.

31. Sample answer: First fi nd the slope of the line through the

two given points.

m = 7 − (−5)

— −1 − 3

= 7 + 5

— − 1 − 3

= 12

— − 4

= −3

Use point slope form to write an equation of the line.

y − y1 = m(x − x1)

y − (−5) = −3(x − 3)

y + 5 = −3(x) − 3(−3)

y + 5 = −3x + 9

y + 5 − 5 = −3x + 9 − 5

y = −3x + 4

An equation that has both (3, −5) and (−1, 7) as solutions is

y = –3x + 4.

5x – 2y = 5(–1) – 2(7)

= –5 – 14

= –19

An equation that has (−1, 7) as a solution but not (3, −5) is

5x − 2y = −19.

So, a system of linear equations that has (−1, 7) as its

solution and (3, −5) as a solution of Equation 1 but not

Equation 2 is

y = −3x + 4

5x − 2y = −19.

32. a. The lines appear to intersect at (4, 5).

b. yes; Write an equation of each line. Then solve the system

of linear equations using substitution.

33. WordsNumber of

pop songs + Number of

rock songs + Number of

hip-hop songs = 272

Number of

pop songs = 3 ⋅ Number of

rock songs

Number of

hip-hop songs = 32 +Number of

rock songs

Variables Let x be the number of pop songs the station plays.

Let y be the number of rock songs the station

plays. Let z be the number of hip-hop songs the

station plays.

System x + y + z = 272 Equation 1

x = 3 ⋅ y Equation 2

z = 32 + y Equation 3

Substitute 3y for x and 32 + y for z in Equation 1 and solve

for y.

x + y + z = 272

3y + y + (32 + y) = 272

5y + 32 = 272

−32 −32

5y = 240

5y — 5 =

240 —

5

y = 48

Substitute 48 for y in Equation 2 and in Equation 3 to solve

for x and z, respectively.

x = 3y z = 32 + y

x = 3(48) z = 32 + 48

x = 144 z = 80

So, the radio station plays 144 pop songs, 48 rock songs, and

80 hip-hop songs.


Chapter 5

34. Sample answer:

Words 0.25 ⋅ Number of

quarters + 0.10 ⋅ Number

of dimes = 2.65

Number of

quarters +Number

of dimes =Number

of coins

Variables Let x be the number of quarters you have, and let y

be the number of dimes you have.

System 0.25x + 0.10y = 2.65 Equation 1

x + y = Number

of coins Equation 2

Let x = 5.

0.25x + 0.10y = 2.65

0.25(5) + 0.10y = 2.65

1.25 + 0.10y = 2.65

−1.25 −1.25

0.10y = 1.40

0.10y — 0.10

= 1.40

— 0.10

y = 14

Then, x + y = 5 + 14 = 19.

So, a system that represent this situation is

0.25x + 0.10y = 2.65.x + y = 19

35. WordsOriginal

tens digit +Original

ones digit = 11

10 ⋅ Original

ones digit +Original

tens digit =

27 + ( 10

Original

tens digit + Original

ones digit )

Variables Let x be the original tens digit, and let y be the

original ones digit.


10y + x = 27 + (10x + y) Equation 2

Step 1 x + y = 11

x − x + y = 11 − x

y = 11 − x

Step 2 10y + x = 27 + 10x + y 10(11 − x) + x = 27 + 10x + (11 − x)

10(11) − 10(x) + x = 27 + 10x + 11 − x

110 − 10x + x = 9x + 38

−9x + 110 = 9x + 38

+9x +9x

110 = 18x + 38

−38 −38

72 = 18x

72

— 18

= 18x

— 18

4 = x

Step 3 x + y = 11

4 + y = 11

−4 −4

y = 7

The solution is (4, 7). So, the original number is 47.


36. (x − 4) + (2x − 7) = x − 4 + 2x − 7

= x + 2x − 4 − 7

= 3x − 11

37. (5y − 12) + (−5y − 1) = 5y − 12 − 5y − 1

= 5y − 5y − 12 − 1

= 0 − 13

= −13

38. (t − 8) − (t + 15) = t − 8 − t − 15

= t − t − 8 − 15

= 0 − 23

= −23

39. (6d + 2) − (3d − 3) = 6d + 2 − 3d + 3

= 6d − 3d + 2 + 3

= 3d + 5

40. 4(m + 2) + 3(6m − 4) = 4(m) + 4(2) + 3(6m) − 3(4)

= 4m + 8 + 18m − 12

= 4m + 18m + 8 − 12

= 22m − 4

41. 2(5v + 6) − 6(−9v + 2) = 2(5v) + 2(6) − 6(−9v) − 6(2)

= 10v + 12 + 54v − 12

= 10v + 54v + 12 − 12

= 64v + 0

= 64v


Chapter 5

Section 5.3


1. a. 1 ⋅ x + 1 ⋅ y = 4.50 ⇒ x + y = 4.5 Equation 1

1 ⋅ x + 5 ⋅ y = 16.50 ⇒ x + 5y = 16.5 Equation 2

b. x + 5y = 16.5 Equation 2

− (x + y = 4.5) Equation 1

0 + 4y = 12.0

4y = 12

Solve this equation to fi nd the value of y and then

substitute this value for y into one of the original

equations and solve to fi nd the value of x.

4y = 12 x + y = 4.5 Equation 1

4y

— 4 =

12 —

4 x + 3 = 4.5

y = 3 −3 −3

x = 1.5

The solution is (1.5, 3). So, one drink costs $1.50 and one

sandwich costs $3.

2. a. Method 1: 3x − y = 6 3x − y = 6

− (3x + y = 0) 3x − (−3) = 6

0 − 2y = 6 3x + 3 = 6

−2y = 6 −3 −3

−2y — −2

= 6 —

−2 3x = 3

y = −3 3x

— 3 =

3 —

3

x = 1

Method 2: 3x − y = 6 3x + y = 0

+ 3x + y = 0 3(1) + y = 0

6x + 0 = 6 3 + y = 0

6x = 6 −3 −3

6x — 6 =

6 —

6 y = −3

x = 1

Using both methods, the solution is (1, −3).

Sample answer: Method 2 is preferred for this problem

because it involved less risk of making a careless error

with negative signs.

b. Method 1: 2x + y = 6 2x + y = 6

− (2x − y = 2) 2x + 2 = 6

0 + 2y = 4 −2 −2

2y = 4 2x = 4

2y — 2 =

4 —

2

2x —

2 =

4 —

2

y = 2 x = 2

Method 2: 2x + y = 6 2x + y = 6

+ 2x − y = 2 2(2) + y = 6

4x + 0 = 8 4 + y = 6

4x = 8 −4 −4

4x — 4 =

8 —

4 y = 2

x = 2

Using both methods, the solution is (2, 2).


because I am less likely to make a mistake adding the

equations.

c. Method 1: x − 2y = −7 x − 2y = −7

− (x + 2y = 5) x − 2(3) = −7

0 − 4y = −12 x − 6 = −7

−4y = −12 +6 +6

−4y — −4

= −12

— −4

x = −1

y = 3

Method 2: x −2y = −7 x − 2y = −7

+ x + 2y = 5 −1 − 2y = −7

2x + 0 = −2 +1 +1

2x = −2 −2y = −6

2x — 2 =

−2 —

2

−2y —

−2 = −6

— −2

x = −1 y = 3

Using both methods, the solution is (−1, 3).


because it has fewer negative numbers. So, I am less likely

to make a careless error.

3. a. no; The coeffi cients of the variables have to be the same

or opposites in order to be eliminated by addition or

subtraction. If you multiply each side of Equation 2 by

−2, then the coeffi cients of the x-terms will be opposites,

and x will be eliminated when you add the equations.

Another option would be to multiply each side of the fi rst

equation by −5.

b. Sample answer:

2x + y = 7 2x + y = 7

x + 5y = 17 Multiply by −2. −2x − 10y = −34

0 − 9y = −27

−9y = −27

−9y — −9

= −27

— −9

y = 3

x + 5y = 17

x + 5(3) = 17

x + 15 = 17

−15 −15

x = 2


Chapter 5

Check 2x + y = 7 x + 5y = 17

2(2) + 3 =?

7 2 + 5(3) =?

17

4 + 3 =?

7 2 + 15 =?

17

7 = 7 ✓ 17 = 17 ✓


4. First, you may have to multiply one or both equations by a

constant so that at least one pair of like terms has the same

or opposite coeffi cients. Then, when you add or subtract

the equations, the resulting equation will not contain that

variable. In other words, the variable will be eliminated.

Solve the resulting equation to fi nd the value of the other

variable. Then, substitute this value into one of the original

equations and solve for the variable that had been eliminated.

5. If at least one pair of like terms has the same coeffi cients,

then you can subtract equations in a system to eliminate one

of the variables. For example, in Exploration 2 part (a), the

x-terms both have a coeffi cient of 3. So, when you subtract

the equations, the x-terms are eliminated.

If at least one pair of like terms has opposite coeffi cients,

then you can add equations in a system to eliminate one of

the variables. For example, in Exploration 2 part (a), the

y-terms have coeffi cients 1 and −1. So, when you add the

equations, the y-terms are eliminated.

If neither pair of like terms has the same or opposite

coeffi cients, then you have to multiply one or both equations

by a constant. For example, in Exploration 3, neither pair

of like terms has the same coeffi cients. So, you multiply

Equation 2 by −2. Then, when you add the equations,

the x-terms are eliminated.

6. By the Multiplication Property of Equality, you can multiply

each side of an equation by the same amount and get an

equivalent equation, which means that it has the same

solution(s).

5.3 Monitoring Progress (pp. 249 –250)

1. Step 2 3x + 2y = 7

− 3x + 4y = 5

0 + 6y = 12

Step 3 6y = 12

6y — 6 =

12 —

6

y = 2

Step 4 3x + 2y = 7 3x + 2(2) = 7

3x + 4 = 7

−4 −4

3x = 3

3x — 3 =

3 —

3

x = 1

Check 3x + 2y = 7 −3x + 4y = 5

3(1) + 2(2) =?

7 −3(1) + 4(2) =?

5

3 + 4 =?

7 −3 + 8 =?

5

7 = 7 ✓ 5 = 5 ✓


2. Step 1 Step 2

x − 3y = 24 x − 3y = 24

3x + y = 12 Multiply by 3. 9x + 3y = 36

10x + 0 = 60

Step 3 10x = 60

10x — 10

= 60

— 10

x = 6

Step 4

3x + y = 12

3(6) + y = 12

18 + y = 12

−18 −18

y = −6

Check x − 3y = 24 3x + y = 12

6 − 3(−6) =?

24 3(6) + (−6) =?

12

6 + 18 =?

24 18 − 6 =?

12

24 = 24 ✓ 12 = 12 ✓


3. Step 1 Step 2

x + 4y = 22 x + 4y = 22

4x + y = 13 Multiply by −4. −16x − 4y = −52

−15x + 0 = −30

Step 3 −15x = −30

−15x — −15

= −30

— −15

x = 2

Step 4

4x + y = 13

4(2) + y = 13

8 + y = 13

−8 −8

y = 5

Check x + 4y = 22 4x + y = 13

2 + 4(5) =?

22 4(2) + 5 =?

13

2 + 20 =?

22 8 + 5 =?

13

22 = 22 ✓ 13 = 13 ✓



Chapter 5

4. Step 1 Step 2

5x + 2y = 235,000 −10x − 4y = −470,000

2x + 3y = 160,000 Multiply by 5. 10x + 15y = 800,000

0 + 11y = 330,000

Step 3 11y = 330,000

11y — 11

= 330,000

— 11

y = 30,000

Step 4

2x + 3y = 160,000

2x + 3(30,000) = 160,000

2x + 90,000 = 160,000

−90,000 −90,000

2x = 70,000

2x — 2 =

70,000 —

2

x = 35,000

The solution (35,000, 30,000) is the same. So, a large van

costs $35,000, and a small van costs $30,000.

5.3 Exercises (pp. 251– 252)


1. Sample answer: Write a system in which at least one pair of

like terms has opposite coeffi cients.

2x − 3y = 2

−5x + 3y = −14

2. Sample answer: First, multiply each side of Equation 1 by 3

so that the coeffi cients of the y-terms are 9 and −9. Then

add the equations to eliminate y. Solve the resulting equation

for x. Then substitute the value for x into one of the original

equations, and solve for y.


3. Step 2

x + 2y = 13

−x + y = 5

0 + 3y = 18

Step 3 3y = 18

3y — 3 =

18 —

3

y = 6

Step 4

x + 2y = 13

x + 2(6) = 13

x + 12 = 13

−12 −12

x = 1

Check x + 2y = 13 −x + y = 5

1 + 2(6) =?

13 −1 + 6 =?

5

1 + 12 =?

13 5 = 5 ✓

13 = 13 ✓


4. Step 2

9x + y = 2

−4x − y = −17

5x + 0 = −15

Step 3 5x = −15

5x — 5 =

−15 —

5

x = −3

Step 4

9x + y = 12

9(−3) + y = 12

−27 + y = 2

+27 +27

y = 29

Check 9x + y = 2 −4x − y = −17

9(−3) + 29 =?

2 −4(−3) − 29 =?

−17

−27 + 29 =?

2 12 − 29 =?

−17

2 = 2 ✓ −17 = −17 ✓


5. Step 2

5x + 6y = 50

x − 6y = −26

6x + 0 = 24

Step 3 6x = 24

6x — 6 =

24 —

6

x = 4

Step 4

x − 6y = −26

4 − 6y = −26

−4 −4

−6y = −30

−6y — −6

= −30

— −6

y = 5

Check 5x + 6y = 50 x − 6y = −26

5(4) + 6(5) =?

50 4 − 6(5) =?

−26

20 + 30 =?

50 4 −30 =?

−26

50 = 50 ✓ −26 = −26 ✓


Multiply by −2.


Chapter 5

6. Step 2

−x + y = 4

x + 3y = 4

0 + 4y = 8

Step 3 4y

— 4 =

8 —

4

y = 2

Step 4

x + 3y = 4

x + 3(2) = 4

x + 6 = 4

−6 −6

x = −2

Check −x + y = 4 x + 3y = 4

−(−2) + 2 =?

4 −2 + 3(2) =?

4

2 + 2 =?

4 −2 + 6 =?

4

4 = 4 ✓ 4 = 4 ✓


7. Step 2

−3x − 5y = −7

−4x + 5y = 14

−7x + 0 = 7

Step 3 −7x = 7

−7x — −7

= 7 —

−7

x = −1

Step 4

−4x + 5y = 14

−4(−1) + 5y = 14

4 + 5y = 14

−4 −4

5y = 10

5y — 5 =

10 —

5

y = 2

Check −3x − 5y = −7 −4x + 5y = 14

−3(−1) − 5(2) =?

−7 −4(−1) + 5(2) =?

14

3 − 10 =?

−7 4 + 10 =?

14

−7 = −7 ✓ 14 = 14 ✓


8. Step 2

4x − 9y = −21

−4x − 3y = 9

0 − 12y = −12

Step 3 −12y = −12

−12y —

−12 =

−12 —

−12

y = 1

Step 4

4x − 9y = −21

4x −9(1) = −21

4x − 9 = −21

+9 +9

4x = −12

4x — 4 =

−12 —

4

x = −3

Check 4x − 9y = −21 −4x − 3y = 9

4(−3) − 9(1) =?

−21 −4(−3) − 3(1) =?

9

−12 − 9 =?

−21 12 − 3 =?

9

−21 = −21 ✓ 9 = 9 ✓


9. Rewrite Equation 1 in the form ax + by = c.

−y − 10 = 6x

−y − 10 + 10 = 6x + 10

−y = 6x + 10

−y − 6x = 6x − 6x + 10

−6x − y = 10

Step 2

−6x − y = 10

5x + y = −10

−x + 0 = 0

Step 3 −x = 0

−x — −1

= 0 —

−1

x = 0

Step 4

5x + y = −10

5(0) + y = −10

0 + y = −10

y = −10

Check −y − 10 = 6x 5x + y = −10

−(−10) − 10 =?

6(0) 5(0) + (−10) =?

−10

10 − 10 =?

0 0 − 10 =?

−10

0 = 0 ✓ −10 = −10 ✓



Chapter 5

10. Rewrite Equation 2 in the form ax − c = by.

7y − 6 = 3x

7y − 7y − 6 = 3x − 7y

−6 = 3x − 7y

−6 − 3x = 3x − 3x − 7y

−3x − 6 = −7y

Step 2

3x − 30 = y

−3x − 6 = −7y

0 − 36 = −6y

Step 3 −36 = −6y

−36

— −6

= −6y

— −6

6 = y

Step 4

3x − 30 = y

3x − 30 = 6

+30 +30

3x = 36

3x — 3 =

36 —

3

x = 12

Check 3x − 30 = y 7y − 6 = 3x

3(12) − 30 =?

6 7(6) − 6 =?

3(12)

36 − 30 =?

6 42 − 6 =?

36

6 = 6 ✓ 36 = 36 ✓


11. Step 1 Step 2

x + y = 2 Multiply by −2. −2x − 2y = −4

2x + 7y = 9 2x + 7y = 9

0 + 5y = 5

Step 3 5y = 5

5y — 5 =

5 —

5

y = 1

Step 4

x + y = 2

x + 1 = 2

−1 −1

x = 1

Check x + y = 2 2x + 7y = 9

1 + 1 =?

2 2(1) + 7(1) =?

9

2 = 2 ✓ 2 + 7 =?

9

9 = 9 ✓


12. Step 1 Step 2

8x − 5y = 11 8x − 5y = 11

4x − 3y = 5 Multiply by −2. −8x + 6y = −10

0 + y = 1

Step 3 y = 1

Step 4

4x − 3y = 5

4x −3(1) = 5

4x − 3 = 5

+3 +3

4x = 8

4x — 4 =

8 —

4

x = 2

Check 8x − 5y = 11 4x − 3y = 5

8(2) − 5(1) =?

11 4(2) − 3(1) =?

5

16 − 5 =?

11 8 − 3 =?

5

11 = 11 ✓ 5 = 5 ✓


13. Step 1 Step 2

11x − 20y = 28 11x − 20y = 28

3x + 4y = 36 Multiply by 5. 15x + 20y = 180

26x + 0 = 208

Step 3 26x = 208

26x — 26

= 208

— 26

x = 8 Step 4

3x + 4y = 36

3(8) + 4y = 36

24 + 4y = 36

−24 −24

4y = 12

4y — 4 =

12 —

4

y = 3

Check 11x − 20y = 28 3x + 4y = 36

11(8) − 20(3) =?

28 3(8) + 4(3) =?

36

88 − 60 =?

28 24 + 12 =?

36

28 = 28 ✓ 36 = 36 ✓



Chapter 5

14. Step 1 Step 2

10x − 9y = 46 10x − 9y = 46

−2x + 3y = 10 Multiply by 5. −10x + 15y = 50

0 + 6y = 96

Step 3 6y = 96

6y — 6 =

96 —

6

y = 16

Step 4

−2x + 3y = 10

−2x + 3(16) = 10

−2x + 48 = 10

−48 −48

−2x = −38

−2x — −2

= −38

— −2

x = 19

Check 10x − 9y = 46 −2x + 3y = 10

10(19) − 9(16) =?

46 −2(19) + 3(16) =?

10

190 − 144 =?

46 −38 + 48 =?

10

46 = 46 ✓ 10 = 10 ✓


15. Step 1 Step 2

4x − 3y = 8 Multiply by 2. 8x −6y = 16

5x − 2y = −11 Multiply by −3. −15x + 6y = 33

−7x + 0 = 49

Step 3 −7x = 49

−7x — −7

= 49

— −7

x = −7

Step 4

4x − 3y = 8

4(−7) − 3y = 8

−28 − 3y = 8

+28 +28

−3y = 36

−3y — −3

= 36 — −3

y = −12

Check 4x − 3y = 8 5x − 2y = −11

4(−7) − 3(−12) =?

8 5(−7) − 2(−12) =?

−11

−28 + 36 =?

8 −35 + 24 =?

−11

8 = 8 ✓ −11 = −11 ✓


16. Step 1 Step 2

−2x − 5y = 9 Multiply by 3. −6x − 15y = 27

3x + 11y = 4 Multiply by 2. 6x + 22y = 8

0 + 7y = 35

Step 3 7y

— 7 =

35 —

7

y = 5

Step 4

−2x − 5y = 9

−2x − 5(5) = 9

−2x − 25 = 9

+25 +25

−2x = 34

−2x — −2

= 34 — −2

x = −17

Check: −2x −5y = 9 3x + 11y = 4

−2(−17) − 5(5) =?

9 3(−17) + 11(5) =?

