Chapter 7whssalisbury.weebly.com/uploads/1/1/2/8/112801805/hscc... · 2018-12-06 · A direct...

Copyright © Big Ideas Learning, LLC Algebra 2 347All rights reserved. Worked-Out Solutions

Chapter 7

Chapter 7 Maintaining Mathematical Profi ciency (p. 357)

1. 3 — 5 +

2 —

3 =

9 —

15 +

10 —

15

= 19

— 15

2. − 4 — 7 +

1 —

6 = − 24

— 42

+ 7 —

42

= − 17

— 42

3. 7 — 9 −

4 —

9 =

3 —

9

= 1 —

3

4. 5 — 12

− ( − 1 — 2 ) = 5 —

12 +

1 —

2

= 5 —

12 +

6 —

12

= 11

— 12

5. 2 —

7 +

1 —

7 −

6 —

7 =

2 —

7 −

5 —

7

= − 3 —

7

6. 3 — 10

− 3 —

4 +

2 —

5 =

6 —

20 −

15 —

20 +

8 —

20

= − 1 —

20

7. 3 — 8 —

5 —

6 =

3 —

8 ÷

5 —

6

= 3 —

8 ⋅

6 —

5

= 18

— 40

= 9 —

20

8. 1 —

4

—

− 5 — 7

= 1 —

4 ÷ ( − 5 —

7 )

= 1 —

4 ⋅ ( − 7 —

5 )

= − 7 —

20

9. 2 —

3 —

2 —

3 +

1 —

4 =

2 —

3 —

8 —

12 +

3 —

12

= 2 —

3 —

11

— 12

= 2 —

3 ÷

11 —

12

= 2 —

3 ⋅

12 —

11

= 24

— 33

= 8 —

11

10. The expression is not defi ned for x = 0 because division by 0

is undefi ned.

Chapter 7 Mathematical Practices (p. 358)

1. 60 mi —

1h ⋅

1h —

60 min ⋅

1min —

60 sec ⋅

1609.34 m —

1mi ≈

26.8 m —

1 sec

2. 200 lb —

1 in2 ⋅

1 in2

— 6.4516 cm2

⋅ 0.454 kg

— 1 lb

≈ 14.07 kg

— 1 cm2

3. Because F = 9 —

5 C + 32, an increase of 3 degrees Celsius is

the same as an increase of 9 —

5 ⋅ 3 = 5.4 degrees Fahrenheit.

5.4°F

— 1 h

⋅ 1 h —

60 min =

0.09°F —

1 min

4. 1 yd3 —

1 min ⋅

60 min —

1h ⋅

27 ft3 —

1 yd3 =

1620 ft3 —

1h

5. 10 gal —

1 min ⋅

1 min —

60 sec ⋅

3.79 L —

1 gal ≈

0.63 L —

1 sec

7.1 Explorations (p. 359)

1. a. As 0.1 kilogram of weight is added, the spring stretches

1.5 centimeters; As the weight increases, the distance the

spring stretches also increases.

b. Weight, x (in kilograms)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

Distance, d (in centimeters)

0 1.5 3 4.5 6 7.5 9 10.5

0.2 0.4 0.6 0.80 x

d

2

3

4

5

6

7

8

9

10

11

10

Weight (kilograms)

Dis

tan

ce (

cen

tim

eter

s)

As the value of x increases, the value of y also increases.

The points lie on a line.

c. d = 15x

d. The weight x is proportional to the distance d.

348 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.

Chapter 7

2. a. x 1 2 4 8 16 32 64

y 64 32 16 8 4 2 1

b. As x increases, y decreases; Because the product of x and

y is a constant, the two variables vary inversely.

c.

10 20 30 40 50 600 x

y

10

15

20

25

30

35

40

45

50

55

60

65

50

Length (inches)

Wid

th (

inch

es)

As the values of x increase, the values of y decrease. The

points lie on a curve.

d. y = 64

— x

3. Two quantities vary directly when the quotient of the

quantities is the same. Two quantities vary inversely when

the product of the quantities is the same.

4. The quantities vary inversely. As the length of a bird’s wings

increase, the fl apping rate of the wings decreases.