4

34 − 25 =?

9 −51 + 55 =?

4

9 = 9 ✓ 4 = 4 ✓


17. Step 1 Step 2

9x + 2y = 39 Multiply by 4. 36x + 8y = 156

6x + 13y = −9 Multiply by −6. −36x − 78y = 54

0 − 70y = 210

Step 3 −70y = 210

−70y — −70

= 210

— −70

y = −3

Step 4

9x + 2y = 39

9x + 2(−3) = 39

9x − 6 = 39

+6 +6

9x = 45

9x — 9 =

45 —

9

x = 5

Check 9x + 2y = 39 6x + 13y = −9

9(5) + 2(−3) =?

39 6(5) + 13(−3) =?

−9

45 − 6 =?

39 30 − 39 =?

−9

39 = 39 ✓ −9 = −9 ✓



Chapter 5

18. Step 1 Step 2

12x − 7y = −2 Multiply by 2. 24x − 14y = −4

8x + 11y = 30 Multiply by−3. −24x − 33y = −90

0 − 47y = −94

Step 3 −47y

— −47

= −94

— −47

y = 2

Step 4

8x + 11y = 30

8x + 11(2) = 30

8x + 22 = 30

−22 −22

8x = 8

8x — 8 =

8 —

8

x = 1

Check 12x − 7y = −2 8x + 11y = 30

12(1) − 7(2) =?

−2 8(1) + 11(2) =?

30

12 − 14 =?

−2 8 + 22 =?

30

−2 = −2 ✓ 30 = 30 ✓


19. The x-terms should have been added, not subtracted.

Step 2

5x − 7y = 16

x + 7y = 8

6x + 0 = 24

Step 3 6x = 24

6x — 6 =

24 —

6

x = 4

20. Each side of Equation 2 should be multiplied by −4,

including −13 on the right.

Step 1 Step 2

4x + 3y = 8 4x + 3y = 8

x − 2y = −13 Multiply by −4. −4x + 8y = 52

0 + 11y = 60

11y — 11

= 60

— 11

y = 60

— 11

21. Words Fee

for oil

change

+Quarts

of oil

used⋅

Cost per

quart of

oil

= Total

cost

Variables Let x be the fee (in dollars) for an oil change, and

let y be the cost per quart of oil used.

System x + 5 ⋅ y = 22.45

x + 7 ⋅ y = 25.45

Step 2 x + 7y = 25.45

−(x + 5y = 22.45)

0 + 2y = 3.00

Step 3 2y = 3

2y

— 2 =

3 —

2

y =1.5

Step 4

x + 7y = 25.45

x + 7(1.5) = 25.45

x + 10.5 = 25.45

−10.5 −10.5

x = 14.95

The solution is (14.95, 1.5). So, the fee for an oil change is

$14.95, and each quart of oil costs $1.50.

22. Words Number

of

individual

songs

⋅ Cost

per

individual

song

+

Number

of

albums

⋅

Cost

per

album

= Total

cost

Variables Let x be the cost per individual song, and let y be

the cost per album.

System 6 ⋅ x + 2 ⋅ y = 25.92

4 ⋅ x + 3 ⋅ y = 33.93

Step 1 Step 2

6x + 2y = 25.92 Multiply by 3. 18x + 6y = 77.76

4x + 3y = 33.93 Multiply by −2. −8x − 6y = −67.86

10x + 0 = 9.90

Step 3 10x = 9.9

10x — 10

= 9.9

— 10

x = 0.99


Chapter 5

Step 4 6x + 2y = 25.92

6(0.99) + 2y = 25.92

5.94 + 2y = 25.92

−5.94 −5.94

2y = 19.98

2y — 2 =

19.98 —

2

y = 9.99

The solution is (0.99, 9.99). So, the website charges $0.99 to

download a song and $9.99 to download an entire album.

23. Step 1 2y = 8 – 5x

2y — 2

= 8 – 5x

— 2

y = 4 – 5 —

2 x

Step 2 3x + 2y = 4

3x + 2 ( 4 – 5 —

2 x ) = 4

3x + 2(4) – 2 ( 5 — 2 x ) = 4

3x + 8 – 5x = 4

–2x + 8 = 4

–8 –8

–2x = –4

–2x — –2

= –4

— –2

x = 2

Step 3 2y = 8 – 5x

2y = 8 – 5(2)

2y = 8 – 10

2y = –2

2y — 2 =

–2 —

2

y = –1

The solution is (2, –1). Sample explanation: I chose

substitution because it took only one step to isolate y in

Equation 2.

24. Step 1 Step 2

−6y + 2 = −4x −6y + 2 = −4x

y − 2 = x Multiply by 4. 4y − 8 = 4x

−2y − 6 = 0

Step 3 −2y − 6 = 0

+6 +6

−2y = 6

−2y

— −2

= 6 —

−2

y = −3

Step 4

y – 2 = x

–3 – 2 = x

–5 = x

The solution is (–5, –3). Sample explanation: I chose

elimination because the original equations had like terms in

the same respective positions.

25. y – x = 2 y = – 1 — 4 x + 7

y – x + x = 2 + x m = – 1 — 4 , b = 7

y = x + 2

m = 1, b = 2

x

y

4

6

8

4 62

y = x + 2

y = − x + 714

(4, 6)

The solution is (4, 6). Sample explanation: I chose graphing

because Equation 2 was in the slope-intercept form, and it

only took one step to rewrite Equation 1 in slope-intercept

form.


Chapter 5

26. Step 1 Step 2

3x + y = 1 —

3 Multiply by 9. 27x + 9y = 3

2x – 3y = 8 —

3 Multiply by 3. 6x – 9y = 8

33x + 0 = 11

Step 3 33x = 11

33x

— 33

= 11

— 33

x = 1 —

3

Step 4

3x + y = 1 —

3

3 ( 1 — 3 ) + y =

1 —

3

1 + y = 1 —

3

–1 –1

y = – 2 — 3

The solution is ( 1 — 3 , – 2 —

3 ) . Sample explanation: I chose

elimination because both equations had like terms in the

same respective positions, and I was able to rewrite both

equations without fractions so that I could minimize how

much I had to work with fractions.

27. If a = 4 or a = −4, you can solve the linear system by

elimination without multiplying fi rst. If the coeffi cients of

the x-terms are both 4, then you can simply subtract the

equations in order to eliminate x. If the coeffi cients of the

x-terms are 4 and −4, then you can simply add the equations

to eliminate x.

28. a. Sample answer: About 10 students chose breakfast, and

about 15 students chose lunch. (Answer does not have to

be exact, but the total should be 25.)

b. Words Students

who chose

breakfast+

Students

who chose

lunch+ 25 = 50

Students

who chose

lunch= 5 +

Students

who chose

breakfast

Variables Let x be the number of students who chose

breakfast, and let y be the number of students

who chose lunch.

System: x + y + 25 = 50

y = 5 + x A system of linear equations that represents the numbers

of students who chose breakfast and lunch is

x + y + 25 = 50.

y = 5 + x

c. Sample answer: Because the y-value is already isolated

in Equation 2, substitution is the most effi cient method

for solving this system. First, substitute the expression

5 + x for y in Equation 1 to get x + (5 + x) + 25 = 50.

Solve this equation for x, and check to see if the solution

matches your guess from part (a) for the number of

students who chose breakfast. Then substitute this value

for x in Equation 2. Solve for y, and check to see if the

solution matches your guess from part (a) for the number

of students who chose lunch.

29. no; Sample answer: If like terms are in the same respective

positions and at least one pair of like terms has the same or

opposite coeffi cients, then elimination is more effi cient than

substitution and will take fewer steps. On the other hand, if

one of the variables in one of the equations is either isolated

already or has a coeffi cient of 1 or –1, then substitution is

more effi cient and will take fewer steps.

30. Sample answer: A system of linear equations that can be

added or subtracted to eliminate a variable is

2x + y = 10

2x − y = 2.

31. a. P = 2ℓ + 2w ⇒ 18 = 2ℓ + 2w

P = 2(3ℓ) + 2(2w) ⇒ 46 = 6ℓ + 4w

Step 1 Step 2

18 = 2ℓ + 2w Multiply by −2. –36 = –4ℓ – 4w

46 = 6ℓ + 4w 46 = 6ℓ + 4w

10 = 2ℓ + 0

Step 3 10 = 2ℓ

10

— 2 =

2ℓ — 2

5 = ℓ Step 4

18 = 2ℓ + 2w

18 = 2(5) + 2w

18 = 10 + 2w

–10 –10

8 = 2w

8 —

2 =

2w —

2

4 = w

The original rectangle is 4 inches wide and 5 inches long.

b. The new rectangle is 2w = 2(4) = 8 inches wide and

3ℓ = 3(5) = 15 inches long.


Chapter 5

32. Sample answer: Begin by multiplying each side of

Equation 2 by 2. By the Multiplication Property of Equality,

2x + 2y = 12. You can rewrite Equation 1 as Equation 3 by

adding 2x + 2y on the left and adding 12 on the right.

You can rewrite Equation 3 as Equation 1 by subtracting

2x + 2y on the left and subtracting 12 on the right. Because

you can rewrite Equation 1 as Equation 3, and you can

rewrite Equation 3 as Equation 1, System 1 and System 2

have the same solution.

33. Words Amount (in quarts)

of 100% fruit juice + Amount (in quarts)

of 20% fruit juice = 6

100% ⋅

Amount

(in quarts)

of 100%

fruit juice

+ 20% ⋅

Amount

(in quarts)

of 20%

fruit juice

= 80% ⋅ 6

Variables Let x be the number of quarts of 100% fruit juice

you should use, and let y be the number of quarts

of 20% fruit juice you should use.

System x + y = 6 ⇒ x + y = 6

1.00x + 0.20y = 0.80(6) ⇒ x + 0.2y = 4.8

Step 2

x + y = 6

– (x + 0.2y = 4.8)

0 + 0.8y = 1.2

Step 3 0.8y = 1.2

0.8y

— 0.8

= 1.2

— 0.8

y = 1.5

Step 4

x + y = 6

x +1.5 = 6

–1.5 –1.5

x = 4.5

The solution is (4.5, 1.5). So, you should mix 4.5 quarts of

100% fruit juice with 1.5 quarts of 20% fruit juice.

34. 40 min ⋅ 1 h —

60 min =

40 —

60 h =

2 —

3 h

60 min ⋅ 1 h —

60 min = 60

— 60

h = 1 h

Words 2 —

3 ⋅ ( speed of boat + speed of current ) = 20

1 ⋅ ( speed of boat – speed of current ) = 20

Variables Let x be the speed (in miles per hour) of the boat,

and let y be the speed (in miles per hour) of the

current.

System 2 —

3 (x + y) = 20 ⇒

2 —

3 x +

2 —

3 y = 20

1 (x – y) = 20 ⇒ x – y = 20

Step 1 Step 2

2 —

3 x +

2 —

3 y = 20 Multiply by 3 — 2 . x + y = 30

x – y = 20 x – y = 20

2x + 0 = 50

Step 3 2x = 50

2x

— 2 =

50 —

2

x = 25

Step 4

x – y = 20

25 – y = 20

–25 –25

–y = –5

–y

— –1

= –5

— –1

y = 5

The solution is (25, 5). So, the speed of the current is

5 miles per hour.

35. Sample answer: Rewrite Equation 2 in standard form.

3z + x – 2y = –7 ⇒ x – 2y + 3z = –7

Subtract Equation 2 from Equation 1.

x + 7y + 3z = 29

– (x – 2y + 3z = –7)

0 + 9y + 0 = 36

9y = 36

Both x and z were eliminated. Solve for y.

9y = 36

9y

— 9 =

36 —

9

y = 4

Substitute 4 for y in Equation 3, and solve for x.

5y = 10 – 2x

5(4) = 10 – 2x

20 = 10 – 2x

–10 –10

10 = –2x

10

— –2

= –2x

— –2

–5 = x


Chapter 5

Substitute –5 for x and 4 for y in Equation 1, and solve for z.

x + 7y + 3z = 29

–5 + 7(4) + 3z = 29

–5 + 28 + 3z = 29

23 + 3z = 29

–23 –23

3z = 6

3z

— 3 =

6 —

3

z = 2

Check x + 7y + 3z = 29 3z + x – 2y = –7

–5 + 7(4) + 3(2) =?

29 3(2) + (–5) – 2(4) =?

–7

–5 + 28 + 6 =?

29 6 – 5 – 8 =?

–7

29 = 29 ✓ –7 = –7 ✓

5y = 10 – 2x

5(4) =?

10 – 2(–5)

20 =?

10 + 10

20 = 20 ✓

So, x = –5, y = 4, and z = 2.


36. 5d – 8 = 1 + 5d

–5d –5d

–8 = 1

Because the statement –8 = 1 is never true, the equation has

no solution.

37. 9 + 4t = 12 – 4t

+4t +4t

9 + 8t = 12

–9 –9

8t = 3

8t

— 8 =

3 —

8

t = 3 —

8

The equation has one solution, which is t = 3 —

8 .

38. 3n + 2 = 2(n – 3)

3n + 2 = 2(n) – 2(3)

3n + 2 = 2n – 6

–2n –2n

n + 2 = –6

–2 –2

n = –8

The equation has one solution, which is n = –8.

39. –3(4 – 2v) = 6v – 12

–3(4) – 3(–2v) = 6v – 12

–12 + 6v = 6v – 12

–6v –6v

–12 = –12

Because the statement –12 = –12 is always true, the

equation has infi nitely many solutions. The solution is all

real numbers.

40. y − y1 = m(x – x1)

y − 1 = –2(x – 4)

y − 1 = –2(x) – 2(−4)

y − 1 = –2x + 8 +1 +1

y = –2x + 9

An equation of the parallel line is y = –2x + 9.

41. y = mx + b

6 = 5(0) + b

6 = 0 + b

6 = b

Using m = 5 and b = 6, an equation of the parallel line is

y = 5x + 6.

42. y – y1 = m(x – x1)

y – (–2) = 2 —

3 (x – (–5))

y + 2 = 2 —

3 (x + 5)

y + 2 = 2 —

3 (x) +

2 —

3 (5)

y + 2 = 2 —

3 x +

10 —

3

–2 –2

y = 2 —

3 x +

4 —

3

An equation of the parallel line is y = 2 —

3 x +

4 —

3 .


Chapter 5

Section 5.4


1. a. Words Cost = Initial

investment + Cost of

materials ⋅

Number of

skateboards

Revenue =

Price per

skateboard ⋅

Number of

skateboards

Equations C = 450 + 20x

R = 20x

x (skateboards) 0 1 2 3 4 5 6 7 8 9 10

C (dollars) 450 470 490 510 530 550 570 590 610 630 650

R (dollars) 0 20 40 60 80 100 120 140 160 180 200

b. Sample answer: Your company will never break even

because for each skateboard you sell, you spend as much

as you make. So, you are spending money at the same rate

as you have money coming in, and you will never recover

your investment or make any money.

2. a. Words Number of

small beads ⋅ Weight per

small bead

+ Number of

large beads ⋅

Weight per

large bead = Total cost

Equations 40 ⋅ x + 6 ⋅ y = 10

20 ⋅ x + 3 ⋅ y = 5

So, a system of linear equations that represents this

situation is

40x + 6y = 10

. 20x + 3y = 5

b. 40x + 6y = 10

40x − 40x + 6y = 10 − 40x

6y = −40x + 10

6y

— 6 =

−40x + 10 —

6

y = − 20

— 3 x +

5 —

3

20x + 3y = 5

20x − 20x + 3y = 5 − 20x

3y = −20x + 5

3y

— 3 =

−20x + 5 —

3

y = − 20

— 3 x +

5 —

3

x

y

1.0

1.5

2.0

0.5

00.2 0.3 0.40.10

40x + 6y = 10

20x + 3y = 5

The two equations describe the same line.

c. no; You cannot fi nd the weight of each type of bead

because both equations describe the same line, and there

are infi nitely many points that are solutions of both

equations. So, there are infi nitely many possibilities for

the weights of the beads.

3. yes; Sample answer: If the linear equations in a system

have the same rate of change, then the lines they describe

are parallel and will never intersect. So, the system has no

solution. If the linear equations in a system describe the

same line, then the system has infi nitely many solutions.

The system y = 3x and y = 3x + 1 has no solution. The

system y = 3x and 2y = 6x has infi nitely many solutions.

4. a. This system has one solution because the lines intersect at

one point.

b. This system has no solution because the lines are parallel

and will never intersect.

c. This system has infi nitely many solutions because both

equations describe the same line.


1. Solve by elimination.

Step 1 Step 2 x + y = 3 Multiply by −2. −2x − 2y = −6

2x + 2y = 6 2x + 2y = 6

0 = 0

The equation 0 = 0 is always true. So, the solutions are

all the points on the line x + y = 3. The system of linear

equations has infi nitely many solutions.

2. Solve by substitution.

2x + 2y = 4

2x + 2(−x + 3) = 4

2x + 2(−x) + 2(3) = 4

2x − 2x + 6 = 4

6 = 4 ✗

The equation 6 = 4 is never true. So, the system of linear

equations has no solution.


Chapter 5


Step 2 −(x + y = 3)

x + 2y = 4

y = −1

Step 3 x + y = 3

x + 1 = 3

−1 −1

x = 2



10x + y = 10

10x + (−10x + 2) = 10

10x − 10x + 2 = 10

2 = 10 ✗



5. The lines still have the same slope, but they now have different

y-intercepts, so they are parallel. So, the system has no solution.

5.4 Exercises (pp. 257 –258)


1. Two lines cannot intersect in exactly two points. So, a system

of linear equations cannot have exactly two solutions.

2. The graph of a linear system that has infi nitely many

solutions is a single line because both equations describe that

same line. The graph of a linear system that has no solution

is two parallel lines that never intersect.


3. Equation 1

Let y = 0. Let x = 0.

−x + y = 1 −x + y = 1

−x + 0 = 1 −0 + y = 1

−x = 1 y = 1

−x

— −1

= 1 —

−1

x = −1

Equation 2

Let y = 0. Let x = 0.

x − y = 1 x − y = 1

x − 0 = 1 0 − y = 1

x = 1 −y = 1

−y

— −1

= 1 —

−1

y = −1

F; Equation 1 has an x-intercept of −1 and a y-intercept

of 1. Equation 2 has an x-intercept of 1 and a y-intercept

of −1. So, graph F matches this system. Because the lines

are parallel, they do not intersect. So, the system of linear



2x − 2y = 4 −x + y = −2

2x − 2x − 2y = 4 − 2x −x + x + y = −2 + x

−2y = −2x + 4 y = x − 2

−2y —

−2 =

−2x + 4 —

−2

y = x − 2

E; The slope-intercept forms of Equations 1 and 2 are the

same, with a slope of 1 and a y-intercept of −2. So, both

equations in this system describe the same line in graph E.

All points on the line are solutions of both equations. So, the

system of linear equations has infi nitely many solutions.


Chapter 5

5. Equation 1

Let x = 0. Let y = 0.

2x + y = 4 2x + y = 4

2(0) + y = 4 2x + 0 = 4

0 + y = 4 2x = 4

y = 4 2x

— 2 =

4 —

2

x = 2

Equation 2

Let x = 0. Let y = 0.

−4x −2y = −8 −4x − 2y = −8

−4(0) − 2y = −8 −4x −2(0) = −8

0 − 2y = −8 −4x − 0 = −8

−2y = −8 −4x = −8

−2y —

−2 =

−8 —

−2

−4x —

−4 = −8

— −4

y = 4 x = 2

B; Equations 1 and 2 both have an x-intercept of 2 and a

y-intercept of 4. So, both equations describe the same line

in graph B. All points on the line are solutions of both

equations. So, the system of linear equations has infi nitely

many solutions.


x − y = 0 5x − 2y = 6

x − x − y = 0 − x 5x − 5x − 2y = 6 − 5x

−y = −x −2y = −5x + 6

−y —

−1 =

−x —

−1

−2y —

−2 =

−5x + 6 —

−2

y = x y = 5 —

2 x −3

C; For Equation 1, the slope is 1 and the y-intercept is 0,

and for Equation 2, the slope 5 —

2 and the y-intercept is −3. So,

graph C matches this system. Because the lines intersect in

one point, the system has one solution.


−2x + 4y = 1 3x − 6y = 9

−2x + 2x + 4y = 1 + 2x 3x − 3x − 6y = 9 − 3x

4y = 2x + 1 −6y = −3x + 9

4y — 4 =

2x + 1 —

4

−6y —

−6 =

−3x + 9 —

−6

y = 1 —

2 x +

1 —

4 y =

1 —

2 x −

3 —

2

D; For Equation 1, the slope is 1 —

2 and the y-intercept is

1 —

4 ,

and for Equation 2, the slope is 1 —

2 and the y-intercept is −

3 — 2 .

So, graph D matches this system. Because the lines have

the same slope and different y-intercepts, they are parallel.

Because parallel lines do not intersect, there is no point

that is a solution of both equations. So, the system of linear



5x + 3y = 17 x − 3y = −2

5x − 5x + 3y = 17 − 5x x − x − 3y = −2 − x

3y = −5x + 17 −3y = −x − 2

3y — 3 =

−5x + 17 —

3

−3y —

−3 =

−x − 2 —

−3

y = − 5 —

3 x +

17 —

3 y =

1 —

3 x +

2 —

3

A; For Equation 1, the slope is − 5 — 3 and the y-intercept is

17 —

3 ,

and for Equation 2, the slope is 1 —

3 and the y-intercept is

2 —

3 . So,

graph A matches this system. Because the lines intersect in

one point, the system has one solution.

9. Solve by graphing.

x

y

−4

4−4

y = 2x − 4y = −2x − 4

(0, −4)

Check y = −2x − 4 y = 2x − 4

−4 =?

−2(0) − 4 −4 =?

2(0) − 4

−4 =?

0 − 4 −4 =?

0 − 4

−4 = −4 ✓ −4 = −4 ✓

The solution of the system of linear equations is (0,−4).

10. Solve by graphing.

x

y8

4

−8

84−4−8

y = −6x − 8

y = −6x + 8

The lines both have a slope of −6, but they have different

y-intercepts. So, they 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.


3x − y = 6

−3x + y = −6

0 = 0


all the points on the line 3x − y = 6. The system of linear



Chapter 5


−x + 2y = 7

x − 2y = 7

0 = 14 ✗




Step 1 4x + 4y = −8 Step 2 4x + 4y = −8

−2x − 2y = 4 Multiply by 2. −4x − 4y = 8 0 = 0

The equation 0 = 0 is always true. So, the solutions are all

the points on the line 4x + 4y = −8. The system of linear



Step 1 15x − 5y = −20 Step 2 15x − 5y = −20

−3x + y = 4 Multiply by 5. −15x + 5y = 20

0 = 0

The equation 0 = 0 is always true. So, the solutions are all

the points on the line −3x + y = 4. The system of linear



Step 1 Step 2

9x − 15y = 24 Multiply by 2. 18x − 30y = 48

6x − 10y = −16 Multiply by −3. −18x + 30y = 48

0 = 96




Step 1 Step 2

3x − 2y = −5 Multiply by 5.

15x − 10y = −25

4x + 5y = 47 Multiply by 2.

8x + 10y = 94

23x + 0 = 69

Step 4 3x − 2y = −5 Step 3 23x = 69

3(3) − 2y = −5 23x

— 23

= 69

— 23

9 − 2y = −5 x = 3

−9 −9

−2y = −14

−2y — −2

= −14

— −2

y = 7

The solution of the system of linear equations is (3, 7).


y = 7x + 13 −21x + 3y = 39

−21x + 21x + 3y = 39 + 21x

3y = 21x + 39

3y — 3 =

21x + 39 —

3

y = 7x + 13

Both equations have a slope of 7 and a y-intercept of 13.

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 infi nitely many solutions.


y = −6x − 2 12x + 2y = −6

12x − 12x + 2y = −6 − 12x

2y = −12x − 6

2y — 2 =

−12x − 6 —

2

y = −6x − 3

The lines both have a slope of −6, but they have different

y-intercepts. So, the lines are parallel. Because parallel lines




4x + 3y = 27 4x − 3y = −27

4x − 4x + 3y = 27 − 4x 4x − 4x − 3y = −27 − 4x

3y = −4x + 27 −3y = −4x − 27

3y — 3 =

−4x + 27 —

3

−3y —

−3 =

−4x − 27 —

−3

y = − 4 — 3 x + 9 y =

4 —

3 x + 9

The slope of Equation 1 is − 4 — 3 , and the slope of Equation 2

is 4 —

3 . So, the lines do not have the same slope, but they both have

a y-intercept of 9. So, they intersect at (0, 9) on the y-axis.

Therefore, the system of linear equations has one solution,

which is (0, 9).


−7x + 7y = 1 2x − 2y = −18

−7x + 7x + 7y = 1 + 7x 2x − 2x −2y = −18 − 2x

7y = 7x + 1 −2y = −2x − 18

7y — 7 =

7x + 1 —

7

−2y —

−2 =

−2x − 18 —

−2

y = x + 1 —

7 y = x + 9

Both lines have a slope of 1, but they have different

y–intercepts. So, the lines are parallel. Because parallel lines




Chapter 5


−18x + 6y = 24 3x − y = −2

−18x + 18x + 6y = 24 + 18x 3x − 3x − y = −2 − 3x

6y = 18x + 24 −y = −3x − 2

6y — 6 =

18x + 24 —

6

−y —

−1 =

−3x − 2 —

−1

y = 3x + 4 y = 3x + 2

Both lines have a slope of 3, but they have different





2x − 2y = 16 3x − 6y = 30

2x − 2x − 2y = 16 − 2x 3x − 3x − 6y = 30 − 3x

−2y = −2x + 16 −6y = −3x + 30

−2y — −2 =

−2x + 16 —

−2 −6y

— −6 =

−3x + 30 —

−6

y = x − 8 y = 1 —

2 x −5

Equation 1 has a slope of 1 and a y-intercept of −8. Equation 2

has a slope of 1 —

2 and a y-intercept of −5. Because the equations

do not have the same slope or y-intercept, they must intersect

in one point that is not on the y-axis. So, the system of linear

equations has one solution.

23. Only lines that have the same slope are parallel and will

never intersect. These lines will intersect at one point if they

are extended. So, the system has one solution.


−4x + y = 4 4x + y = 12

−4x + 4x + y = 4 + 4x 4x − 4x + y = 12 − 4x

y = 4x + 4 y = −4x + 12

x

4

8

12

2 4−2−4

y

−4x + y = 4

4x + y = 12

(1, 8)

Check −4x + y = 4 4x + y = 12

−4(1) + 8 =?

4 4(1) + 8 =?

12

−4 + 8 =?

4 4 + 8 =?

12

4 = 4 ✓ 12 = 12 ✓

The lines intersect at (1, 8), which is the solution of the

system.

24. The lines do have the same slope of 3, but they have different




25. Words

Amount

(in cups)

of dried

fruit

⋅

Cost

per

cup of

dried

fruit

+

Amount

(in cups)

of

almonds

⋅

Cost per

cup of

almonds

= Total

Cost

System 3 ⋅ x + 4 ⋅ y = 6

4 1 —

2 ⋅ x + 6 ⋅ y = 9

Solve by elimination.

Step 1 Step 2

3x + 4y = 6 Multiply by 3. 9x + 12y = 18

9 —

2 x + 6y = 9 Multiply by −2. −9x − 12y = −18

0 = 0

The equation 0 = 0 is always true. In this context, x and y

must be positive. So, the solutions are all points on the line

3x + 4y = 6 in Quadrant I. The system of linear equations

has infi nitely many solutions.

26. Words Team A: Distance

(in miles)

traveled

= Team A’s

speed ⋅

Time

(in hours)

it takes to

fi nish

− 2

Team B: Distance

(in miles)

traveled

= Team B’s

speed ⋅

Time

(in hours)

it takes to

fi nish

Variables Let x be how long (in hours) it takes to fi nish

the race, and let y be how far (in miles) the team

travels for the remainder of the race.

System Team A: y = 6x − 2

Team B: y = 6x

Because the two teams are traveling at the same speed,

Team B will not catch up to Team A. The graphs of these

lines are parallel because they have the same slope. So,

there is no point that is a solution of both equations, and the

system of linear equations has no solution.

27. Words Number

of

coach

tickets

⋅

Cost

per

coach

ticket

+

Number

of

business

class

tickets

⋅

Cost per

business

class

ticket

= Money

collected

Variables Let x be the cost (in dollars) of one coach ticket,

and let y be the cost (in dollars) of one business

class ticket.

System 150 ⋅ x + 80 ⋅ y = 22,860

170 ⋅ x + 100 ⋅ y = 27,280

A system of equations is 150x + 80y = 22,860 and

170x + 100y = 27,280.


Chapter 5

Equation 1

150x + 80y = 22,860

150x −150x + 80y = 22,860 − 150x

80y = 22,860 −150x

y = − 15

— 8 x +

1143 —

4

Equation 2

170x + 100y = 27,280

170x −170x + 100y = 27,280 − 170x

100y = −170x + 27,280

y = − 17

— 10

x + 1364

— 5

Equation 1 has a slope of − 15 —

8 and a y-intercept of

1143 —

4 .

Equation 2 has a slope of − 17 —

10 and a y-intercept of

1364 —

5 .

Because the equations do not have the same slope or

y-intercept, they must intersect in one point that is not

on the y-axis. So, the system of linear equations has

one solution.

28. Sample answer: y = x + 1 Equation 1

y = −x + 3 Equation 2

y = 1 —

2 x − 1 Equation 3

As shown in the graph, each pair of lines intersects in one

point, but there is no point that is a solution of all three

equations.

x

y4

2

−2

2−4

y = x + 1

y = −x + 3

y = x − 112

83

13( , )

(1, 2)

(−4, −3)

29. The system has one solution. If two lines do not have the

same slope, then the lines are neither parallel nor the same.

So, the lines must intersect in one point. This point of

intersection is the one and only solution of the system.

30. a. Sample answer: Team C’s runner passed Team B’s runner

at about 40 meters.

b. yes; Team C’s runner would have passed Team A’s runner

eventually. The lines that represent Team A’s and Team C’s

runners have different slopes, so they will intersect.

c. no; Team B’s runner could not have passed Team A’s

runner. Because Team B’s runner and Team A’s runner are

running at the same speed, the lines that represent them

have the same slope and are therefore parallel. So, the

lines will never intersect.

31. a. never; The y-intercept of y = ax + 4 is 4, and the

y-intercept of y = bx − 2 is −2. Because the lines do not have the same y-intercept, they are not the same line. So,

they cannot have infi nitely many solutions.

b. sometimes; If a = b, then the lines will have the same

slope and therefore be parallel. So, the system of linear

equations would have no solution.

c. always; When a < b, the lines do not have the same slope.

So, the graphs are neither parallel lines nor the same line.

Therefore, they must intersect in exactly one point, which

means the system always has one solution.

32. no; You cannot determine the exact costs with the

information given.

Words Number

of

admissions

⋅

Cost per

ad mission

+

Number

of skate

rentals

⋅

Cost

per

skate

rental

=

Total

cost

System 3 ⋅ x + 2 ⋅ y = 38

15 ⋅ x + 10 ⋅ y = 190


Step 1 Step 2

3x + 2y = 38 Multiply by −5. –15x – 10y = –190

15x + 10y = 190 15x + 10y = 190

0 = 0

The equation 0 = 0 is always true. So, the system of linear

equations has infi nitely many solutions. In this context, x and

y must be positive. So, the solutions are all points on the line

3x + 2y = 38 in Quadrant I. Therefore, there are infi nitely

many possibilities for the exact cost of one admission and

one skate rental.


33. ∣ 2x + 6 ∣ = ∣ x ∣

2x + 6 = x or 2x + 6 = –x

–2x –2x –2x –2x

6 = –x 6 = –3x

6 —

–1 =

–x — –1

6 —

–3 =

–3x —

–3

–6 = x –2 = x

Check ∣ 2x + 6 ∣ = ∣ x ∣ ∣ 2x + 6 ∣ = ∣ x ∣

∣ 2 ( –6 ) + 6 ∣ =? ∣ –6 ∣ ∣ 2 ( –2 ) + 6 ∣ =? ∣ –2 ∣

∣ –12 + 6 ∣ =? 6 ∣ –4 + 6 ∣ =? 2

∣ –6 ∣ =? 6 ∣ 2 ∣ =? 2

6 = 6 ✓ 2 = 2 ✓

The solutions are x = –6 and x = –2.


Chapter 5

34. ∣ 3x – 45 ∣ = ∣ 12x ∣

3x – 45 = 12x or 3x – 45 = –12x

–3x –3x –3x –3x

–45 = 9x –45 = –15x

–45

— 9 =

9x —

9

–45 —

–15 =

–15x —

–15

–5 = x 3 = x

Check ∣ 3x – 45 ∣ = ∣ 12x ∣ ∣ 3x – 45 ∣ = ∣ 12x ∣

∣ 3 ( –5 ) – 45 ∣ =? ∣ 12 ( –5 ) ∣ ∣ 3 ( 3 ) – 45 ∣ =? ∣ 12 ( 3 ) ∣

∣ –15 – 45 ∣ =? ∣ –60 ∣ ∣ 9 – 45 ∣ =? ∣ 36 ∣

∣ –60 ∣ =? 60 ∣ –36 ∣ =? 36

60 = 60 ✓ 36 = 36 ✓

The solutions are x = –5 and x = 3.

35. ∣ x – 7 ∣ = ∣ 2x – 8 ∣

x – 7 = 2x – 8 or x – 7 = − ( 2x – 8 )

–x –x x – 7 = –2x + 8

–7 = x – 8 +2x +2x

+8 +8 3x – 7 = 8

1 = x + 7 +7

3x = 15

3x

— 3 =

15 —

3

x = 5

Check ∣ x – 7 ∣ = ∣ 2x – 8 ∣ ∣ x – 7 ∣ = ∣ 2x – 8 ∣

∣ 1 – 7 ∣ =? ∣ 2 ( 1 ) – 8 ∣ ∣ 5 – 7 ∣ =? ∣ 2 ( 5 ) – 8 ∣

∣ –6 ∣ =? ∣ 2 – 8 ∣ ∣ –2 ∣ =? ∣ 10 – 8 ∣

6 =?

∣ –6 ∣ 2 = ∣ 2 ∣ 6 = 6 ✓ 2 = 2 ✓

The solutions are x = 1 and x = 5.

36. ∣ 2x + 1 ∣ = ∣ 3x –11 ∣

2x + 1 = 3x – 11 or 2x + 1 = – ( 3x – 11 )

–2x –2x 2x + 1 = –3x + 11

1 = x – 11 +3x +3x

+11 +11 5x + 1 = 11

12 = x –1 –1

5x = 10

5x

— 5 =

10 —

5

x = 2

Check ∣ 2x + 1 ∣ = ∣ 3x – 11 ∣ ∣ 2x + 1 ∣ = ∣ 3x – 11 ∣

∣ 2 ( 12 ) + 1 ∣ =? ∣ 3 ( 12 ) – 11 ∣ ∣ 2 ( 2 ) + 1 ∣ =? ∣ 3 ( 2 ) – 11 ∣

∣ 24 + 1 ∣ =? ∣ 36 – 11 ∣ ∣ 4 + 1 ∣ =? ∣ 6 – 11 ∣

∣ 25 ∣ =? ∣ 25 ∣ ∣ 5 ∣ =? ∣ –5 ∣ 25 = 25 ✓ 5 = 5 ✓

The solutions are x = 12 and x = 2.

5.1–5.4 What Did You Learn? (p. 259)

1. You know the total number of songs played, the relationship

between the number of pop songs played and the number

of rock songs played, and the relationship between the

number of hip-hop songs played and the number of rock

songs played. The solution can be found by writing a system

of three linear equations in three variables that represents

the problem. Then, substitute an expression for x and an

expression for z into the fi rst equation that contains all three

variables. Solve this equation for y. Substitute the value of

y into each of the other two equations and solve for x and z, respectively.

2. Sample answer: An Internet site offers commercial-free

viewing of individual episodes of a TV show for one price

or access to an entire season of a TV show for another price.

If you knew how many individual shows and seasons you

purchased in Month 1 and Month 2 and the total charge for

each of those months, you could write a system of linear

equations, similar to the one in Exercise 22, that could be

solved to fi nd the cost of viewing one episode and the cost

for access to an entire season.

3. Sample answer: What are the slope and y-intercept of the

line that describes the fi rst receipt? the second receipt? How

are these two equations related? What does that tell you

about the system of linear equations?

5.1–5.4 Quiz (p. 260)


Check y = − 1 — 3 x + 2 y = x − 2

1 =?

− 1 — 3 (3) + 2 1 =?

3 − 2

1 =?

−1 + 2 1 = 1 ✓

1 = 1 ✓


2. The lines appear to intersect at (−2, −2).

Check y = 1 —

2 x − 1 y = 4x + 6

−2 =?

1 — 2 (−2) − 1 −2 =

? 4(−2) + 6

−2 =?

−1 − 1 −2 =?

−8 + 6

−2 = −2 ✓ −2 = −2 ✓



Chapter 5


Check y = 1 y = 2x + 1

1 = 1 ✓ 1 =?

2(0) + 1

1 =?

0 + 1

1 = 1 ✓


4. Substitute x − 4 for y in Equation 2 and solve for x.

−2x + y = 18 Step 3 y = x − 4

−2x + (x − 4) = 18 y = −22 − 4

−2x + x − 4 = 18 y = −26

−x − 4 = 18

+ 4 + 4

−x = 22

−x — −1

= 22

— −1

x = −22

Check y = x − 4 −2x + y = 18

−26 =?

−22 − 4 −2(−22) + (−26) =?

18

−26 = −26 ✓ 44 − 26 =?

18

18 = 18 ✓


5. Step 1 y − x = −5

y − x + x = −5 + x

y = x − 5

Step 2 2y + x = −4 Step 3 y − x = −5

2(x − 5) + x = −4 y − 2 = −5

2(x) − 2(5) + x = −4 +2 +2

2x − 10 + x = −4 y = −3

3x − 10 = −4

+10 +10

3x = 6

3x

— 3 = 6 —

3

x = 2

Check 2y + x = −4 y − x = −5

2(−3) + 2 =?

−4 −3 −2 =?

−5

−6 + 2 =?

−4 −5 = −5 ✓

−4 = −4 ✓


6. Step 1 x + 4y = 10

x + 4y − 4y= 10 − 4y

x = 10 − 4y

Step 2 3x − 5y = 13 Step 3 x + 4y = 10

3(10 − 4y) − 5y = 13 x + 4(1) = 10

3(10) − 3(4y) − 5y = 13 x + 4 = 10

30 − 12y − 5y = 13 −4 −4

30 − 17y = 13 x = 6

−30 −30

−17y = −17

−17y

— −17

= −17

— −17

y = 1

Check 3x − 5y = 13 x + 4y = 10

3(6) − 5(1) =?

13 6 + 4(1) =?

10

18−5 =?

13 6 + 4 =?

10

13 = 13 ✓ 10 = 10 ✓


7. Step 2 x + y = 4

−3x − y = −8

−2x + 0 = −4

Step 3 −2x = −4

−2x

— −2

= −4

— −2

x = 2

Step 4 x + y = 4

2 + y = 4

−2 −2

y = 2

Check x + y = 4 −3x − y = −8

2 + 2 =?

4 −3(2) − 2 =?

−8

4 = 4 ✓ −6 − 2 =?

−8

−8 = −8 ✓



Chapter 5

8. Step 1 Step 2

x + 3y = 1 Multiply by −2. −2x − 6y = −2

5x + 6y = 14 5x + 6y = 14

3x + 0 = 12

Step 4 x + 3y = 1 Step 3 3x = 12

4 + 3y = 1 3x

— 3 =

12 —

3

−4 −4 x = 4

3y = −3

3y

— 3 =

−3 —

3

y = −1

Check x + 3y = 1 5x + 6y = 14

4 + 3(−1) =?

1 5(4) + 6(−1) =?

14

4 − 3 =?

1 20 − 6 =?

14

1 = 1 ✓ 14 = 14 ✓


9. Step 1 Step 2

2x − 3y = −5 Multiply by 2. 4x − 6y = −10

5x + 2y = 16 Multiply by 3. 15x + 6y = 48

19x + 0 = 38

Step 4 5x + 2y = 16 Step 3 19x = 38

5(2) + 2y = 16 19x

— 19

= 38

— 19

10 + 2y = 16 x = 2

−10 −10

2y = 6

2y

— 2 =

6 —

2

y = 3

Check 2x − 3y = −5 5x + 2y = 16

2(2) − 3(3) =?

−5 5(2) + 2(3) =?

16

4 − 9 =?

−5 10 + 6 =?