7.1 Monitoring Progress (pp. 360–362)

1. Equation Solved for y Type of Variation

6x = y y = 6x direct


xy = −0.25 y = −0.25

— x inverse


y + x = 10 y = −x + 10 neither

4. Find the products xy and ratios y —

x .

xy −80 −60 −20 −5

y — x

26 —

−4 = −5

15 —

−3 = −5

16 —

−2 = −5

5 —

−1 = −5

So, x and y show direct variation.


x .

xy 60 60 60 60

y — x

60 —

1 = 60

30 —

2 = 15

20 —

3

15 —

4

So, x and y show inverse variation.

6. y = a —

x

5 = a — 4

20 = a

The inverse variation equation is y = 20

— x . When x = 2,

y = 20

— 2 = 10.

7. y = a —

x

−1 = a —

6

−6 = a

The inverse variation equation is y = −6

— x . When x = 2,

y = −6

— 2 = −3.

8. y = a —

x

16 = a

— 1 —

2

8 = a

The inverse variation equation is y = 8 —

x . When x = 2,

y = 8 —

2 = 4.

9. t = a —

n

12 = a —

10

120 = a

The inverse variation equation is t = 120

— n . When n = 18,

t = 120

— 15

= 8 hours.

7.1 Exercises (pp. 363–364)

Vocabulary and Core Concept Check

1. A direct variation equation has the quotient of two variables

equal to a constant and an inverse variation has the product

of two variables equal to a constant.

2. “What is an equation for which the ratios y —

x are constant

and a = 4?” is different than the other three with the answer

y = 4x. The other questions have the answer y = 4 —

x .

Monitoring Progress and Modeling with Mathematics

3. Given Equation Solved for y Type of Variation

y = 2 —

x y =

2 —

x inverse


xy = 12 y = 12

— x inverse


Chapter 7


y —

x = 8 y = 8x direct


4x = y y = 4x direct


y = x + 4 y = x + 4 neither


x + y = 6 y = −x + 6 neither


8y = x y = x —

8 direct


xy = 1 —

5 y =

1 —

5x inverse


x .

xy 1584 3564 5819 9251 12,716

y — x

132 —

12 = 11

198 —

18 = 11

253 —

23 = 11

319 —

29 = 11

374 —

34 = 11



x .

xy 20.25 56.25 144 506.25 900

y — x

13.5 —

1.5 = 9

22.5 —

2.5 = 9

36 —

4 = 9

67.5 —

7.5 = 9

90 —

10 = 9



x .

xy 84 84 84 84 84

y — x

21 —

4 = 5.25

14 —

6 ≈ 2.33

10.5 —

8 ≈ 1.31

10 —

8.4 ≈ 1.19

7 —

12 ≈ 0.58



x .

xy 64 55 62 63 66

y — x

16 —

4 = 4

11 —

5 = 2.2

10 —

6.2 ≈ 1.61

9 —

7 ≈ 1.29

6 —

11 ≈ 0.55

So, x and y show neither direct nor inverse variation.

15. y = a —

x

−4 = a —

5

−20 = a


— x . When x = 3,

y = −20

— 3 .

16. y = a —

x

9 = a —

1

9 = a


x . When x = 3,

y = 9 —

3 = 3.

17. y = a —

x

8 = a —

−3

−24 = a


— x . When x = 3,

y = −24

— 3 = −8.

18. y = a —

x

2 = a —

7

14 = a


— x . When x = 3,

y = 14

— 3 .

19. y = a —

x

28 = a

— 3 —

4

21 = a


— x . When x = 3,

y = 21

— 3 = 7.

20. y = a —

x

−5 —

4 =

a —

−4

5 = a


x . When x = 3, y =

5 —

3 .

21. y = a —

x

− 1 —

6 =

a —

−12

2 = a


x . When x = 3, y =

2 —

3 .


Chapter 7

22. y = a —

x

−7 = a

— 5 —

3

− 35 —

3 = a

The inverse variation equation is y = − 35 —

3x . When x = 3,

y = − 35 —

3(3) = − 35

— 9 .

23. The error is that the direct variation form was used rather

inverse variation.

y = a —

x

5 = a —

8

40 = a

So, y = 40

— x .

24. The error is that y was solved for incorrectly.

xy = a

5 ⋅ 2 = a

10 = a

So, y = 10

— x .

25. a. Because 2500 songs fi t when the average size is 4 MB,

you have

y = a —

x

2500 = a —

4

10,000 = a

So, the equation that models the relationship between

song size and number of songs is y = 10,000

— x .

Average Size (MB) Number of Songs

2

2.5

3

5

10,000

— 2 = 5,000

10,000

— 2.5

= 4,000

10,000

— 3 ≈ 3,333

10,000

— 5 = 2,000

b. The number of songs that it will hold decreases as the

average song size increases.

26. P = a —

A

0.43 = a —

360

154.8 = a

The equation is P = 154.8

— A

. When A = 60,

P = 154.8

— 60

= 2.58 pounds per square inch.

27. First, fi nd the products Ac and ratios c —

A .

Ac 25,984 26,288 25,872 26,320

c —

A

448 —

58 ≈ 7.72

424 —

62 ≈ 6.84

392 —

66 ≈ 5.94

375 —

70 ≈ 5.37

The relationship appears to be an approximately inverse

relationship. So, a model is c = 26,000

— A

. When A = 81,

c = 26,000

— 81

≈ 321 chips per wafer.

28. Because the graph is linear and passes through the origin, the

function f represents direct variation.

29. Your friend is correct, the product of the number of hats and

the price per hat is $50, which is constant.

30. w = a —

d2

210 = a —

39782

3,323,141,640 = a

When d = 3978 + 200 = 4178,

w = 3,323,141,640

—— 41782

≈ 190 pounds.

31. Sample answer: As the speed of your car increases, the

number of minutes per mile decreases.

32. When xy = a and yz = b, y = a —

x and y =

b —

z . Then

a —

x =

b —

z or

az = bx which gives z = b —

a x. So, x varies directly with z.

33. The distance the dog is from the fulcrum is 6 − d. So,

two equations are d = a —

7 and 6 − d =

a —

14 , or 7d = a and

84 − 14d = a.

7d = 84 − 14d

21d = 84

d = 4

So, the cat is 4 feet from the fulcrum and the dog is

6 − 4 = 2 feet from the fulcrum.


Chapter 7

Maintaining Mathematical Profi ciency

34. x − 9

x + 11 ) ‾‾ x2 + 2x − 99

x2 + 11x

−9x − 99

−9x − 99

0

So, ( x2 + 2x − 99 ) ÷ (x + 11) = x − 9.

35. x2 −6

3x2 − x + 5 ) ‾‾‾ 3x4 − x3 − 13x2 + 6x − 30

3x4 − x3 + 5x2

−18x2 + 6x − 30

−18x2 + 6x − 30

0

So, ( 3x4 − 13x2 − x3 + 6x − 30 ) ÷ ( 3x2 − x + 5 ) = x2 − 6

36. y

42−2−4−6

2

4

6

8

x

The domain is all real

numbers and the range

is y > 4.

37. y

42−2−4

2

4

6

8

x

The domain is all real

numbers and the range

is y > 0.

38. y

4 6 82 x

−4

−6

−8

−2

The domain is x > 0

and the range is all

real numbers.

39. y

84−4

8

x

−8

−4

The domain is x > −9

and the range is all real

numbers.


1. a. The graph is C because the denominator is x − 1, and

so the function is undefi ned at x = 1. The graph of g is a

translation 1 unit to the right of the graph of f.

b. The graph is B because the denominator is x − 1, and so

the function is undefi ned at x = 1, and a refl ection of the

graph of f. The graph of g is a refl ection and a translation

1 unit to the right of the graph of f.

c. The graph is E because the denominator is x − 1, and

so the function is undefi ned at x = 1, and when x = 0,

y = −1. The graph of g is a translation 1 unit to the right

and 1 unit up from the graph of f.

d. The graph is A because the denominator is x + 1, and

so the function is undefi ned at x = −1, and when x = 0,

y = −2. The graph of g is a refl ection and a translation

1 unit to the left and 1 unit up from the graph of f.

e. The graph is F because the denominator is x + 2, and so

the function is undefi ned at x = −2 and when x > 0,

g(x) > 0. The graph of g is a refl ection and a translation

2 units to the left and 1 unit up from the graph of f.

f. The graph is D because the denominator is x + 2,

and so the function is undefi ned at x = −2, and when

x > 0, g(x) < 0. The graph of g is a translation 2 units to

the left and 1 unit down from the graph of f.