16

−5 = −5 ✓ 16 = 16 ✓



Step 2

x − y = 1

−(x − y = 6)

0 = −5

The equation 0 = −5 is never true. So, the system of linear



Step 1 6x + 2y = 16 Step 2 6x + 2y = 16

2x − y = 2 Multiply by 2. 4x − 2y = 4

10x + 0 = 20

Step 4 6x + 2y = 16 Step 3 10x = 20

6(2) + 2y = 16 10x

— 10

= 20

— 10

12 + 2y = 16 x = 2

−12 −12

2y = 4

2y — 2 =

4 —

2

y = 2

The solution is (2,2).


Step 1 Step 2

3x − 3y = −2 Multiply by 2. 6x − 6y = −4

−6x + 6y = 4 −6x + 6y = 4

0 = 0


all points on the line 3x − 3y = −2. The system of linear


13. a. Words14 + 4 ⋅

Growing

time

(in years)= Height

(in inches)

8 + 6 ⋅Growing

time

(in years)= Height

(in inches)

Variables Let x be how long (in years) the trees are

growing, and let y be the height (in inches) of

the trees.

System 14 + 4x = y

8 + 6x = y

A system of linear equations that represents this situation

is y = 4x + 14 and y = 6x + 8.


Chapter 5

b.

1 2 3 40 x

y

20

30

40

10

0

Time (years)

Hei

gh

t (i

nch

es)

y = 4x + 14

y = 6x + 8

(3, 26)

Check y = 4x + 14 y = 6x + 8

26 =?

4(3) + 14 26 =?

6(3) + 8

26 =?

12 + 14 26 =?

18 + 8

26 = 26 ✓ 26 = 26 ✓

The solution is (3, 26). So, in 3 years, both trees will be

26 inches tall.

14. a. Words Time

(in hours)

on highway+

Time

(in hours)

on other roads= 3

55 ⋅Time

(in hours)

on highway

+ 40 ⋅Time

(in hours)

on other roads= 135

Variables Let x be how much time (in hours) you spend

driving at 55 miles per hour on highways, and

let y be how much time (in hours) you spend

driving at 40 miles per hour on the rest of

the roads.

System x + y = 3

55x + 40y = 135

Solve by substitution.

Step 1 x + y = 3

x − x + y = 3 − x

y = 3 − x

Step 2 55x + 40y = 135

55x + 40(3−x) = 135

55x + 40(3) − 40(x) = 135

55x + 120 − 40x = 135

15x + 120 = 135

−120 −120

15x = 15

15x

— 15

= 15

— 15

x = 1

Step 3 x + y = 3

1 + y = 3

−1 −1

y = 2

The solution is (1,2). So, you spend 1 hour driving at

55 miles per hour on highways, and you spend 2 hours

driving at 40 miles per hour on the rest of the roads.

b. You drive 55x = 55(1) = 55 miles on highways and

40y = 40(2) = 80 miles on the rest of the roads.

15. Words Number of

touchdowns+

Number of

fi eld goals= 6

7 ⋅ Number of

touchdowns+ 3 ⋅ Number of

fi eld goals= 26

Variables Let x be the number of touchdowns the home

team scores, and let y be the number of fi eld goals

the home team scores.

System x + y = 6

7x + 3y = 26


Step 1 Step 2

x + y = 6 Multiply by −3. −3x − 3y = −18

7x + 3y = 26 7x + 3y = 26

4x + 0 = 8

Step 4 x + y = 6 Step 3 4x = 8

2 + y = 6 4x

— 4 =

8 —

4

−2 −2 x = 2

y = 4

The solution is (2, 4). So, the home team scores

2 touchdowns and 4 fi eld goals.


Chapter 5

Section 5.5


1. a. A linear equation that uses the left side is y = 2x − 1.

A linear equation that uses the right side is y = − 1 — 2 x + 4.

b.

x

y

2

6

4 62−2

y = 2x − 1

(2, 3)y =− x + 41

2

The x-value of the point of intersection is 2.

Check 2x − 1 = − 1 — 2 x + 4

2(2) − 1 =?

− 1 — 2 (2) + 4

4 − 1 =?

−1 + 4

3 = 3 ✓

c. The two sides of the equation are equal to each other.

If you set one side of the equation equal to y, the transitive

property allows you to set the other side of the equation

equal to y.

2. a. Method 1 Method 2

1 —

2 x + 4 = −

1 — 4 x + 1

x

y6

2

−2

−2−4−6

y = x + 412

y = − x + 114

(−4, 2)

+ 1 — 4 x +

1 — 4 x

3 —

4 x + 4 = 1

−4 −4

3 — 4 x = −3

4 — 3 ⋅

3 —

4 x =

4 —

3 ⋅ (−3)

x = −4

Using either method, the solution is x = −4.

b. Method 1 Method 2

2 — 3 x + 4 = 1 —

3 x + 3

x

y6

2

−2

2−2−4

y = x + 423

y = x + 313

(−3, 2)

− 1 — 3 x −

1 — 3 x

1 — 3 x + 4 = 3

−4 −4

1 — 3 x = −1

3 ⋅ 1 —

3 x = 3 ⋅ (−1)

x = −3

Using either method, the solution is x = −3.

c. Method 1 Method 2

− 2 — 3 x − 1 =

1 —

3 x − 4

x

y

−6

−2

4 62−2

y = x − 413

y = − x − 123

(3, −3) +

2 — 3 x +

2 — 3 x

−1 = x − 4 +4 +4

3 = x

Using either method, the solution is x = 3.

d. Method 1 Method 2

4 — 5 x +

7 —

5 = 3x − 3

x

y4

−2

42−4

y = x + 45

75

y = 3x − 3

(2, 3) − 4 — 5 x −

4 — 5 x

7 — 5 =

11 —

5 x − 3

+3 +3

22 —

5 =

11 —

5 x

5 — 11

⋅ 22

— 5 =

5 —

11 ⋅

11 —

5 x

2 = x


e. Method 1 Method 2

−x + 2.5 = 2x − 0.5

x

y

2

1

21−1

y = −x + 2.5

y = 2x − 0.5

(1, 1.5) +x +x

2.5 = 3x − 0.5

+0.5 +0.5

3.0 = 3x

3 —

3 =

3x —

3

1 = x


f. Method 1 Method 2

−3x + 1.5 = x + 1.5

x

y3

1

−1

21−1

y = −3x + 1.5

y = x + 1.5 (0, 1.5)

+3x +3x

1.5 = 4x + 1.5

−1.5 −1.5

0 = 4x

0 —

4 =

4x —

4

0 = x



Chapter 5

3. Sample answer: First, use the left side of the original

equation to write a linear equation. Then use the right side

of the original equation to write a second linear equation.

Graph these two linear equations, and fi nd the x-value of

the point of intersection. This value is the solution of the

original equation.

4. Sample answer: When you use the algebraic method to solve

an equation with variables on both sides, you will always

be able to get an exact answer, even if the solution is not

a whole number. However, you sometimes have to work

with fractions and decimals, which can be tedious, and it is

possible to make careless errors. When you use the graphical

method to solve an equation with variables on both sides,

you see a visual representation of how the value of each

expression changes as the value of x changes. However, it

can be tedious to fi nd the appropriate scale for the axes, and

the graphical method may only provide an estimate of the

solution, especially if the solution is not a whole number.


1. 1 — 2 x − 3 = 2x

Graph the system.

x

y

2

−4

2−2−4

y = 2x

(−2, −4)

y = x − 312

y = 1 —

2 x − 3

y = 2x

The graphs intersect at (−2, −4).

Check 1 —

2 x − 3 = 2x

1 — 2 (−2) − 3 =

? 2(−2)

−1 − 3 =?

−4

−4 = −4 ✓

So, the solution of the equation is x = −2.

2. −4 + 9x = −3x + 2

Graph the system.

x

y4

2

−4

−2

2 3−1

y = −4 + 9x

y = −3x + 2

(0.5, 0.5)

y = −4 + 9x

y = −3x + 2

The graphs intersect

at (0.5, 0.5).

Check −4 + 9x = −3x +2

−4 + 9(0.5) =?

−3(0.5) + 2

−4 + 4.5 =?

−1.5 + 2

0.5 = 0.5 ✓

So, the solution of the equation is x = 0.5.

3. ∣ 2x + 2 ∣ = ∣ x − 2 ∣ Equation 1 Equation 2

2x + 2 = x − 2 2x + 2 = −(x − 2)

2x + 2 = −x + 2

System 1 System 2

y = 2x + 2 y = 2x + 2

y = x − 2 y = −x + 2

x

y

−4

2−2−4

y = 2x + 2

y = x − 2

(−4, −6)

x

y

4

−2

42−2−4

y = 2x + 2

y = −x + 2 (0, 2)

The graphs intersect The graphs intersect

at (−4, −6). at (0, 2).

Check Check

∣ 2x + 2 ∣ = ∣ x − 2 ∣ ∣ 2x + 2 ∣ = ∣ x − 2 ∣ ∣ 2(−4) + 2 ∣ =? ∣ −4 − 2 ∣ ∣ 2(0) + 2 ∣ =? ∣ 0 − 2 ∣ ∣ −8 + 2 ∣ =? ∣ −6 ∣ ∣ 0 + 2 ∣ =? ∣ −2 ∣ ∣ −6 ∣ =? 6 ∣ 2 ∣ =? 2

6 = 6 ✓ 2 = 2 ✓

So, the solutions are x = −4 and x = 0.


Chapter 5

4. ∣ x − 6 ∣ = ∣ −x + 4 ∣ Equation 1 Equation 2

x − 6 = −x + 4 x − 6 = −(−x + 4)

x − 6 = x − 4

System 1 System 2

y = x − 6 y = x − 6

y = −x + 4 y = x − 4

x

y

2

−4

−2

42

y = −x + 4

y = x − 6

(5, −1)

x

y

−8

−2

42−2−4

y = x − 4

y = x − 6

The graphs intersect The lines are parallel and

at (5, −1). do not intersect.

Check

∣ x − 6 ∣ = ∣ −x + 4 ∣

∣ 5 − 6 ∣ =? ∣ −5 + 4 ∣

∣ −1 ∣ =? ∣ −1 ∣ 1 = 1 ✓

So, the solution is x = 5.

5. Words Company A Company C

Cost

per

mile⋅ Miles +

Flat

fee =Cost

per

mile⋅ Miles +

Flat

fee

Variable Let x be the number of miles traveled.

Equation 3.25x + 125 = 3.3x + 115

Use a graphing calculator to graph the system.

y = 3.25x + 125

y = 3.3x + 115

2500

0

900


Because the graphs intersect at (200, 775), the solution of

the equation is x = 200. So, the total costs are the same after

200 miles.



1. The solution of the equation is x = 6.

2. For the fi rst system, set y equal to both expressions inside the

absolute value symbols, 2x − 4 and −5x + 1. For the second

system, set y equal to the expression inside the fi rst absolute

value symbol, 2x − 4, and the opposite of the expression

inside the second absolute value symbol, 5x − 1.


3. The graphs of y = −2x + 3 and y = x intersect at (1, 1).

Check

−2x + 3 = x

−2(1) + 3 =?

1

−2 + 3 =?

1

1 = 1 ✓

So, the solution of the original equation is x = 1.

4. The graphs of y = −3 and y = 4x + 1 intersect at (−1, −3).

Check

−3 = 4x + 1

−3 =?

4(−1) + 1

−3 =?

−4 + 1

−3 = −3 ✓

So, the solution of the original equation is x = −1.

5. The graphs of y = −x − 1 and y = 1 —

3 x + 3 intersect at

(−3, 2).

Check

−x − 1 = 1 —

3 x + 3

−(−3) − 1 =?

1 —

3 (−3) + 3

3 − 1 =?

−1 + 3

2 = 2 ✓


6. The graphs of y = − 3 — 2 x – 2 and y = −4x + 3 intersect at

(2, −5).

Check

− 3 — 2 x − 2 = −4x + 3

− 3 — 2 (2) − 2 =

? −4(2) + 3

−3 − 2 =?

−8 + 3

−5 = −5 ✓

So, the solution of the original equation is x = 2.


Chapter 5

7. x + 4 = −x

Graph the system.

x

y

2

6

−2

42−2

y = −x

y = x + 4

(−2, 2)

y = x + 4

y = −x

The graphs intersect Check at (−2, 2). x + 4 = −x

−2 + 4 =?

−(−2)

2 = 2 ✓


8. 4x = x + 3

Graph the system.

x

y

4

42−2

y = 4x

y = x + 3 (1, 4) y = 4x

y = x + 3

The graphs intersect Check at (1, 4). 4x = x + 3

4(1) =?

1 + 3

4 = 4 ✓

So, the solution of the equation is x = 1.

9. x + 5 = −2x − 4

Graph the system.

x

y

4

−4

y = x + 5

y = −2x − 4

(−3, 2)

y = x + 5

y = −2x − 4

The graphs intersect Check at (−3, 2). x + 5 = −2x − 4

−3 + 5 =?

−2(−3) − 4

2 =?

6 − 4

2 = 2 ✓


10. −2x + 6= 5x − 1

Graph the system.

x

y

4

2

4 62−2

y = 5x − 1

y = −2x + 6

(1, 4)

y = −2x + 6

y = 5x − 1

The graphs intersect Check at (1, 4). −2x + 6 = 5x − 1

−2(1) + 6 =?

5(1) − 1

−2 + 6 =?

5 − 1

4 = 4 ✓


11. 1 — 2 x – 2 = 9 − 5x

Graph the system.

x

y4

2

−4

4 6−2

y = 9 − 5x

y = x − 212

(2, −1)

y = 1 —

2 x − 2

y = 9 − 5x

The graphs intersect Check

at (2, −1). 1 —

2 x − 2 = 9 − 5x

1 —

2 (2) − 2 =

? 9 − 5(2)

1 − 2 =?

9 − 10

−1 = −1 ✓



Chapter 5

12. −5 + 1 — 4 x = 3x + 6

Graph the system.

x

y

−8

−4

84−4−8

y = 3x + 6

y = −5 + x14

(−4, −6)

y = −5 + 1 — 4 x

y = 3x + 6

The graphs intersect Check

at (−4, −6). −5 + 1 — 4 x = 3x + 6

−5 + 1 — 4 (−4) =?

3(−4) + 6

−5 − 1 =?

−12 + 6

−6 = −6 ✓


13. 5x − 7 = 2(x + 1)

5x − 7 = 2(x) + 2(1)

5x − 7 = 2x + 2

Graph the system.

x

y

8

−4

84−4−8

y = 2x + 2

y = 5x − 7

(3, 8) y = 5x − 7

y = 2x + 2

The graphs intersect Check at (3, 8). 5x − 7 = 2(x + 1)

5(3) − 7 =?

2(3 + 1)

15 − 7 =?

2(4)

8 = 8 ✓


14. −6(x + 4) = −3x − 6

−6(x) − 6(4) = −3x − 6

−6x − 24 = −3x − 6

Graph the system.

x

y24

−24

2412−12−24

y = −6x − 24y = −3x − 6

(−6, 12)

y = −6x − 24

y =−3x − 6

The graphs intersect Check at (−6, 12). −6(x + 4) = −3x − 6

−6(−6 + 4) =?

−3(−6) − 6

−6(−2) =?

18 − 6

12 = 12 ✓


15. 3x − 1 = −x + 7

Graph the system.

x

y

4

6

2

4 62

y = 3x − 1

y = −x + 7(2, 5)

y = 3x − 1

y = −x + 7

The graphs intersect Check at (2, 5). 3x − 1 = −x + 7

3(2) − 1 =?

−2 + 7

6 − 1 =?

5

5 = 5 ✓

So, the equation has one solution, which is x = 2.

16. 5x − 4 = 5x + 1

Graph the system.

x

y

2

42−2−4

y = 5x − 4

y = 5x + 1

y = 5x − 4

y = 5x + 1

The lines have the same slope

but different y-intercepts. So,

they are parallel. Because

parallel lines never intersect,

there is no point that is

a solution of both linear

equations. So, the original

equation has no solution.


Chapter 5

17. −4(2 − x) = 4x − 8

−4(2) − 4(−x) = 4x − 8

−8 + 4x = 4x − 8

Graph the system.

x

y

−4

−6

−8

−2

4 6−2

y = 4x − 8

y = −4(2 − x)

y = −8 + 4x

y = 4x − 8

Because the lines have the

same slope and the same

y-intercept, the lines are the

same. So, all points on the

line are solutions of both

equations, which means

the original equation has

infi nitely many solutions.

18. −2x − 3 = 2(x − 2)

−2x − 3 = 2(x) − 2(2)

−2x − 3 = 2x − 4

Graph the system.

x

y2

−4

4−2−4

y = 2x − 4y = −2x − 3

(0.25, −3.5)

y = −2x − 3

y = 2x − 4

The graphs intersect Check at (0.25, −3.5). −2x − 3 = 2(x − 2)

−2(0.25) − 3 =?

2(0.25 − 2)

−0.5 − 3 =?

2(−1.75)

−3.5 = −3.5 ✓

So, the original equation has one solution, which is x = 0.25.

19. −x − 5 = − 1 — 3 (3x + 5)

−x − 5 = − 1 — 3 (3x) − 1 — 3 (5)

−x − 5 = −x − 5 —

3

Graph the system.

x

y

2

−2

2−2−6

y = −x − 5

y = −x − 53

y = −x − 5

y = −x − 5 —

3


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 linear



20. 1 — 2 (8x + 3) = 4x +

3 —

2

1 — 2 (8x) +

1 —

2 (3) = 4x +

3 —

2

4x + 3 —

2 = 4x +

3 —

2

Graph the system.

x

y

4

42−2−4

y = 4x + 32

y = (8x + 3)12

y = 4x + 3 —

2

y = 4x + 3 —

2


and the same y-intercept. So,

the lines are the same, which

means all points on the line

are solutions of both linear


equation has infi nitely many

solutions.

21. The lines intersect at (−2, −6) in the fi rst graph, and in the

second graph, the lines intersect at (1, −3).

Check ∣ x − 4 ∣ = ∣ 3x ∣ ∣ x − 4 ∣ = ∣ 3x ∣

∣ −2 − 4 ∣ =? ∣ 3(−2) ∣ ∣ 1 − 4 ∣ =? ∣ 3(1) ∣

∣ −6 ∣ =? ∣ −6 ∣ ∣ −3 ∣ =? ∣ 3 ∣ 6 = 6 ✓ 3 = 3 ✓


22. In the fi rst graph, the lines intersect at (−5, −6), and in the

second graph, the lines intersect at (−1, 2).

Check

∣ 2x + 4 ∣ = ∣ x − 1 ∣ ∣ 2x + 4 ∣ = ∣ x − 1 ∣

∣ 2(−5) + 4 ∣ =? ∣ −5 − 1 ∣ ∣ 2(−1) + 4 ∣ =? ∣ −1 − 1 ∣

∣ −10 + 4 ∣ =? ∣ −6 ∣ ∣ −2 + 4 ∣ =? ∣ −2 ∣

∣ −6 ∣ =? 6 ∣ 2 ∣ =? 2

6 = 6 ✓ 2 = 2 ✓

So, the solutions are x = −5 and x = −1.


Chapter 5

23. ∣ 2x ∣ = ∣ x + 3 ∣ Equation 1 Equation 2

2x = x + 3 2x = −(x + 3)

2x = −x − 3

System 1 System 2

y = 2x y = 2x

y = x + 3 y = −x − 3

x

y

4

6

2

42−2

y = 2x

y = x + 3

(3, 6)

x

y4

2

−4

42−4

y = 2x

y = −x − 3

(−1, −2)


at (3, 6). at (−1, −2).

Check ∣ 2x ∣ = ∣ x + 3 ∣ Check ∣ 2x ∣ = ∣ x + 3 ∣

∣ 2(3) ∣ =? ∣ 3 + 3 ∣ ∣ 2(−1) ∣ =? ∣ −1 + 3 ∣

∣ 6 ∣ =? ∣ 6 ∣ ∣ −2 ∣ =? ∣ 2 ∣ 6 = 6 ✓ 2 = 2 ✓

So, the solutions are x = 3 and x = −1.

24. ∣ 2x − 6 ∣ = ∣ x ∣ Equation 1 Equation 2

2x − 6 = x 2x − 6 = −x

System 1 System 2

y = 2x − 6 y = 2x − 6

y = x y = −x

x

y8

4

−4

84−4−8

y = x

y = 2x − 6

(6, 6)

x

y

−4

−6

−2

4 62−2

y = −x

y = 2x − 6

(2, −2)


at (6, 6). at (2, −2).

Check ∣ 2x − 6 ∣ = ∣ x ∣ Check ∣ 2x − 6 ∣ = ∣ x ∣

∣ 2(6) − 6 ∣ =? ∣ 6 ∣ ∣ 2(2) − 6 ∣ =? ∣ 2 ∣

∣ 12 − 6 ∣ =? 6 ∣ 4 − 6 ∣ =? 2

∣ 6 ∣ =? 6 ∣ −2 ∣ =? 2 6 = 6 ✓ 2 = 2 ✓

So, the solutions are x = 6 and x = 2.

25. ∣ −x + 4 ∣ = ∣ 2x − 2 ∣ Equation 1 Equation 2

−x + 4 = 2x − 2 −x + 4 = −(2x − 2)

−x + 4 = −2x + 2

System 1 System 2

y = −x + 4 y = −x + 4

y = 2x − 2 y = −2x + 2

x

y

2

−2

4 62−2

y = −x + 4

y = 2x − 2

(2, 2)

x

y

6

2

42−2−4

y = −x + 4

y = −2x + 2

(−2, 6)


at (2, 2). at (−2, 6).

Check Check

∣ −x + 4 ∣ = ∣ 2x − 2 ∣ ∣ −x + 4 ∣ = ∣ 2x − 2 ∣

∣ −2 + 4 ∣ =? ∣ 2(2) − 2 ∣ ∣ −(−2) + 4 ∣ =? ∣ 2(−2) − 2 ∣

∣ 2 ∣ =? ∣ 4 − 2 ∣ ∣ 2 + 4 ∣ =? ∣ −4 − 2 ∣

2 =?

∣ 2 ∣ ∣ 6 ∣ =? ∣ −6 ∣ 2 = 2 ✓ 6 = 6 ✓

So, the solutions are x = 2 and x = −2.


Chapter 5

26. ∣ x + 2 ∣ = ∣ −3x + 6 ∣ Equation 1 Equation 2

x + 2 = −3x + 6 x + 2 = −(−3x + 6)

x + 2 = 3x − 6

System 1 System 2

y = x + 2 y = x + 2

y = −3x + 6 y = 3x − 6

x

y

4

4−4

y = x + 2

y = −3x + 6

(1, 3)

x

y8

−4

84−8

y = x + 2

y = 3x − 6

(4, 6)


at (1, 3). at (4, 6).

Check Check

∣ x + 2 ∣ = ∣ −3x + 6 ∣ ∣ x + 2 ∣ = ∣ −3x + 6 ∣

∣ 1 + 2 ∣ =? ∣ −3(1) + 6 ∣ ∣ 4 + 2 ∣ =? ∣ −3(4) + 6 ∣

∣ 3 ∣ =? ∣ −3 + 6 ∣ ∣ 6 ∣ =? ∣ −12 + 6 ∣

3 =?

∣ 3 ∣ 6 =?

∣ −6 ∣ 3 = 3 ✓ 6 = 6 ✓

So, the solutions are x = 1 and x = 4.

27. ∣ x + 1 ∣ = ∣ x − 5 ∣ Equation 1 Equation 2

x + 1 = x − 5 x + 1 = −(x − 5)

x + 1 = −x + 5

System 1 System 2

y = x + 1 y = x + 1

y = x − 5 y = −x + 5

x

y2

−4

−2

2−4

y = x + 1

y = x − 5

x

y

4

2

42

y = x + 1

y = −x + 5

(2, 3)

The graphs do not The graphs intersect

intersect. So, this at (2, 3).

system has no solution.

Check ∣ x + 1 ∣ = ∣ x − 5 ∣

∣ 2 + 1 ∣ =? ∣ 2 − 5 ∣

∣ 3 ∣ =? ∣ −3 ∣ 3 = 3 ✓

So, the solution is x = 2.


Chapter 5

28. ∣ 2x + 5 ∣ = ∣ −2x + 1 ∣ Equation 1 Equation 2

2x + 5 = −2x + 1 2x + 5 = −(−2x + 1)

2x + 5 = 2x − 1

System 1 System 2

y = 2x + 5 y = 2x + 5

y = −2x + 1 y = 2x − 1

x

y

6

42−2−4

y = 2x + 5

y = −2x + 1(−1, 3)

x

y

6

2

42−2−4

y = 2x + 5

y = 2x − 1

The graphs intersect The graphs do not intersect.

at (−1, 3). So, this system has no

solution.

Check ∣ 2x + 5 ∣ = ∣ −2x + 1 ∣

∣ 2(−1) + 5 ∣ =? ∣ −2(−1) + 1 ∣

∣ −2 + 5 ∣ =? ∣ 2 + 1 ∣

∣ 3 ∣ =? ∣ 3 ∣ 3 = 3 ✓

So, the solution is x = −1.

29. ∣ x − 3 ∣ = 2 ∣ x ∣ Equation 1 Equation 2

x − 3 = 2(x) x − 3 = 2(−x)

x − 3 = 2x x − 3 = −2x

System 1 System 2

y = x − 3 y = x − 3

y = 2x y = −2x

x

y

−6

4−2−4

y = 2x

y = x − 3

(−3, −6)

x

y4

2

−2

4−2−4

y = x − 3

y = −2x

(1, −2)

The graphs intersect at The graphs intersect at

(−3, −6). (1, −2).

Check ∣ x − 3 ∣ = 2 ∣ x ∣ Check ∣ x − 3 ∣ = 2 ∣ x ∣

∣ −3 − 3 ∣ =? 2 ∣ −3 ∣ ∣ 1 − 3 ∣ =? 2 ∣ 1 ∣

∣ −6 ∣ =? 2(3) ∣ −2 ∣ =? 2(1)

6 = 6 ✓ 2 = 2 ✓


30. 4 ∣ x + 2 ∣ = ∣ 2x + 7 ∣ Equation 1 Equation 2

4(x + 2) = 2x + 7 4(x + 2) = −(2x + 7)

4(x) + 4(2) = 2x + 7 4(x) + 4(2) = −2x − 7

4x + 8 = 2x + 7 4x + 8 = −2x − 7

System 1 System 2

y = 4x + 8 y = 4x + 8

y = 2x + 7 y = −2x − 7

x

y

4

2

2−6

y = 4x + 8

y = 2x + 7

(−0.5, 6)

x

y

4

−8

−4

−4−6

y = 4x + 8

y = −2x − 7

(−2.5, −2)


at (−0.5, 6). at (−2.5, −2).

Check 4 ∣ x + 2 ∣ = ∣ 2x + 7 ∣ 4 ∣ −0.5 + 2 ∣ =? ∣ 2(−0.5) + 7 ∣ 4 ∣ 1.5 ∣ =? ∣ −1 + 7 ∣ 4(1.5) =

? ∣ 6 ∣

6 = 6 ✓

Check 4 ∣ x + 2 ∣ = ∣ 2x + 7 ∣ 4 ∣ −2.5 + 2 ∣ =? ∣ 2(−2.5) + 7 ∣ 4 ∣ −0.5 ∣ =? ∣ −5 + 7 ∣ 4(0.5) =

? ∣ 2 ∣

2 = 2 ✓

The solutions are x = −0.5 and x = −2.5.

31. 0.7x + 0.5 = −0.2x − 1.3

Graph the system.

6

−4

−6

4

IntersectionX=-2 Y=-.9

y = 0.7x + 0.5

y = −0.2x − 1.3

The solution is x = −2.

32. 2.1x + 0.6 = −1.4x + 6.9

Graph the system.

6

−2

−6

6

IntersectionX=1.8 Y=4.38

y = 2.1x + 0.6

y = −1.4x + 6.9

The solution is x = 1.8.


Chapter 5

33. Words

Company A Company B

Cost

per

guest⋅ Guests +

Flat

fee =Cost

per

guest⋅ Guests +

Flat

fee

Variable Let x be the number of guests at the wedding

reception.

Equation 20x + 500 = 16x + 800

Use a graphing calculator

1000

0

3000


to graph the system.

y = 20x + 500

y = 16x + 800

Because the graphs intersect at (75, 2000), the solution of

the equation is x = 75. So, the total costs are the same for

75 guests.

34. Words

Current

age in

dog

years

Human

years

Current

age

in cat

years

Human

years+ 7 ⋅ + 4 ⋅ =

Variable Let x be how many human years have passed.

Equation 16 + 7x = 28 + 4x

Use a graphing calculator

100

0

100


to graph the system.

y = 16 + 7x

y = 28 + 4x

Because the graphs intersect at (4, 44), the solution of the

equation is x = 4. So, after 4 human years, your dog will be

44 in dog years, and your cat will be the same age of 44 in

cat years.

35. Use a graphing calculator to graph the system.

d = ∣ −5t + 100 ∣

d = ∣ − 10

— 3 t + 50 ∣

500

0

100


The graphs intersect between 15 and 20 seconds, but that is

when you and your friend are running in opposite directions.

The intersection at (30, 50) represents when you catch up to

your friend. So, the solution is x = 30, meaning you catch up

to your friend after 30 seconds.

36. no; The x-value of the coordinate pair is the solution of the

equation. So, the solution of the equation −x + 4 = 2x −8

is x = 4. When x = 4, the equation is true because each side

of the equation is equal to 0.

37. Sample answer: mx + b = −2x − 1

m(−3) + b = −2(−3) − 1

−3m + b = 6 − 1

−3m + b = 5

Let m = 1. −3(1) + b = 5

−3 + b = 5

+3 +3

b = 8

So, if m = 1 and b = 8, then the solution of

x + 8 = −2x − 1 is x = −3.

38. a. Sample answer: The graphs intersect at approximately

(7, 5.7).

b. Sample answer: The company will break even after being

open for about 7 years. In other words, the company

will be open for business for about 7 years before it

has generated enough revenue to recover the start-up

expenses.

39. P = x + (x − 2) + 6 = 2x + 4

A = 1 —

2 (x − 2)(6) = 3(x − 2) = 3x − 6

Graph the system.

x

y

16

24

8

8 124

y = 2x + 4

y = 3x − 6

(10, 24) y = 2x + 4

y = 3x − 6

The graphs intersect

at (10, 24). So, x = 10.

40. Sample answer: Let b = 15,000.

b + mx = 20,000 − 1500x

15,000 + m(5) = 20,000 − 1500(5)

15,000 + 5m = 20,000 − 7500

15,000 + 5m = 12,500

−15,000 −15,000

5m = −2500

5m — 5 =

−2500 —

5

m = −500

So, a car with an initial value of $15,000 that decreases in

value at a rate of $500 per year will also have a value of

$12,500 in exactly 5 years.


Chapter 5

41. a. The solution will be negative. The lines will intersect to

the left of the y-axis as in the example shown, where

x = −2 is the solution of the equation 2x + 4 = 1 —

2 x + 1.

x

y

4

2

−2

42−4

y = 2x + 4,y = cx + d

y = x + 1,

y = ax + b

12

(−2, 0)

b. The solution will be positive. The lines will intersect to

the right of the y-axis as in the example shown, where

x = 2 is the solution of the equation 2x − 6 = 1 —

2 x − 3.

x

y

−4

−6

−2

2−2−4

y = 2x − 6,y = cx + d

y = x − 3,

y = ax + b

12 (2, −2)


42. y > 5 0 2 4 6 8

5

43. x ≤ −2 0 2−2−4−6

44. n ≥ 9 0 4 8 12 16

9

45. c < −6 0−4−8−12−16

−6

46. x

y

−2

−2−4

g(x) = f(x + 2)

f(x) = x − 5

The function g is of the form y = f (x − h), where h = −2.

So, the graph of g is a horizontal translation 2 units left of

the graph of f.

47. x −1 0 1

−6x 6 0 −6

−f (x) −6 0 6

x

y

2

84−4−8

g(x) = −f(x)

f(x) = 6x

The function g is of the form y = −f (x). So, the graph of g is

a refl ection in the x-axis of the graph of f.

48. x −1 0 1

4x −4 0 4

f (4x) 9 1 −7

x

y

−8

−4

84−4−8

g(x) = f(4x)

f(x) = −2x + 1

The function g is of the form y = f (ax), where a = 4. So, the

graph of g is a horizontal shrink of the graph of f by a factor

of 1 ÷ 4 = 1 —

4 .

49. x −1 1 3

x − 1 −2 0 2

f (x − 1) −3 −2 −1

x

y2

−4

−6

−2−4

g(x) = f(x − 1)

f(x) = x − 212

The function g is of the form y = f (x − h), where h = 1.

So, the graph of g is a horizontal translation 1 unit right of

the graph of f.


Chapter 5

Section 5.6


1. a. The dashed line crosses the y-axis at (0, −3), and the

slope of the line is 1. So, an equation represented by the

dashed line is y = x − 3.

y = mx + b

y = 1x + (−3)

y = x − 3 b. The solutions are all the points below the line y = x − 3.

c. An inequality represented by the graph is y < x − 3.

Sample answer: The point (4, 0) is in the shaded region,

and to make the inequality true for that point, the < symbol

is needed.

2. a. Check students’ work.

b. Check y ≥ 1 — 4 x − 3

0 ≥? 1 — 4 (0) − 3

0 ≥?

0 − 3

0 ≥ − 3 ✓

The point (0, 0) is a solution of the inequality y ≥ 1 — 4 x − 3.

3. a.

x

2

4

6

−2

2−2−4

y

Test (0, 0).

y > x + 5

0 >? 0 + 5

0 > 5 ✗

Sample answer: Graph y = x + 5 with a dashed line

because the inequality symbol > indicates that the

points on the line are not solutions. Test the point (0, 0)

to determine whether it is a solution of the inequality.

Because the point (0, 0) is not a solution, shade the half-

plane that does not contain (0, 0).

b.

x

2

−2

−4

4

42−2−4

y

Test (0, 0).

y ≤ − 1 — 2 x + 1

0 ≤? − 1 — 2 (0) + 1

0 ≤? 0 + 1

0 ≤ 1 ✓

Sample answer: Graph y = − 1 — 2 x + 1 with a solid line

because the inequality symbol ≤ indicates that the points

on the line are solutions. Test the point (0, 0) to determine

whether it is a solution of the inequality. Because the point

(0, 0) is a solution, shade the half-plane that contains (0, 0).

c.

x

y2

−6

−2

2−2−6

Test (0, 0).

y ≥ −x − 5

0 ≥? −0 − 5

0 ≥ −5 ✓

Sample answer: Graph y = −x − 5 with a solid line

because the inequality symbol ≥ indicates that the points

on the line are solutions. Test the point (0, 0) to determine

whether it is a solution of the inequality. Because the point

(0, 0) is a solution, shade the half-plane that contains (0, 0).

4. To graph a linear inequality in two variables, fi rst graph the

boundary line for the inequality. Use a dashed line for < or >.

Use a solid line for ≤ or ≥. Next, test a point that is not on

the boundary line to determine whether it is a solution of

the inequality. When the test point is a solution, shade the

half-plane that contains the test point. When the test point is

not a solution, shade the half-plane that does not contain the

test point.


Chapter 5

5. Sample answer: You are selling tickets for your band’s fi rst

show. Adult tickets cost $10 each, and child tickets cost $5

each. You and the other band members set a goal of selling at

least $500 worth of tickets. You can write and graph a linear

inequality to represent how many of each type of ticket you

must sell in order to reach your goal. Let x be how many

adult tickets you sell, and let y be how many child tickets you

sell. Then, a linear inequality that represents this situation is

10x + 5y ≥ 500.


1. x + y > 0

−2 + 2 >?

0

0 > 0 ✗

So, (−2, 2) is not a solution of the inequality.

2. 4x − y ≥ 5

4(0) − 0 ≥? 5

0 − 0 ≥? 5

0 ≥ 5 ✗

So, (0, 0) is not a solution of the inequality.

3. 5x − 2y ≤ −1

5(−4) − 2(−1) ≤?

−1

−20 + 2 ≤?

−1

−18 ≤ −1 ✓

So, (−4, −1) is a solution of the inequality.

4. −2x − 3y < 15

−2(5) − 3(−7) <?

15

−10 + 21 ≤?

15

11 ≤ 15 ✓

So, (5, −7) is a solution of the inequality.

5.

x

2

−2

−4

4

42−2−4

y Test (0, 0).

y > −1

0 > −1 ✓

6.

x

2

−2

−4

4

−2

y

−6−8

Test (0, 0).

x ≤ −4

0 ≤ −4 ✗

7. x + y ≤ −4

x

y2

−6

−2

2−2−6

x − x + y ≤ −4 − x y ≤ −x − 4 Test (0, 0).

x + y ≤ −4

0 + 0 ≤?

−4

0 ≤ −4 ✗

8. x − 2y < 0

x

2

−2

−4

4

42−2−4

y

x − x − 2y < 0 − x −2y < − x

−2y — −2

> −x

— −2

y > 1 —

2 x

Test (0, 2).

x − 2y < 0

0 − 2(2) <? 0

0 − 4 <? 0

−4 < 0 ✓

9. Words

Cost per

pound

of red

peppers

Pounds

of red

peppers

Cost per

pound of

tomatoes

Pounds

of

tomatoes

Amount

you can

spend⋅ + ⋅ ≤

Variables Let x be the weight (in pounds) of red peppers,

and let y be the weight (in pounds) of tomatoes.

Inequality 4 ⋅ x + 3 ⋅ y ≤ 12

4x + 3y ≤ 12

4x − 4x + 3y ≤ 12 − 4x

3y ≤ −4x + 12

3y

— 3 ≤

−4x + 12 —

3

y ≤ − 4 — 3 x + 4

2

3

4

1

00

2 3 41

Red peppers (pounds)

Tom

ato

es (

po

un

ds)

x

y Test (0, 0).

4x + 3y ≤ 12

4(0) + 3(0) ≤? 12

0 + 0 ≤? 12

0 ≤ 12 ✓


Chapter 5

Check 4x + 3y ≤ 12 4x + 3y ≤ 12

4(1) + 3(1) ≤? 12 4(1.5) + 3(2) ≤

? 12

4 + 3 ≤? 12 6 + 6 ≤

? 12

7 ≤ 12 ✓ 12 ≤ 12 ✓

Sample answer: One possible solution is (1, 1) because it lies

in the shaded half-plane. Another possible solution is (1.5, 2)

because it lies on the solid line. So, you can buy 1 pound of

red peppers and 1 pound of tomatoes, or 1.5 pounds of red

peppers and 2 pounds of tomatoes.



1. To tell whether an ordered pair is a solution of a linear

inequality, substitute the values into the linear inequality.

If they make the inequality true, then the ordered pair is a

solution of the linear inequality. If they make the inequality

false, then the ordered pair is not a solution of the linear

inequality.

2. The graph of a linear inequality in two variables and a linear

equation in two variables both have lines. The graph of a

linear equation in two variables is always a solid line, and

only the points on the line are solutions of the equation. The

graph of a linear inequality in two variables has a boundary

line that is either solid or dashed, and you shade on one side

or the other of the boundary line to indicate all of the points

that are solutions of the inequality.


3. x + y < 7

2 + 3 <?

7

5 < 7 ✓

So, (2, 3) is a solution of the inequality.

4. x − y ≤ 0

5 − 2 ≤?

0

3 ≤ 0 ✗

So, (5, 2) is not a solution of the inequality.

5. x + 3y ≥ −2

−9 + 3(2) ≥?

−2

−9 + 6 ≥?

−2

−3 ≥ −2 ✗


6. 8x + y > −6

8(−1) + 2 >? −6

−8 + 2 >? −6

−6 > −6 ✗


7. −6x + 4y ≤ 6

−6(−3) + 4(−3) ≤?

6

18 − 12 ≤?

6

6 ≤ 6 ✓


8. 3x − 5y ≥ 2

3(−1) − 5(−1) ≥?

2

−3 + 5 ≥?

2

2 ≥ 2 ✓


9. −x − 6y > 12

−(−8) − 6(2) >? 12

8 − 12 >? 12

−4 > 12 ✗


10. −4x − 8y < 15

−4(−6) − 8(3) <? 15

24 − 24 <? 15

0 < 15 ✓

So, (−6, 3) is a solution of the inequality.

11. no; (0, −1) is not a solution because it lies in the half-plane

that is not shaded.

12. yes; (−1, 3) is a solution because it lies in the shaded

half-plane.

13. yes; (1, 4) is a solution because it lies in the shaded half-plane.

14. no; (0, 0) is not a solution because it lies on the dashed line.

15. no; (3, 3) is not a solution because it lies on the dashed line.

16. no; (2, 1) is not a solution because it lies in the half-plane that

is not shaded.

17. Test (12, 14). 8x + 12y ≤ 250

8(12) + 12(14) ≤? 250

96 + 168 ≤? 250

264 ≤ 250 ✗

no; (12, 14) is not a solution of the inequality 8x + 12y ≤ 250.

So, the carpenter cannot buy twelve 2-by-8 boards and fourteen

4-by-4 boards.


Chapter 5

18. Test (20, 18). 3x + 2y ≥ 93

3(20) + 2(18) ≥? 93

60 + 36 ≥? 93

96 ≥ 93 ✓

yes; (20, 18) is a solution of the inequality 3x + 2y ≥ 93.