2. The graph of a rational function may have two branches and

vertical or horizontal asymptotes.

3. The x-intercept is x = a, the y-intercept is y = a —

b , the

asymptotes are x = b and y = 1, the domain is all real

numbers except for b, and the range is all real numbers

except for 1.


Chapter 7


1. Step 1 The function is of the form g(x) = a —

x , so the

asymptotes are x = 0 and y = 0. Draw the

asymptotes.

Step 2 Make a table of values and plot the points. Include

both positive and negative values of x.

x −3 −2 −1 1 2 3

y 2 3 6 −6 −3 −2

Step 3 Draw the two branches of the hyperbola so that they

pass through the plotted points and approach the

asymptotes.

y

2−4−8

4

x

−8

−4g

f

The graph of g lies farther from the axes and is refl ected over

the x-axis. Both graphs have the same asymptotes, domain,

and range.

2. Step 1 Draw the asymptotes x = 0 and y = −2.

Step 2 Plot points to the left of the vertical asymptote, such

as (−3, −3), ( −2, − 7 — 2 ) , and (−1, −5). Plot points to

the right of the vertical asymptote, such as (1, 1),

( 2, − 1 — 2 ) , and (3, −1).



asymptotes.

y

4−2−4

2

x

−4

The domain is all real numbers except 0 and the range is all

real numbers except −2.

3. Step 1 Draw the asymptotes x = −4 and y = 0.

Step 2 Plot points to the left of the vertical asymptote,

such as ( −7, 1 —

3 ) , ( −6,

1 —

2 ) , and (−5, 1). Plot points

to the right of the vertical asymptote, such as

(−3, −1), ( −2, − 1 — 2 ) , and ( −1, −

1 — 3 ) .



asymptotes.

y

2−6−8

2

4

6

x

−4

−2

−6

−10

The domain is all real numbers except −4 and the range is

all real numbers except 0.

4. Step 1 Draw the asymptotes x = 1 and y = 5.


as ( −2, 14

— 3 ) , ( −1,

9 —

2 ) , and (0, 4). Plot points to the

right of the vertical asymptote, such as (2, 6), ( 3, 4 —

2 ) ,

and ( 4, 16

— 3 ) .



asymptotes.

y

2 4 6−4 −2

2

4

6

8

x

10


real numbers except 5.


Chapter 7

5. Step 1 Draw the asymptotes. Solve x + 3 = 0 for x to fi nd

the vertical asymptote, x = −3. The horizontal

asymptote is the line y = a —

c = 1 —

1 = 1.


as (−7, 2), ( −6, 7 —

3 ) , and (−5, 3). Plot points to the

right of the vertical asymptote, such as (−2, −3),

(−1, −1), and ( 0, – 1 — 3 ) .



asymptotes.

y

2 4−4−6−8 −2

2

4

6

x

−4

−2

The domain is all real numbers except –3 and the range is all


6. Step 1 Draw the asymptotes. Solve 4x – 2 = 0 for x to

fi nd the vertical asymptote, x = 1 —

2 . The horizontal


c =

2 —

4 =

1 —

2 .


as ( −2, 3 —

10 ) , ( −1,

1 —

6 ) , and ( 0, −

1 —

2 ) . Plot points to the

right of the vertical asymptote, such as ( 1, 3 —

2 ) , ( 2,

5 —

6 ) ,

and ( 3, 7 —

10 ) .



asymptotes.

y

2 4 6−4−6 −2

2

4

6

x

−4

−6

−2

The domain is all real numbers except 1 —

2 and the range is all

real numbers except 1 —

2 .

7. Step 1 Draw the asymptotes. Solve –x – 1 = 0 for x to fi nd

the vertical asymptote, x = –1. The horizontal


c =

−3 —

−1 = 3.


as (–2, 8), ( –3, 11

— 2 ) , and ( –4,

14 —


right of the vertical asymptote, such as (0, –2), ( 1, 1 —

2 ) ,

and ( 2, 4 —

3 ) .



asymptote.

y

2 4−4−6 −2

2

4

6

8

x

−2



8. Rewrite the function by using long division:

2 x + 1 ) ‾ 2x + 3

2x + 2

1

The rewritten function is g(x) = 1 —

x + 1 + 2.

y

2 4−4−6 −2

2

6

8

x

−2

−4

The graph of g is a translation 1 unit to the left and 2 units up

of the graph of f.

9. c = (Unit cost)(Number printed) + (Cost of printer)

———— Number printed

= 50m + 800 —

m

Use a graphing calculator to graph the function. Using the

trace feature, the average cost falls to $90 per model after

20 models are printed, but the average cost still approaches

$50 as more models are printed.


Chapter 7



1. The function y = 7 —

x + 4 + 3 has a range of all real numbers

except 3 and a domain of all real numbers except −4.

2. The function is not a rational function because the

denominator is not a polynomial.



x , so the


asymptotes.



x –3 –2 –1 1 2 3

y −1 − 3 —

2 −3 3

3 —

2 1



asymptotes.

y

84−8

8

4

x

−8

g

f

The graph of g lies farther from the axes. Both graphs lie in

the fi rst and third quadrants and have the same asymptotes,

domain, and range.


x , so the


asymptotes.



x –3 –2 –1 1 2 3

y − 10

— 3 –5 –10 10 5

10 —

3



asymptotes.

y

84−4

8

4

x

−4

gf

The graph of g lies farther from the axes. Both graphs lie in


domain, and range.


x , so the


asymptotes.



x –3 –2 –1 1 2 3

y 5 —

3

5 —

2 5 –5 – 5 — 2 –

5 —

3



asymptotes.

y

2−8 −4

2

x

−8

−4g

f



and range.


Chapter 7


x , so the


asymptotes.



x –3 –2 –1 1 2 3

y 3 9 —

2 9 –9 – 9 — 2 –3



asymptotes.

y

4−8 −4

4

x

−8

−4 g

f



and range.


x , so the


asymptotes.



x –3 –2 –1 1 2 3

y –5 – 15 — 2 –15 15

15 —

2 5



asymptotes.

y

84−8 −4

8

4

x

−4

g

f

The graph of g lies farther from the axis. Both graphs lie in


domain, and range.

8. Step 1 The function is of the form g(x) = a

— x , so the


asymptotes.



x –3 –2 –1 1 2 3

y 4 6 12 –12 6 4



asymptotes.

4−8 −4

8

4

x

−4

−8

y

g

f



and range.

9. Step 1 The function is of the form g(x) = a

— x , so the


asymptotes.



x –3 –2 –1 1 2 3

y 0.1667 0.25 0.5 –0.5 –0.25 –0.1667



asymptotes.

y

4−4

4

2

x

−4

g

f

The graph of g lies closer to the axes and is refl ected over


and range.


Chapter 7


x , so the


asymptotes.



x –3 –2 –1 1 2 3

y –0.033 –0.05 –0.1 0.1 0.05 0.033



asymptotes.

y

42−4

4

2

x

−4

gf

The graph of g lies closer to the axes. Both graphs lie in


domain, and range.



as ( –3, 5 —

3 ) , (–2, 1), and (–1, –1). Plot points to the right

of the vertical asymptote, such as (1, 7), (2, 5),

and ( 3, 13

— 3 ) .



asymptotes.

y

4 62−4−6 −2

4

6

8

2

x



12. Step 1 Draw the asymptotes x = 0 and y = –3.


as ( –3, – 11 — 3 ) , (–2, –4), and (–1, –5). Plot points to the

right of the vertical asymptote, such as (1, –1),

(2, –2), and ( 3, – 7 — 3 ) . Step 2 Draw the two branches of the hyperbola so that they


asymptotes.

y

4 6−4−6 −2

4

2

x

−4

−2

2


real numbers except –3.



as (–2, –2), (–1, –3), and (0, –6). Plot points to the

right of the vertical asymptote, such as (2, 6), (3, 3),

and (4, 2).



asymptotes.

4 6 82−4−6 −2

4

6

2

x

−4

−2

−6

−8

y




Chapter 7

14. Step 1 Draw the asymptotes x = –2 and y = 0.


as ( –5, – 1 — 3 ) , ( –4, – 1 — 2 ) , and (–3, –1). Plot points to the

right of the vertical asymptote, such as (–1, 1), ( 0, 1 —

2 ) ,

and ( 1, 1 —

3 ) .



asymptotes.