So, you earn an A on the test.

19.

x

2

4

6

8

2 4−2−4

y Test (0, 0).

y ≤ 5

0 ≤ 5 ✓

20.

x

2

4

8

2 4−2−4

y Test (0, 0).

y > 6

0 > 6 ✗

21.

x

2

−2

−4

4

4−2−4

y Test (0, 0).

x < 2

0 < 2 ✓

22.

x

2

−2

−4

4

2−2−4−6

y Test (0, 0).

x ≥ −3

0 ≥ −3 ✓

23. x

−4

−8

−12

−16

42−2−4

y Test (0, 0).

y > −7

0 > −7 ✓

24.

x

2

−2

−4

4

84 12 16

y Test (0, 0).

x < 9

0 < 9 ✓

25.

x

2

−4

−6

42−4

y Test (0, 0).

y > −2x − 4

0 >? −2(0) − 4

0 >? 0 − 4

0 > −4 ✓

26.

x

2

4

42−2−4

y Test (0, 0).

y ≤ 3x − 1

0 ≤? 3(0) − 1

0 ≤? 0 − 1

0 ≤ −1 ✗

27. −4x + y < −7

−4x + 4x + y < −7 + 4x

y < 4x −7

x

−2

−4

−6

4−4 −2

y Test (0, 0).

−4x + y < −7

−4(0) + 0 <? −7

0 < −7 ✗

28. 3x − y ≥ 5

3x − 3x − y ≥ 5 − 3x

−y ≥ −3x + 5

−y —

−1 ≤

−3x + 5 —

−1

y ≤ 3x − 5

x

−2

−4

42−2 6

y Test (0, 0).

3x − y ≥ 5

3(0) − 0 ≥? 5

0 ≥ 5


Chapter 5

29. 5x − 2y ≤ 6

5x − 5x − 2y ≤ 6 − 5x

−2y ≤ −5x + 6

−2y —

−2 ≥

−5x + 6 —

−2

y ≥ 5 —

2 x − 3

x

2

−2

4

42−2−4

y Test (0, 0).

5x − 2y ≤ 6

5(0) − 2(0) ≤? 6

0 ≤ 6 ✓

30. −x + 4y > −12

−x + x + 4y > −12 + x

4y > x −12

4y

— 4 >

x −12 —

4

y > 1 —

4 x − 3

x

2

−4

−2

−6

42−2−4

y Test (0, 0).

−x + 4y > −12

−0 + 4(0) >? −12

0 > −12 ✓

31. The line should be dashed for <.

x

2

−2

4

42−2

y

32. The half-plane on the other side of the boundary line should

be shaded.

x

2

−2

4

42−2

y Test (0, 0).

y ≤ 3x − 2

0 ≤? 3(0) − 2

0 ≤? 0 − 2

0 ≤ −2 ✗

33. Words

Cost per

arcade

game

Number

of arcade

games

Cost

per

snack

Number

of

snacks

Amount

you can

spend⋅ ⋅+ ≤

Variables Let x be the number of arcade games you can play,

and let y be the number of snacks you can buy.

Inequality 0.75 ⋅ x + 2.25 ⋅ y ≤ 20

0.75x + 2.25y ≤ 20

0.75x − 0.75x + 2.25y ≤ 20 − 0.75x

2.25y ≤ −0.75x + 20

2.25y — 2.25

≤ −0.75x + 20

—— 2.25

y ≤ − 1 — 3 x +

80 —

9

Test (0, 0).

0.75x + 2.25y ≤ 20

0.75(0) + 2.25(0) ≤? 20

0 + 0 ≤? 20

0 ≤ 20 ✓

8

12

16

4

00 16 24 328

Number of games

Nu

mb

er o

f sn

acks

x

y

Check

0.75x + 2.25y ≤ 20 0.75x + 2.25y ≤ 20

0.75(8) + 2.25(4) ≤? 20 0.75(16) + 2.25(2) ≤

? 20

6 + 9 ≤? 20 12 + 4.5 ≤

? 20

15 ≤ 20 ✓ 16.5 ≤ 20 ✓

Sample answer: Two possible solutions are (8, 4) and (16, 2)

because they lie in the shaded half-plane. So, you can play

8 games and buy 4 snacks for a total of $15.00, or you can

play 16 games and buy 2 snacks for a total of $16.50.


Chapter 5

34. Words

Cost per

adult

ticket

Number

of adult

tickets

Cost per

student

ticket

Number

of student

tickets

1500⋅ ⋅+ ≥

Variables Let x be the number of adult tickets the drama

club must sell, and let y be the number of student

tickets the club must sell.

Inequality 10 ⋅ x + 6 ⋅ y ≥ 1500

10x + 6y ≥ 1500

10x − 10x + 6y ≥ 1500 − 10x

6y ≥ − 10x + 1500

6y

— 6 ≥

−10x + 1500 ——

6

y ≥ − 5 — 3 x + 250

200

300

400

100

100 150 20050

Adult tickets

Stu

den

t ti

cket

s

00 x

y Test (0, 0).

10x + 6y ≥ 1500

10(0) + 6(0) ≥? 1500

0 + 0 ≥? 1500

0 ≥ 1500 ✗

Check

10x + 6y ≥ 1500 10x + 6y ≥ 1500

10(75) + 6(200) ≥? 1500 10(150) + 6(300) ≥

? 1500

750 + 1200 ≥? 1500 1500 + 1800 ≥

? 1500

1950 ≥ 1500 ✓ 3300 ≥ 1500 ✓

Sample answer: Two possible solutions are (75, 200) and

(150, 300) because they lie in the shaded half-plane. So, you

can sell 75 adult tickets and 200 student tickets for a total

of $1950, or you can sell 150 adult tickets and 300 student

tickets for a total of $3300.

35. The slope of the boundary line through (0, 1) and (1, 3) is

m = 3 − 1

— 1 − 0

= 2 — 1 , or 2.

The boundary line crosses the y-axis at (0, 1). So, b = 1.

y = mx + b

y = 2x + 1

Test (0, 4).

y > 2x + 1

4 >? 2(0) + 1

4 >? 0 + 1

4 > 1 ✓

So, an inequality that represents the graph is y > 2x + 1.

36. The slope of the boundary line through (0, 2) and (2, 3)

is m = 3−2

— 2−0

= 1 —

2 .

The boundary line crosses the y-axis at (0, 2). So, b = 2.

y = mx + b

y = 1 —

2 x + 2

Test (0, 4). y > 1 —

2 x + 2

4 >? 1 —

2 (0) + 2

4 >? 0 + 2

4 > 2 ✓

So, an inequality that represents the graph is y > 1 —

2 x + 2.

37. The boundary line passes through (0, −2) and (2, −3).

The slope is m = −3 − (−2)

— 2 − 0

= −3 + 2

— 2 − 0

= −1

— 2 .

The boundary line crosses the y-axis at (0, −2). So, b = −2.

y = mx + b

y = − 1 —

2 x − 2

Test (0, −4). y ≤ − 1 —

2 x − 2

−4 ≤? −

1 —

2 (0) − 2

−4 ≤? 0 − 2

−4 ≤ −2 ✓

So, an inequality that represents this graph is y = − 1 — 2 x − 2.

38. The boundary line passes through (0, −3) and (1, 0).

The slope is m = 0 − (−3)

— 1 − 0

= 0 + 3

— 1 − 0

= 3 —

1 , or 3.

The boundary line crosses the y-axis at (0, −3). So, b = −3.

y = mx + b

y = 3x − 3

Test (2, 0), y ≤ 3x − 3

0 ≤? 3(2) − 3

0 ≤? 6 − 3

0 ≤ 3 ✓

So, an inequality that represents this graph is y ≤ 3x − 3.


Chapter 5

39. a. Words Weight of

delivery person + Weight per

small box ⋅

Number of

small boxes

+ Weight per

large box ⋅

Number of

large boxes ≤

Weight

limit

Variables Let x be the number of small boxes the

delivery person can take on the elevator, and

let y be the number of large boxes.

Inequality 200 + 75 ⋅ x + 40 ⋅ y ≤ 2000

200 + 75x + 40y ≤ 2000

200 − 200 + 75x + 40y ≤ 2000 − 200

75x + 40y ≤ 1800

75x − 75x + 40y ≤ 1800 − 75x

40y ≤ −75x + 1800

40y

— 40

≤ −75x + 1800

—— 40

y ≤ − 15 —

8 x + 45

Test (0, 0).

200 + 75x + 40y ≤ 2000

200 + 75(0) + 40(0) ≤? 2000

200 + 0 ≤? 2000

200 ≤ 2000 ✓

40

60

80

20

00

20 30 4010

Large boxes

Smal

l bo

xes

x

y

b. Sample answer: The shaded region contains points whose

coordinates are not whole numbers, but it is not possible

to load only part of a box on the elevator. Also, it is

possible that even though a certain number of boxes are

allowed on the elevator based on their weight, they may

be too big in size to fi t inside the elevator.

40. a. C; Test (0, 0). 3x − 2y ≤ 6

3(0) − 2(0) ≤? 6

0 ≤ 6 ✓

Graph C has (0, 0) in the shaded half-plane and has a solid

line for ≤. b. A; Test (0, 0). 3x − 2y < 6

3(0) − 2(0) <? 6

0 < 6 ✓

Graph A has (0, 0) in the shaded half-plane and has a

dashed line for <.

c. D; Test (0, 0). 3x − 2y > 6

3(0) − 2(0) >? 6

0 > 6 ✗

The shaded half-plane of Graph D does not contain (0, 0).

Also, the boundary line is dashed for >.

d. B; Test (0,0). 3x − 2y ≥ 6

3(0) − 2(0) ≥? 6

0 ≥ 6 ✗

The shaded half-plane of Graph B does not contain (0, 0).

Also, the boundary line is solid for ≥.

41. Sample answer: You have to know which side of the

boundary line to shade. So, you need to choose a point on

one side or the other of the boundary line to know which

region contains the solutions of the inequality.

42. Sample answer: m = −1 − 1

— 3 − 1

= −2

— 2 , or −1

The line passes through (0, 2). So, b = 2.

y = mx + b

y = −x + 2

Test (0, 0). y ≥ −x + 2

0 ≥?

− 0 + 2

0 ≥?

0 + 2

0 ≥ 2 ✗

So, y ≥ −x + 2 is an inequality in which (1, 1), (3, −1), and

(−1, 3) are solutions that lie on the line. Also, (0, 0), (0, −1),

and (0, 1) are not solutions because they lie in the unshaded

half-plane.

x

y4

−4

42−2

y ≥ −x + 2

(−1, 3)(1, 1)

(3, −1)(0, −1)

(0, 1)(0, 0)

43. no; If the point (0, 0) is on the boundary line, then you have

to choose a different test point that is not on the boundary

line.


Chapter 5

44. The slope of the boundary line is

m = 5 − (−5)

— 2 − (−3)

= 5 + 5

— 2 + 3

= 10

— 5 , or 2.

y − y1 = m (x − x1)

x

4

−4

−8

8

84−4−8

y

(2, 5)

(−3, −5)

y − 5 = 2(x − 2)

y − 5 = 2(x) − 2(2)

y − 5 = 2x − 4

+5 +5

y = 2x + 1

So, an equation of the boundary line is y = 2x + 1.

An inequality is y ≤ 2x + 1.

Check

Test (−2, −3). y ≤ 2x + 1 Test (6, 5). y ≤ 2x +1

−3 ≤? 2(−2) + 1 5 ≤

? 2(6) + 1

−3 ≤? −4 + 1 5 ≤

? 12 + 1

−3 ≤ −3 ✓ 5 ≤ 13 ✓ So, (6, 5) and (−2, −3) are solutions of y ≤ 2x + 1.

45. The slope of the boundary line is

m = 8 − (−16)

— 1 − (−7)

= 8 + 16

— 1 + 7

= 24

— 8 , or 3.

y − y1 = m(x − x1)

x

8

−8

−16

16

84−4−8

y

(1, 8)

(−7, −16)

y − 8 = 3(x − 1)

y − 8 = 3(x) − 3(1)

y − 8 = 3x − 3

+8 +8

y = 3x + 5

So, an equation of the boundary line is y = 3x + 5.

An inequality is y < 3x + 5.

Check

Test (−7, 0). y < 3x + 5 Test (3, 14). y < 3x + 5

0 <?

3(−7) + 5 14 <?

3(3) + 5

0 <?

−21 + 5 14 <?

9 + 5

0 < −16 ✗ 14 < 14 ✗

So, (−7, 0) and (3, 14) are not solutions of y < 3x + 5.


46. d = 8 − 0 = 8

Position 1 2 3 4 5 6 7 8

Term 0 8 16 24 32 40 48 56

+8 +8 +8

The next three terms are 40, 48, and 56.

47. d = −8 − (−5) = −8 + 5 = −3

Position 1 2 3 4 5 6 7 8

Term −5 −8 −11 −14 −17 −20 −23 −26

+(−3) +(−3) +(−3)

The next three terms are −20, −23, and −26.

48. d = 3 —

2 − 1 — 2 =

2 —

2 , or 1

Position 1 2 3 4 5 6 7 8

Term − 3 — 2 − 1 — 2 1 —

2

3 —

2

5 —

2

7 —

2

9 —

2

11 —

2

+ 2 — 2 + 2 — 2 + 2 — 2

The next three terms are 7 —

2 ,

9 —

2 , and

11 —

2 .

Section 5.7

5.7 Explorations (p.273)

1. Inequality 1 Inequality 2

Test (1,0). 2x + y ≤ 4 Test (1,0). 2x − y ≤ 0

2(1) + 0 ≤? 4 2(1) − 0 ≤

? 0

2 + 0 ≤? 4 2 − 0 ≤

? 0

2 ≤ 4 ✓ 2 ≤ 0 ✗

A; The point (1,0) is in

the shaded half-plane

of Graph A, and it is a

solution of Inequality 1.

B; The point (1,0) is not in the shaded region of

Graph B, and it is not a

solution of Inequality 2.

2. a. Inequality 1 Inequality 2

2x + y ≤ 4 2x − y ≤ 0 2x − 2x + y ≤ 4 − 2x 2x − 2x − y ≤ 0 − 2x

y ≤ −2x + 4 −y ≤ −2x

−y — −1

≥ −2x — −1

y ≥ 2x

x

2

−4

4

42−2−4

y

The shaded half-planes overlap as shown.

b. Sample answer: The two regions that are shaded with

only one color contain points whose coordinates are

the solutions of one inequality but not the other. The

region that is shaded with both colors contains points

whose coordinates are solutions of both inequalities. The

unshaded region contains points whose coordinates are

not solutions of either inequality.


Chapter 5

3. Sample answer: In order to graph a system of linear

inequalities, graph each inequality in the same coordinate

plane. Look for the intersection, or overlapping portion of

the shaded half-planes that are solutions of the inequalities.

The point in this intersection have coordinates that are

solutions of the system.

4. The solution of the system is represented by the region

where the shaded half-planes of the inequalities overlap.

5. no; If the boundary lines are parallel and their half-planes do

not overlap, then the system has no solution.

6. The blue line is vertical and passes through the point (2, 0).

So, an equation for this line is x = 2. Because the shaded

region is to the left of this solid boundary line, an inequality

is x ≤ 2. The red line is horizontal and passes through the

point (0, 3). So, an equation for this line is y = 3. Because

the shaded region is below this solid boundary line, an

inequality is y ≤ 3. So, a system of linear inequalities

represented by the graph is x ≤ 2 and y ≤ 3.


1. y < 5 y > x − 4

5 < 5 ✗ 5 >?

−1 − 4 5 > −5 ✓

Because (−1, 5) is not a solution of each inequality, it is not a solution of the system.

2. y ≥ 3x + 1 y > x −1

4 ≥?

3(1) + 1 4 >?

1 − 1

4 ≥?

3 + 1 4 > 0 ✓

4 ≥ 4 ✓

Because the ordered pair (1, 4) is a solution of each

inequality, it is a solution of the system.

3. y ≥ − x + 4 x + y ≤ 0 x − x + y ≤ 0 − x

y ≤ −x

x

4

−4

−8

8

84−4−8

y

4.

x

2

−2

4

42−2−4

y y > 2x − 3 y ≥ 1 —

2 x + 1

5. −2x + y < 4 2x + y > 4

−2x + 2x + y < 4 + 2x 2x − 2x + y > 4 − 2x

y < 2x + 4 y > −2x + 4

x

2

−2

4

6

4 62

y

6. Inequality 1: The vertical boundary line passes through (3, 0).

So, an equation of the line is x = 3. Because the shaded region

is to the left of the boundary line, the inequality is x < 3.

Inequality 2: The slope of the other boundary line is −1,

and the y-intercept is 2. So, an equation of the line is

y = −x + 2. Because the shaded region is below the dashed

boundary line, the inequality is y < −x + 2.

So, the system of linear inequalities represented by the graph

is x < 3 and y < −x + 2.

7. Inequality 1: One of the lines has a slope of − 1 — 2 and a

y-intercept of 1. So, an equation of the line is y = − 1 — 2 x + 1.

Because the shaded region is above this solid boundary

line, the inequality is y ≥ − 1 — 2 x + 1.

Inequality 2: The slope of the other boundary line is 2,

and the y-intercept is −3. So, an equation of this line is

y = 2x − 3. Because the shaded region is below this solid

boundary line, the inequality is y ≤ 2x − 3.

So, the system of linear inequalities represented by the

graph is y ≥ − 1 — 2 x + 1 and y ≤ 2x − 3.

8. Sample answer: Another ordered pair in the solution region

is (3, 4.5). So, you can spend 3 hours at the mall and

4.5 hours at the beach.

Check x + y ≤ 8 x ≥ 2 y > 4

3 + 4.5 ≤? 8 3 ≥ 2 ✓ 4.5 > 4 ✓

7.5 ≤ 8 ✓

9. The boundary line at x = 2 will now be at x = 3. So, the

shaded region will be one small triangular region of the

graph between this line and the other two existing lines.

Because the solution (2.5, 5) will no longer be inside the

shaded region, it will no longer be a solution.


Chapter 5



1. You can substitute the values into each inequality of the

system and verify that the values make the inequalities true.

2. The ordered pair that does not belong is (1, −2). This ordered

pair is on both of the boundary lines, and because one of the

boundary lines is dashed, this ordered pair is a solution of only

one of the inequalities. So, it is not a solution of the system.

The other three ordered pairs are solutions of the system

because each one is in the shaded region.


3. no; The ordered pair (−4, 3) is a solution of one inequality

but not the other because it is on the solid boundary line, but

is not in the shaded region below the dashed boundary line.

So, it is not a solution of the system.

4. yes; The ordered pair (−3, −1) is a solution of each

inequality because it is in the shaded region below both

boundary lines. So, it is a solution of the system.

5. no; The ordered pair (−2, 5) is not in the shaded region. So,

it is not a solution of the system.

6. no; The ordered pair (1, 1) is not in the shaded region. So, it

is not a solution of the system.

7. y < 4 y > x + 3

2 < 4 ✓ 2 >? −5 + 3

2 > −2 ✓

Because the ordered pair (−5, 2) is a solution of each


8. y > −2 y > x − 5

−1 > −2 ✓ −1 >? 1 − 5

−1 > −4 ✓



9. y ≤ x + 7 y ≥ 2x + 3

0 ≤? 0 + 7 0 ≥

? 2(0) + 3

0 ≤ 7 ✓ 0 ≥? 0 + 3

0 ≥ 3 ✗

Because (0, 0) is not a solution of each inequality, it is not a

solution of the system.

10. y ≤ −x + 1 y ≤ 5x − 2

−3 ≤? −4 + 1 −3 ≤

? 5(4) − 2

−3 ≤ −3 ✓ −3 ≤? 20 − 2

−3 ≤ 18 ✓



11. Graph the system.

x

4

8

2−2−4−6

y

y ≥ −3

y ≥ 5x


x

2

4

2 6 8

y

y < −1

x > 4


x

4

−4

42−2−4

y

y < −2

y > 2


x

2

−4

4

42−4

y

y < x − 1

y ≥ x + 1


x

2

−4

−6

42−2−4

y

y ≥ −5

y − 1 < 3x

y − 1 + 1 < 3x + 1

y < 3x + 1


x

2

−2

−4

4

2 64 8

y

x + y > 4

x − x + y > 4 − x

y > − x + 4

y ≥ 3 —

2 x − 9


Chapter 5


x + y > 1 −x − y < − 3

x − x + y > 1 − x −x + x − y < −3 + x

y > −x + 1 −y < x − 3

x

4

2

−4

−2

4 62−2

y −y

— −1

> x − 3

— −1

y > −x + 3


2x + y ≤ 5 y + 2 ≥ −2x

2x − 2x + y ≤ 5 − 2x y + 2 − 2 ≥ −2x − 2

y ≤ −2x + 5 y ≥ −2x − 2

x

2

4

−2

42−2−4

y


x

4

2

−4

−2

62−2

y

x < 4

y > 1

y ≥ −x + 1


x

4

6

8

2 4 6 8

y

x + y ≤ 10

x − x + y ≤ 10 − x

y ≤ −x + 10

x − y ≥ 2

x − x − y ≥ 2 − x

− y ≥ −x + 2

−y

— −1

≤ −x + 2

— −1

y ≤ x − 2

y > 2

21. Inequality 1: The vertical boundary line passes through

(−1, 0). So, an equation of the line is x = −1. Because the

shaded region is to the right of the solid boundary line, the

inequality is x ≥ −1.