42−8

4

6

2

x

−4

−2

−6

y



15. Step 1 Draw the asymptotes x = –2 and y = 0.


as (–5, 1), ( –4, 3 —

2 ) , and (–3, 3). Plot points to the right

of the vertical asymptote, such as (–1, –3), ( 0, – 3 — 2 ) , and (1, –1).



asymptotes.

1−8 −6 −4

4

6

2

x

−4

−6

y





as ( 4, 2 —

3 ) , (5, 1), and (6, 2). Plot points to the right of

the vertical asymptote, such as (8, –2), (9, –1), and

( 10, – 2 — 3 ) . Step 3 Draw the two branches of the hyperbola so that they


asymptotes.

2 4 6 8

4

6

2

x

−4

−2

−6

y



17. Step 1 Draw the asymptotes x = 4 and y = –1.


as (1, 0), ( 2, 1 —

2 ) , and (3, 2). Plot points to the right of

the vertical asymptote, such as (5, –4), ( 6, – 5 — 2 ) , and (7, –2).



asymptotes.

2 6 8 10

4

6

2

−4

−2

−6

−8

y

x




Chapter 7

18. Step 1 Draw the asymptotes x = –7 and y = –5.


as ( –10, – 25 — 3 ) , (–9, –10), and (–8, –15). Plot points to

the right of the vertical asymptote, such as (–6, 5),

(–5, 0), and ( –4, – 5 — 3 ) . Step 3 Draw the two branches of the hyperbola so that they


asymptotes.

2

−4

−2

−6

−8

y

x

−10

−12

−2−4−6−8−10−12−14



19. The error is that the graph should lie in the second and fourth

quadrants instead of the fi rst and third quadrants.

4−8 −4

4

x

−4

−8

y

20. The error is that the vertical asymptote should be x = 1.

42−4−6 −2

4

2

x

−4

−6

y

21. The graph is A because the asymptotes are x = 3 and y = 1.

22. The graph is C because the asymptotes are x = –3 and y = 1.

23. The graph is B because the asymptotes are x = 3 and y = –1.

24. The graph is D because the asymptotes are x = –3 and

y = –1.

25. Step 1 Draw the asymptotes. Solve x – 3 = 0 for x to fi nd

the vertical asymptote x = 3. The horizontal


c =

1 —

1 = 1.


as ( 0, – 4 — 3 ) , ( 1, – 5 — 2 ) , and (2, −6). Plot points to the

right of the vertical asymptote, such as (4, 8), ( 5, 9 —

2 ) ,

and ( 6, 10

— 3 ) .



asymptotes.

4 6 8 102−4

4

6

8

2

x

−4

−2

−6

y



26. Step 1 Draw the asymptotes. Solve x + 5 = 0 for x to fi nd

the vertical asymptote x = –5. The horizontal


c =

1 —

1 = 1.


as (–8, 3), (–7, 4), and (–6, 7). Plot points to the right

of the vertical asymptote, such as (–4, –5), (–3, –2),

and (–2, –1).



asymptotes.

2−4 −2−6−8

4

6

8

2

x

−4

−2

−6

y

−10−12




Chapter 7

27. Step 1 Draw the asymptotes. Solve 4x − 8 = 0 for x to fi nd



c =

1 —

4 .


as ( –1, – 5 — 12

) , ( 0, – 3 — 4 ) , and ( 1, – 7 —


right of the vertical asymptote, such as ( 3, 9 —

4 ) , ( 4,

5 —

4 ) ,

and ( 5, 11

— 12

) . Step 3 Draw the two branches of the hyperbola so that they


asymptotes.

4 6 8−4

4

6

2

x

−4

−2

−6

y


real numbers except 1 —

4 .

28. Step 1 Draw the asymptotes. Solve 2x – 6 = 0 for x to fi nd



c =

8 —

2 = 4.


as ( 0, – 1 — 2 ) , ( 1, – 11

— 4 ) , and ( 2, – 19

— 2 ) . Plot points to the

right of the vertical asymptote, such as ( 4, 35

— 2 ) ,

( 5, 43

— 4 ) , and ( 6,

49 —

6 ) .



asymptotes.