Inequality 2: The horizontal boundary line passes through

(0, 3). So, an equation of the line is y = 3. Because the

shaded region is below this dashed boundary line, the

inequality is y < 3.


is x ≥ −1 and y < 3.

22. One vertical line passes through (2,0), and the other passes

through (4,0). So, equations for the lines are x = 2 and

x = 4, respectively. Because the shaded region is to the

right of the dashed boundary line x = 2 and to the left of the

dashed boundary line x = 4, the system of linear inequalities

represented by the graph is x > 2 and x < 4.

23. Inequality 1: One of the lines has a slope of −3 and a

y-intercept of 2. So, an equation of the line is y = −3x + 2.

Because the shaded region is above this solid boundary line,

the inequality is y ≥ −3x + 2.

Inequality 2: The slope of the other boundary line is 2 —

3 ,

and the y-intercept −2. So, an equation of this line is

y = 2 —

3 x − 2. Because the shaded region is above this

solid boundary line, the inequality is y ≥ 2 —

3 x − 2.


is y ≥ −3x + 2 and y ≥ 2 —

3 x − 2.

24. Inequality 1: One of the lines has a slope of 5 and a

y-intercept of 1. So, an equation of the line is y = 5x + 1.

Because the shaded region is below this solid boundary line,

the inequality is y ≤ 5x + 1.

Inequality 2: The slope of the other boundary line is 1,

and the y-intercept is −2. So, an equation of this line is

y = x − 2. Because the shaded region is above this dashed

boundary line, the inequality is y > x − 2.


is y ≤ 5x + 1 and y > x − 2.

25. This system has no solution. Both lines have a slope of

−2. One line has a y-intercept of −1, and the other has a

y-intercept of −3. So, the equations of the boundary lines

are y = −2x − 1 and y = −2x − 3. Because the graph has

no shaded region, the half-planes of the inequalities must

not intersect. So, the solutions of the inequality bounded

by y = −2x − 1 must be above the dashed line, and the

inequality must be y > −2x − 1. Also, the solutions of the

inequality bounded by y = −2x − 3 must be below the

dashed line, and the inequality must be y < −2x − 3. So,

the system of linear inequalities represented by the graph is

y > −2x − 1 and y < −2x − 3.


Chapter 5

26. Both of the boundary lines in this graph have a slope of 1 —

3 .

One line has a y-intercept of 0, and the other has a y-intercept

of −2. So, the equations of the boundary lines are y = 1 —

3 x and

y = 1 —

3 x − 2. Because the shaded region lies above both solid

lines, the inequalities must be y ≥ 1 —

3 x and y ≥ 1 —

3 x − 2.


is y ≥ 1 —

3 x and y ≥

1 —

3 x − 2.

27. The system should have no solution. The solutions of y ≤ x − 1

should be below the line on the graph, and the solutions of

y ≥ x + 3 should be above the line on the graph. Because

the lines have the same slope of 1, they are parallel. So, the

half-planes will not intersect.

x

2

−4

4

42−4 −2

y

28. The shaded region is correctly below the line y ≤ 3x + 4, but

it should be above the line y > 1 —

2 x + 2.

x

4

−2

2−2−4

y

29. a. Words Blueberries’

cost per

pound

⋅

Pounds of

blueberries

+ Strawberries’

cost per

pound

⋅ Pounds of

strawberries ≤ Amount you

can spend

Pounds of

blueberries +

Pounds of

strawberries ≥

Total

amount

Variables Let x be the weight (in pounds) of blueberries

you can buy, and let y be the weight

(in pounds) of strawberries you can buy.

System 4 ⋅ x + 3 ⋅ y ≤ 21

x + y ≥ 3

4x + 3y ≤ 21 x + y ≥ 3

4x −4x + 3y ≤ 21 − 4x x − x + y ≥ 3 − x

3y ≤ − 4x + 21 y ≥ − x + 3

3y — 3 ≤

−4x + 21 —

3

y ≤ − 4 —

3 x + 7

4

6

8

2

4 6 82

Blueberries (pounds)

Stra

wb

erri

es (

po

un

ds)

00 x

y

b. Sample Answer: One ordered pair in the solution region

is (3, 2). So, you can buy 3 pounds of blueberries and

2 pounds of strawberries.

c. yes; The ordered pair (4, 1) is in the solution region.

30. a. Words Grocery store

earnings

(in dollars)

per hour

⋅

Grocery

store hours

worked

+

Music lesson

earnings

(in dollars)

per hour

⋅

Music

lesson

hours

worked

≥

120

Grocery

store hours

worked

≥

8

Grocery

store hours

worked

+

Music

lesson

hours

worked

≤ 20

Variables Let x be the number of hours you work at the

grocery store, and let y be the number of hours

you work teaching music lessons.

System 10x + 15y ≥ 120

x ≥ 8

x + y ≤ 20

10x + 15y ≥ 120 x + y ≤ 20

10x −10x + 15y ≥ 120 − 10x x − x + y ≤ 20 − x

15y ≥ −10x + 120 y ≤ − x + 20

15y

— 15

≥ −10x + 120

— 15

y ≥ − 2 — 3 x + 8

8

12

16

20

4

8 12 16 204

Hours at grocery store

Ho

urs

tea

chin

g m

usi

c

00 x

y


Chapter 5

b. Sample answer: One ordered pair in the solution region

is (12, 4). So, you can work 12 hours at the grocery store

and 4 hours teaching music lessons.

c. no; The ordered pair (8, 1) is not in the shaded solution

region.

31. a. Words Number of

surfperch ≤ 15

Number of

rockfi sh ≤ 10

Number of

surfperch + Number of

rockfi sh ≤ 20

Variables Let x be the number of surfperch you can

catch, and let y be the number of rockfi sh you

can catch.

System x ≤ 15

y ≤ 10

x + y ≤ 20

x − x + y ≤ 20 − x

y ≤ − x + 20

8

12

16

20

4

00 8 12 16 204

Surfperch caughtx

y

b. yes; The ordered pair (11, 9) is on one of the solid

boundary lines of the shaded solution region. So, it is a

solution of the system.

32. x − y ≤ 4 x − y ≥ 4

x − x − y ≤ 4 − x x − x − y ≥ 4 − x −y ≤ −x + 4 −y ≥ − x + 4

−y — −1

≥ −x + 4 — −1

−y

— −1

≤ −x + 4

— −1

y ≥ x − 4 y ≤ x − 4

The equations each describe a half-plane with the same solid

boundary line, but the solutions are on opposite sides of the

line. So, the intersection of the regions is the line y = x − 4

only. In other words, only the points on the line y = x − 4

are solutions of both inequalities.

33.

x

y

2

42−2

(6, −3)(−1, −3)

(−1, 1) (6, 1)

a. A system of linear inequalities represented by the shaded

rectangle is y ≥ −3, y ≤ 1, x ≥ −1, and x ≤ 6.

b. The area of the rectangle is

A =ℓw = 7(4) = 28 square units.

34.

x

y

4

2

−2

4 62−2

(6, −3)(−2, −3)

(2, 5)

a. The base of the triangle is defi ned by the line y = −3. The

shaded solution region is above this line. So, an inequality

is y ≥ −3.

m = 5 − (−3)

— 2 − (−2)

= 5 + 3

— 2 + 2

= 8 —

3 m =

−3 − 5 —

6 − 2 = −8

— 4 , or −2

y − y1 = m(x − x1) y − y1 = m(x − x1)

y − 5 = 2(x − 2) y − 5 = −2(x − 2)

y − 5 = 2(x) − 2(2) y − 5 = −2(x)−2(−2)

y − 5 = 2x − 4 y − 5 = −2x + 4

+5 +5 +5 +5

y = 2x + 1 y = −2x + 9

So, the equations of the lines that defi ne the other two

sides of the triangle are y = 2x + 1 and y = −2x + 9.

Because the shaded area is below each of these lines,

the inequalities are y ≤ 2x + 1 and y ≤ −2x + 9.

The system of linear inequalities represented by the

shaded triangle is y ≤ 2x + 1, y ≤ −2x + 9, and y ≥ −3.

b. The area of the triangle is

A = 1 —

2 bh =

1 —

2 (8)(8) = 32 square units.


Chapter 5

35. a. Words Amount

spent on

housing

+

Amount

spent on

savings

≤ 1 — 2 (2000)

Amount spent

on savings ≥ 10% ⋅ 2000

Amount spent

on housing ≤ 30% ⋅ 2000

Variables Let x be how much you spend on housing and

let y be how much you spend on savings.

System x + y ≤ 1000

y ≥ 0.1 (2000) x ≤ 0.3 (2000)

x + y ≤ 1000

x − x + y ≤ 1000 − x

y ≤ − x + 1000

y ≥ 0.1(2000)y ≥ 200

x ≤ 0.3(2000)

x ≤ 600

400 800 12000 x

y

800

1200

400

0

Money spent onhousing (in dollars)

Mo

ney

sp

ent

on

savi

ng

s (i

n d

olla

rs)

Sample answer: One ordered pair in the solution region is

(400, 400). So, you can spend $400 each on housing and

savings.

36. Words Hours you

drive

+

Hours friend

drives

< 15

70 ⋅ Hours you

drive + 60 ⋅

Hours friend

drives ≥ 600

Hours friend

drives >

Hours you

drive

Variables Let x be how many hours you drive, and

let y be how many hours your friend drives.

System x + y < 15

70x + 6y ≥ 600

y > x

x + y < 15 70x + 60y ≥ 600 y > x

x − x + y < 15 − x 70x − 70x + 60y ≥ 600 − 70x

y < − x + 15 60y ≥ −70x + 600

60y

— 60

≥ −70x + 600

— 60

y ≥ − 7 — 6 x + 10

4 8 12 160 x

y

8

12

16

4

0

Hours you drive

Ho

urs

yo

ur

frie

nd

dri

ves

Sample answer: One ordered pair in the shaded solution

region is (4, 8). So, one possibility is for you to drive 4 hours

and your friend to drive 8 hours in one day.

37. Sample answer: When you solve systems of linear

inequalities, you are fi nding coordinate pairs of at least two

variables that make multiple inequalities true. The same is

true for systems of linear equations in that you are fi nding

coordinate pairs of at least two variables for at least two

equations. Both types of systems can have infi nitely many

solutions or no solution. Graphing can be used to solve

both kinds of systems, and the graphs of both kinds of

systems involve straight lines. Graphing is the only method

that can be used to solve systems of linear inequalities,

however, and there are two other methods (substitution and

elimination) for solving systems of linear equations. When

you solve a system of linear inequalities, you are looking

for an overlapping region of coordinate pairs that make

the inequalities true. When you solve a system of linear

equations, you are usually looking for a single coordinate

pair that makes the equations true. The only time you get

more than one solution for a system of linear equations is

when the equations in the system describe the same line.

In this case, all of the points on the line are solutions of

the system.

38. The systems that have point C as a solution, but not points

A, B, and D are y > −3x + 4 and y ≤ 2x + 1 or y > −3x + 4

and y < 2x + 1. Sample answer: Point C is below the line

y = 2x + 1 and above the line y = −3x + 4. So, one

inequality must be either y < 2x + 1 or y ≤ 2x + 1, and the

other must be either y > −3x + 4 or y ≥ −3x + 4. Because

point B is on the line y = −3x + 4 and point B cannot be a

solution, it must be that y > −3x + 4. The other inequality

could be either y < 2x + 1 or y ≤ 2x + 1.

39. A system of linear inequalities that is equivalent to

∣ y ∣ < x is ⎧⎨y < x, if x > 0

⎩y > −x, if x > 0.

x

−4

−2

62 4−2

y4

2


Chapter 5

40. no; It is possible for a system of linear inequalities in which

the boundary lines are parallel to have infi nitely many

solutions. Sample answer: Examples of systems with parallel

boundary lines and infi nitely many solutions are Exercises 22

and 26, where the shaded solution region is either between

the lines or above (or below) both of the lines.

41. no, The solution of each inequality is a half-plane, and so the

intersection can be at most a half-plane.

42. Sample answer: A system of linear inequalities whose

solutions are all in Quadrant I is y > 0 and x > 0.

43. Sample answer: A system of linear inequalities whose

solutions have one positive coordinate and one negative

coordinate is x > 0 and y < 0.

44. Sample answer: A system of linear inequalities that has no

solution is y > x and y < x.

45 a. Sample answer: In order for the system to have no

solution, another inequality is −4x + 2y < 6. b. Sample answer: In order for the system to have infi nitely

many solutions, another inequality is −2x + y > 3.

46. Sample answer: If the T-shirts cost $8 each, and the

sweatshirts are $16 each, and your gift card is for $100 and

can be used to buy up to 10 clothing items, then you can buy

9 T-shirts and 1 sweatshirt, but you cannot buy 3 T-shirts and

8 sweatshirts.

Words 8 ⋅ Number of

T-shirts + 16 ⋅

Number of

sweatshirts ≤ 100

Number of

T-shirts +

Number of

sweatshirts ≤ 10

Variables Let x be how many T-shirts you can buy, and let y

be how many sweatshirts you can buy.

System 8x + 16y ≤ 100

x + y ≤ 10

8x + 16y ≤ 100 x + y = ≤ 10

8x − 8x + 16y ≤ 100 − 8x x − x + y ≤ 10 − x

16y ≤ −8x + 100 y ≤ − x + 10

16y

— 16

≤ − 8x + 100

— 16

y ≤ − 1 —

2 x + 6.25

4 8 120 x

y

8

12

4

0

T-shirts

Swea

tsh

irts

(9, 1)

(3, 8)

Check 8x + 16y ≤ 100 x + y ≤ 10

8(9) + 16(1) ≤? 100 9 + 1 ≤

? 10

72 + 16 ≤? 100 10 ≤ 10 ✓

88 ≤ 100 ✓

8x + 16y ≤ 100 x + y ≤ 10

8(3) + 16(8) ≤? 100 3 + 8 ≤

? 10

24 + 128 ≤? 100 11 ≤ 10 ✗

152 ≤ 100 ✗

So, the coordinate pair (9, 1) is in the solution region,

but (3, 8) is not.

47. Sample answer: A system of linear equations that has exactly

one solution is y ≥ ∣ x ∣ and y ≤ 0. The solution is (0, 0).

48. a. A system for this situation is

0.5x + 0.25y ≤ 20, 2x + 3y ≤ 120, x ≥ 0, and y ≥ 0.

0.5x + 0.25y ≤ 20 2x + 3y ≤ 120

0.5x − 0.5x + 0.25y ≤ 20 − 0.5x 2x − 2x + 3y ≤ 120 − 2x

0.25y ≤ −0.5x + 20 3y ≤ −2x + 120

0.25y

— 0.25

≤ −0.5x + 20

— 0.25

3y

— 3 ≤

−2x + 120 —

3

y ≤ −2x + 80 y ≤ − 2 —

3 x + 40

20

30

40

50

10

20 30 40 5010

Necklaces

Key

ch

ain

s

00 x

y

b. The boundary lines x = 0 and y = 0 intersect at (0, 0).

The boundary lines x = 0 and y = − 2 — 3 x + 40 intersect at

(0, 40) because 40 = − 2 — 3 (0) + 40.

The boundary lines y = 0 and y = −2x + 80 intersect at

(40, 0) because 0 = −2(40) + 80.

The boundary lines y = −2x + 80 and y = − 2 — 3 x + 40

intersect at (30, 20) because 20 = −2(30) + 80 and

20 = − 2 — 3 (30) + 40.

So, the vertices of the graph of the system are (0, 0),

(0, 40), (40, 0), and (30, 20).


Chapter 5

c. For (0, 0): R = 10x + 8y

R = 10(0) + 8(0)

R = $0

For (0, 40): R = 10x + 8y

R = 10(0) + 8(40)

R = 0 + 320

R = $320

For (40, 0): R = 10x + 8y

R = 10(40) + 8(0)

R = 400 + 0

R = $400

For (30, 20): R = 10x + 8y

R = 10(30) + 8(20)

R = 300 + 160

R = $460

The vertex (30, 20) results in the maximum revenue. So, you

will make the most money if you make and sell 30 necklaces

and 20 key chains.


49. 4 ⋅ 4 ⋅ 4 ⋅ 4 ⋅ 4 = 45

50. (−13) ⋅ (−13) ⋅ (−13) = (−13)3

51. x ⋅ x ⋅ x ⋅ x ⋅ x ⋅ x = x6

52. y = mx + b

y = 1x + (−6)

y = x − 6

So, an equation is y = x − 6.

53. y = mx + b

y = −3x + 5

So, an equation is y = −3x + 5.

54. y = mx + b

y = − 1 — 4 x + (−1)

y = − 1 — 4 x − 1

So, an equation is y = − 1 — 4 x − 1.

55. y = mx + b

y = 4 —

3 x + 0

y = 4 —

3 x

So, an equation is y = 4 —

3 x.

5.5 –5.7 What Did You Learn? (p. 281)

1. Sample answer: You and your friend are running to a fence

and back. So, for every point on your way toward the fence,

you are the same distance from the fence as another point on

your return from the fence. Every output of an absolute value

function is paired with two different input values. So, it can

be a good model for your distance from the fence during

the race.

2. Sample answer: If your answer is inaccurate, the delivery

person might overload the elevator, which could lead to a

dangerous situation and put the delivery person or others

at risk.

3. Sample answer: Use a verbal model to write inequalities

to represent each part of the problem, and designate two

variables to represent the two unknown values. Simplify the

inequalities, and rewrite one so that it is in slope-intercept

form. Graph all three inequalities on the same coordinate

plane and shade the intersection. Finally, choose a point in

the shaded region as a solution of the problem.

Chapter 5 Review (pp. 282–284)

1. x

y

−4

−6

−2

42−2

y = x − 7

y = −3x + 1

(2, −5)


y = −3x + 1 y = x − 7

−5 =?

−3(2) + 1 −5 =?

2 − 7

−5 =?

−6 + 1 −5 = −5 ✓

−5 = −5 ✓


2. 4x − 2y = 6

x

y

−2

42−2−4

y = 2x − 3

y = −4x + 3

(1, −1)

4x − 4x − 2y = 6 − 4x

−2y = −4x + 6

−2y

— −2

= −4x + 6

— −2

y = 2x − 3


y = −4x + 3 4x − 2y = 6

−1 =?

−4(1) + 3 4(1) − 2(−1) =?

6

−1 =?

−4 + 3 4 + 2 =?

6

−1 = −1 ✓ 6 = 6 ✓



Chapter 5

3. 5x + 5y = 15 2x − 2y =10

5x − 5x + 5y = 15 − 5x 2x − 2x − 2y = 10 − 2x

5y = −5x + 15 −2y = −2x + 10

5y

— 5 =

−5x + 15 —

5

−2y —

−2 =

−2x + 10 —

−2

y = −x + 3 y = x − 5

x

y

2

−4

−2

2

y = x − 5

y = −x + 3

(4, −1)


5x + 5y = 15 2x − 2y = 10

5(4) + 5(−1) =?

15 2(4) − 2(−1) =?

10

20 − 5 =?

15 8 + 2 =?

10

15 = 15 ✓ 10 = 10 ✓


4. Substitute 5x + 7 for y in Equation 1 and solve for x.

3x + y = −9

3x + (5x + 7) = −9

8x + 7 = −9

−7 −7

8x = −16

8x — 8 =

−16 —

8

x = −2

Substitute −2 for x in Equation 2 and solve for y.

y = 5x + y

y = 5(−2) + 7

y = −10 + 7 y = −3

Check 3x + y = −9 y = 5x + 7

3(−2) + (−3) =?

−9 −3 =?

5(−2) + 7

−6 − 3 =?

−9 −3 =?

−10 + 7

−9 = −9 ✓ −3 = −3 ✓


5. Step 1 Step 2 Step 3

x − y = 1 x + 4y = 6 x − y = 1

x − y + y = 1 + y (y + 1) + 4y = 6 x − 1 = 1

x = y +1 5y + 1 = 6 +1 +1

−1 −1 x = 2

5y = 5

5y

— 5 =

5 —

5

y = 1

Check x + 4y = 6 x − y = 1

2 + 4(1) =?

6 2 − 1 =?

1

2 + 4 =?

6 1 = 1 ✓

6 = 6 ✓

The solution is (2, 1)

6. Step 1

y + 3x = 6

y + 3x − 3x = 6 − 3x

y = 6 − 3x

Step 2

2x + 3y = 4

2x + 3(6 − 3x) = 4

2x + 3(6) − 3(3x) = 4

2x + 18 − 9x = 4

−7x + 18 = 4

−18 −18

−7x = −14

−7x — −7

= −14

— −7

x = 2

Step 3

y + 3x = 6

y + 3(2) = 6

y + 6 = 6

−6 −6

y = 0

Check 2x + 3y = 4 y + 3x = 6

2(2) + 3(0) =?

4 0 + 3(2) =?

6

4 + 0 =?

4 0 + 6 =?

6

4 = 4 ✓ 6 = 6 ✓



Chapter 5

7. Words 4 ⋅ Tubes

of paint + 0.50 ⋅

Paint

brushes = 20

Paint

brushes = 2 ∙

Tubes

of paint

Variables Let x be the number of tubes of paint you

purchase, and let y be the number of paint brushes

you purchase.

System 4x + 0.5y = 20 Equation 1

y = 2x Equation 2

Substitute 2x for y in Equation 1.

4x + 0.5y = 20

4x + 0.5(2x) = 20

4x + x = 20

5x = 20

5x — 5 =

20 —

5

x = 4

Substitute 4 for x in Equation 2.

y = 2x

y = 2(4)

y = 8

The solution is (4, 8). So, you purchase 4 tubes of paint and

8 paint brushes.

8. Step 2

9x − 2y = 34

5x + 2y = −6

Step 3 14x = 28

14x

— 14

= 28

— 14

x = 2

Step 4

5x + 2y = −6

5(2) + 2y = −6

10 + 2y = −6

−10 −10

2y = −16

2y

— 2 =

−16 —

2

y = −8

Check

9x − 2y = 34 5x + 2y = −6

9(2) − 2(−8) =?

34 5(2) + 2(−8) =?

−6

18 + 16 =?

34 10 − 16 =?

−6

34 = 34 ✓ −6 = −6 ✓


9. Step 1 Step 2

x + 6y = 28 x + 6y = 28

2x − 3y = −19 Multiply by 2. 4x − 6y = −38

5x + 0 = −10

Step 3 5x = −10

5x — 5 =

−10 —

5

x = −2

Step 4

x + 6y = 28

−2 + 6y = 28

+2 +2

6y = 30

6y

— 6 =

30 —

6

y = 5

Check x + 6y = 28 2x − 3y = −19

−2 + 6(5) =?

28 2(−2) − 3(5) =?

−19

−2 + 30 =?

28 −4 − 15 =?

−19

28 = 28 ✓ −19 = −19 ✓

So, the solution is (−2, 5).

10. Step 1 Step 2

8x − 7y = −3 Multiply by 3. 24x − 21y = −9

6x − 5y = −1 Multiply by −4. −24x + 20y = 4

0 − y = −5

Step 3 −y = −5

−y

— −1

= −5

— −1

y = 5

Step 4 6x − 5y = −1

6x − 5(5) = −1

6x − 25 = −1

+25 +25

6x = 24

6x — 6 =

24 —

6

x = 4

Check 8x − 7y = −3 6x − 5y = −1

8(4) − 7(5) =?

−3 6(4) − 5(5) =?

−1

32 − 35 =?

−3 24 − 25 =?

−1

−3 = −3 ✓ −1 = −1 ✓

So, the solution is (4, 5).


Chapter 5


Substitute y + 2 for x in Equation 2.

−3x + 3y = 6

−3( y + 2) + 3y = 6

−3( y) − 3(2) + 3y = 6

−3y − 6 + 3y = 6

0 − 6 = 6

−6 = 6 ✗

The equation −6 = 6 is never true. So, the system has

no solution.


Step 1 Step 2

3x − 6y = −9 Multiply by 5. 15x − 30y = −45

−5x + 10y = 10 Multiply by 3. − 15x + 30y = 30

0 = −15

The equation 0 = −15 is never true. So, the system has

no solution.


Step 1

3y = 3x + 24

3y — 3 =

3x + 24 —

3

y = x + 8

Step 2

−4x + 4y = 32

−4x + 4(x + 8) = 32

−4x + 4(x) + 4(8) = 32

−4x + 4x + 32 = 32

0 + 32 = 32

32 = 32

Because the equation 32 = 32 is always true, every solution

of −4x + 4y = 32 is also a solution of 3x + 24 = 3y. So,

the system has infi nitely many solutions.


x

y

4

6

2

2−2−4

y = −2x − 2

y = x + 513

(−3, 4) y =

1 —

3 x + 5

y = −2x − 2

The graphs intersect at (−3, 4).

Check

1 — 3 x + 5 = −2x − 2

1 — 3 (−3) + 5 =

? −2(−3) − 2

−1 + 5 =?

6 − 2

4 = 4 ✓


15. ∣ x + 1 ∣ = ∣ −x − 9 ∣ Equation 1 Equation 2

x + 1 = −x − 9 x + 1 = −(−x − 9)

x + 1 = x + 9

System 1 System 2

y = x + 1 y = x + 1

y = −x − 9 y = x + 9

x

y

−6

−2

−4−6

y = x + 1

y = −x − 9

(−5, −4)

x

y

12

4

84−4−8

y = x + 9

y = x + 1

The graphs intersect The lines are parallel. So,

at (−5, −4). they will never intersect, and

this system has no solution.

Check

∣ x + 1 ∣ = ∣ −x − 9 ∣

∣ −5 + 1 ∣ =? ∣ −(−5) − 9 ∣

∣ −4 ∣ =? ∣ 5−9 ∣

4 =?

∣ −4 ∣ 4 = 4 ✓

So, the solution is x = −5.


Chapter 5

16. ∣ 2x − 8 ∣ = ∣ x + 5 ∣ Equation 1 Equation 2

2x − 8 = x + 5 2x − 8 = −(x + 5)

2x − 8 = −x − 5

System 1 System 2

y = 2x − 8 y = 2x − 8

y = x + 5 y = −x − 5

x

y

16

24

8

16 248

y = x + 5

y = 2x − 8

(13, 18)

x

y

−6

−2

42−2

y = −x − 5

y = 2x − 8

(1, −6)


at (13, 18) at (1, −6).

Check Check

∣ 2x − 8 ∣ = ∣ x + 5 ∣ ∣ 2x − 8 ∣ = ∣ x + 5 ∣

∣ 2(13) − 8 ∣ =? ∣ 13 + 5 ∣ ∣ 2(1) − 8 ∣ =? ∣ 1 + 5 ∣

∣ 26 − 8 ∣ =? ∣ 18 ∣ ∣ 2 − 8 ∣ =? ∣ 6 ∣

∣ 18 ∣ =? 18 ∣ −6 ∣ =? 6

18 = 18 ✓ 6 = 6 ✓

So, the solutions of the equation are x = 13 and x = 1.

17.

x

2

−2

−6

42−2−4

y Test (0, 0). y > −4

0 > −4 ✓

18. −9x + 3y ≥ 3

x

2

−4

4

42−4 −2

y

−9x + 9x + 3y ≥ 3 + 9x

3y ≥ 9x + 3

3y — 3 ≥

9x + 3 —

3

y ≥ 3x + 1

Test (0, 0). −9x + 3y ≥ 3

−9(0) + 3(0) ≥?

3

0 ≥ 3 ✗

19. 5x + 10y < 40

x

2

6

−2

42−2−4

y

5x − 5x + 10y < 40 − 5x

10y < −5x + 40

10y — 10

< −5x + 40

— 10

y < − 1 — 2 x + 4

Test (0,0). 5x + 10y < 40

5(0) + 10(0) <?

40

0 < 40 ✓


x

2

−2

4

4−4

y

y ≤ x − 3

y ≥ x + 1


x

2

−2

6

4 6

y

y > −2x + 3

y ≥ 1 —

4 x − 1

22. x + 3y > 6

x

4

6

−2

2−2−4

y

x − x + 3y > 6 − x

3y > −x + 6

3y — 3 >

−x + 6 —

3

y > − 1 —

3 x + 2

2x + y < 7

2x − 2x + y < 7 − 2x

y < −2x + 7


Chapter 5

Chapter 5 Test (p. 285)

1. Sample answer: Solve by elimination because the

coeffi cients of the x-terms are opposites.

Step 2

8x + 3y = –9

–8x + y = 29

0 + 4y = 20

Step 3 4y = 20

4y

— 4 =

20 —

4

y = 5

Step 4

–8x + y = 29

–8x + 5 = 29

–5 –5

–8x = 24

–8x

— –8

= 24

— –8

x = –3

Check 8x + 3y = –9 –8x + y = 29

8(–3) + 3(5) =?

–9 –8(–3) + 5 =?

29

–24 + 15 =?

–9 24 + 5 =?

29

–9 = –9 ✓ 29 = 29 ✓

The solution is (–3, 5).

2. Sample answer: Solve by graphing because one equation

is in slope-intercept form, and the other will be after only

one step.

1 —

2 x + y = –6

x

y8

−8

−4

−4

y = x + 535

y = − x − 612

(−10, −1)

1 — 2 x –

1 —

2 x + y = –6 – 1 — 2 x

y = – 1 — 2 x – 6

y = 3 —

5 x + 5

The graphs appear to intersect at (–10, –1).

Check 1 —

2 x + y = –6 y =

3 —

5 x + 5

1 —

2 (–10) + (–1) =

? –6 –1 =

? 3 —

5 (–10) + 5

–5 – 1 =?

–6 –1 =?

–6 + 5

–6 = –6 ✓ –1 = –1 ✓

The solution is (–10, –1).

3. Sample answer: Solve by substitution because y is already

isolated on one side of the equation.

Substitute 4x + 4 for y in Equation 2.

–8x + 2y = 8

–8x + 2(4x + 4) = 8

–8x + 2(4x) + 2(4) = 8

–8x + 8x + 8 = 8

0 + 8 = 8

8 = 8

Because the equation 8 = 8 is always true, the solutions

of the system are all points on the line y = 4x + 4. So, the

system has infi nitely many solutions.

4. Sample answer: Solve by substitution because x is already


Substitute y – 11 for x Substitute –6 for y

in Equation 2. in Equation 1.

x − 3y = 1 x = y − 11

(y − 11) − 3y = 1 x = –6 − 11

–2y − 11 = 1 x = –17

+11 +11

–2y = 12

–2y — −2

= 12

— –2

y = –6

Check x = y – 11 x – 3y = 1

–17 =?

–6 – 11 –17 – 3(–6) =?

1

–17 = –17 ✓ –17 + 18 =?

1

1 = 1 ✓

The solution is (–17, –6).

5. Sample answer: Solve by elimination because both equations

are in standard form, and none of the terms have a coeffi cient

of 1 or –1.

Step 1 Step 2

6x – 4y = 9 Multiply by 3. 18x – 12y = 27

9x – 6y = 15 Multiply by −2. –18x + 12y = –30

0 + 0 = –3

0 = –3 ✗

The equation 0 = –3 is never true. So, the system has

no solution.


Chapter 5

6. Sample answer: Solve by substitution because y is already


Substitute 5x – 7 for y in Equation 2.

–4x + y = –1

–4x + (5x – 7) = –1

x – 7 = – 1

+7 +7

x = 6

Substitute 6 for x in Equation 1.

y = 5x – 7

y = 5(6) – 7

y = 30 – 7

y = 23

Check y = 5x – 7 –4x + y = –1

23 =?

5(6) – 7 –4(6) + 23 =?

–1

23 =?

30 – 7 –24 + 23 =?

–1

23 = 23 ✓ –1 = –1 ✓


7. Sample answer:

x

y

−8

−4

84−8

y ≤ x + 4y ≥ −x

(−2, 8)

(1, 2)

(4, −3)

So, a system of inequalities that has the points (1, 2) and

(4, −3) in its solution region, but does not have (−2, 8) as a

solution is y ≤ x + 4 and y ≥ −x.

8. Sample answer: When solving ∣ 2x + 1 ∣ = ∣ x − 7 ∣ , you write

two equivalent equations, and you solve each of those equations

in the same way that you solve 4x + 3 = −2x + 9. You write

and graph a linear equation for each side of the equation. The

x-value of the point where the two linear equations intersect

is the solution of the original equation. The absolute value

equation has two solutions, one for each system, whereas the

equation 4x + 3 = −2x + 9 only has one solution.

9. 2y ≤ x + 4

x

6

−2

2−2−4

y

2y

— 2 ≤

x + 4 —

2

y ≤ 1 —

2 x + 2

10. x + y < 1

x

y2

−6

−2

−4

2 4 6−2

x − x + y < 1 − x

y < − x + 1

5x + y > 4

5x − 5x + y > 4 − 5x

y > −5x + 4

11. −3x + y > −2

x

6

4

2

−2

−2−4

y

−3x + 3x +y > −2 + 3x

y > 3x − 2

12. a. Words10 ⋅

Cost per

gallon of

gasoline+ 2 ⋅

Cost per

quart of

oil= 45.50

5 ⋅

Cost per

gallon of

gasoline+ 1 ⋅

Cost per

quart of

oil= 22.75

Variables Let x be the cost of 1 gallon of gasoline, and

let y be the cost of 1 quart of oil.

System 10x + 2y = 45.5 Equation 1

5x + y = 22.75 Equation 2


Step 1

10x + 2y = 45.5 10x + 2y = 45.5

5x + y = 22.75 Multiply by −2. −10x − 2y = −45.5

0 + 0 = 0

0 = 0

Because the equation 0 = 0 is always true, every solution

of 10x + 2y = 45.5 is also a solution of 5x + y = 22.75.

So, the system has infi nitely many solutions, and you do

not have enough information to fi nd the answer.

b. Words 8 ⋅

Cost per

gallon of

gasoline+ 2 ⋅

Cost per

quart of

oil= 38.40

Equation 8x + 2y = 38.40

This equation is not equivalent to either of the equations

from part (a). So, you can write a system of equations

with this equation and one of the equations from part (a),

and the solution of the system will be the cost of 1 gallon

of gasoline and 1 quart of oil.


Chapter 5

c. Solve by elimination.

Step 2 Step 3

10x + 2y = 45.5 10x +2y = 45.5

−(8x + 2y = 38.40) 10(3.55) + 2y = 45.5

2x + 0 = 7.1 35.5 + 2y = 45.5

2x

— 2 =

7.1 —

2 −35.5 −35.5

x = 3.55 2y = 10

2y

— 2 =

10 —

2

y = 5

The solution is (3.55, 5). So, 1 gallon of gasoline costs $3.55,

and 1 quart of oil costs $5.

13. Sample answer: Solving a system of linear equations by

graphing gives you a visual picture of how the output values

of each equation change as the input values change and can

be quick when an estimate is suffi cient. However, it can be

tedious to fi nd the best scale for the axes that will allow you

to see where the solution occurs, and if one or more of the

coordinates of the solution is not a whole number, then the

graphing method may only give you an estimate.

14. a. Words12 ⋅ Number of

trophies+ 3 ⋅ Number of

medals≤ 60

Variables Let x be the number of trophies you can buy,

and let y be the number of medals you can buy.

Inequality 12x + 3y ≤ 60

12x + 3y ≤ 60

12x − 12x + 3y ≤ 60 − 12x

3y ≤ −12x + 60

3y

— 3 ≤

−12x + 60 —

3

y ≤ −4x + 20

8

12

16

20

4

2 3 4 51

Trophies

Med

als

00 x

y

Sample answer: One of the points in the shaded region is

(2, 8). So, you can buy 2 trophies and 8 medals.

b. Words Number of

trophies+

Number of

medals ≥ 6

Inequality x + y ≥ 6 System 12x + 3y ≤ 60

x + y ≥ 6 x + y ≥ 6 x − x + y ≥ 6 − x

y ≥ −x + 6

Graph the system.

y ≥ −x + 6

y ≤ −4x +20

8

12

16

20

4

2 3 4 51

TrophiesM

edal

s0

0 x

y

Sample answer: One of the points in the shaded region is

(3, 4). So, you can buy 3 trophies and 4 medals.

15. 8x + 4y = 12 3y = −6x − 15

8x − 8x + 4y = 12 − 8x 3y

— 3 =

−6x − 15 —

3

4y = −8x +12 y = −2x − 5

4y

— 4 =

−8x + 12 —

4

y = −2x + 3

The slopes of both graphs are −2, but the graph of

8x + 4y = 12 has a y-intercept of 3, and the graph of

3y = −6x −15 has a y-intercept of −5. So, the graphs

are parallel and will never intersect. So, the system has

no solution.


Chapter 5

Chapter 5 Standards Assessment (pp. 286 –287)

1. D; To fi nd the x-intercept, let y = 0.

−9x + 2y = −18

−9x + 2(0) = −18

−9x + 0 = −18

−9x = −18

−9x — −9

= −18

— −9

x = 2

To fi nd the y-intercept, let x = 0.

−9x + 2y = −18

−9(0) + 2y = −18

0 + 2y = −18

2y = −18

2y

— 2 =

−18 —

2

y = −9

So, the intercepts of −9x + 2y = −18 are (2, 0) and (0, −9).

2. C(6) = 100 +5(6) = 100 + 30 = 130

C(8) = 100 + 5(8) = 100 + 40 = 140

C(12) = 100 +5(12) = 100 + 60 = 160

C(16) = 100 +5(16) = 100 + 80 = 180

So, the numbers in the range of the function are 130, 140,

160, and 180.

3. Because the shaded region is above both boundary lines,

one of which is solid and one of which is dashed, the system

is y ≥ 3x − 2 and y > −x + 5.

4. no; If you were to write a system of equations representing

the two sides of this equation, the lines would have the same

slope. So, they are either the same line and the equation has

infi nitely many solutions, or they are parallel lines and the


5. The phrases you should use are “refl ection in the x-axis,”

“horizontal translation,” and “vertical translation.”

6. The two equations that form a system of linear equations

that has no solution are y = 3x + 2 and y = 3x + 1 —

2 . These

equations have the same slope (m = 3) and different

y-intercepts (b = 2 and b = 1 —

2 , respectively). So, they are

parallel lines and will never intersect.

7. a. ax − 8 = 4 − x

a(−2) − 8 = 4 − (−2)

−2a − 8 = 4 + 2

−2a − 8 = 6

+8 +8

−2a = 14

−2a — −2

= 14

— −2

a = −7

When a = −7, the solution is x = −2.

b. ax − 8 = 4 − x

a(12) − 8 = 4 − 12

12a − 8 = −8

+8 +8

12a = 0

12a — 12

= 0 —

12

a = 0

When a = 0, the solution is x = 12.

c. ax − 8 = 4 − x

a(3) − 8 = 4 − 3

3a − 8 = 1

+8 +8

3a = 9

3a — 3 =

9 —

3

a = 3

When a = 3, the solution is x = 3.

8. C; The only ordered pair that is in the shaded half-plane is

(−1, −1). The point (−1, 1) is on the dashed line. So, it is

not a solution of the linear inequality. The other two points

are in the unshaded half-plane. So, they are not solutions

either.

9. The systems

4x − 5y = 3 4x − 5y = 3 12x − 15y = 9

2x + 15y = −1 −4x − 30y = 2 2x + 15y = −1

are equivalent.

If you multiply Equation 1 of the fi rst system by 3, you get

3(4x − 5y = 3) ⇒ 12x − 15y = 9. If you multiply Equation 2

of the fi rst system by −2, you get

−2(2x + 15y = −1) ⇒ −4x − 30y = 2.

10. P = x + 16 + 13 A = 1 —

2 bh

A = 1 —

2 (x)(13)

A > 1 —

2 (9)(13)

A > 58.5

P > 9 + 16 + 13

P > 38

So, P > 38 and A > 58.5

describe the perimeter and

area of the triangle.

Chapter 5 - MDHS-MathChapter 5 5.1 Monitoring Progress (pp. 236–238)4. a. x −3 −2 −1 01 y =...

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Transcript of Chapter 5 - MDHS-MathChapter 5 5.1 Monitoring Progress (pp. 236–238)4. a. x −3 −2 −1 01 y =...