84 12−8 −4

8

x

−4

y16

12



29. Step 1 Draw the asymptotes. Solve 4x + 5 = 0 for x to

fi nd the vertical asymptote x = – 5 —

4 . The horizontal


c =

−5 —

4 .


as (−4, −2), ( −3, − 17 —

7 ) , and (−2, −4). Plot points

to the right of the vertical asymptote, such as (−1, 7),

( 0, 2 —

5 ) , and ( 1, − 1 —

3 ) .



asymptotes.

4 6−4 −2−6−8

4

6

x

−4

−2

−6

−8

y

The domain is all real numbers except − 5 —

4 and the range is

all real numbers except − 5 —

4 .

30. Step 1 Draw the asymptotes. Solve 3x − 1 = 0 for x to fi nd

the vertical asymptote x = 1 —

3 . The horizontal


c =

6 —

3 = 2.


as ( −3, 19

— 10

) , ( −2, 13

— 7 ) , and ( −1,

7 —

4 ) . Plot points to

the right of the vertical asymptote, such as ( 1, 5 —

2 ) ,

( 2, 11

— 5 ) , and ( 3,

17 —

8 ) .


Chapter 7



asymptotes.

42 6−4−6 −2

4

6

8

x

−4

−2

y

The domain is all real numbers except 1 —

3 and the range is all


31. Step 1 Draw the asymptotes. Solve −2x − 3 = 0 for x to

fi nd the vertical asymptote x = − 3 —

2 . The horizontal


c =

−5 —

−2 =

5 —

2 .


as (−4, 4), (−3, 5), and (−2, 10). Plot points to the

right of the vertical asymptote, such as (−1, −5),

(0, 0), and (1, 1).



asymptotes.

42−4−6−8 −2

4

2

6

8

x

−4

y

The domain is all real numbers except − 3 —

2 and the range is

all real numbers except 5 —

2 .

32. Step 1 Draw the asymptotes. Solve −x + 10 = 0 for x to

fi nd the vertical asymptote x = 10. The horizontal


c =

−2 —

−1 = 2.

Step 2 Plot points to the left of the vertical asymptote,

such as ( 7, − 11

— 3 ) , ( 8, −

13 —

2 ) , and (9, −15). Plot

points to the right of the vertical asymptote, such as

(11, 19), ( 12, 21

— 2 ) , and ( 13,

23 —

3 ) .



asymptotes.

8 1612 20−4

8

4

x

−8

−4

y

12




5

x + 1 ) ‾ 5x + 6

5x + 5

1


x + 1 + 5.

42−4−6 −2

4

2

6

x

y

The graph of g is a translation 1 unit to the left and 5 units up

of the graph of f.


Chapter 7


7

x − 3 ) ‾ 7x + 4

7x − 21

25


x − 3 + 7.

168−8

8

x

−8

y

16

−16

−16

The graph of g is a translation 3 units to the right and 7 units

up of the graph of f.


2

x − 5 ) ‾ 2x − 4

2x − 10

6


x − 5 + 2.

42 86 10−2

4

2

6

8

x

−4

−2

y




4

x − 2 ) ‾ 4x − 11

4x − 8

−3

The rewritten function is g(x) = −3

— x − 2

+ 4.

4 86−2−4

2

6

8

x

−2

y

10




1

x − 6 ) ‾ x + 18

x − 6

24


x − 6 + 1.

84 1612

4

8

x

−4

−8

y12

−12

The graph of g is a translation 6 units to the right and 1 unit



Chapter 7


1

x − 8 ) ‾ x + 2

x − 8

10


x − 8 + 1.

204 1612

4

8

x

−4

−8

y12

−12

−4

The graph of g is a translation 8 units to the right and 1 unit



7

x + 13 ) ‾ 7x − 20

7x + 91

−111


— x + 13

+ 7.

8−8

8

x

−8

y

16

−16−32

−16

−24

24

32

The graph of g is a translation 13 units to the left and 7 units



9

x + 7 ) ‾ 9x − 3

9x + 63

−66


— x + 7

+ 9.

8−8

8

x

−8

y

16

−16−24

24



41. a. Let c be the average cost (in dollars) and s be the number

of students.

c = (cost per student)(number of students) + (initial cost)

————— number of students

= 20s + 500

— s


trace feature, the average cost falls to $30 per student after

about 50 students.

b. Because the horizontal asymptote is c = 20, the average

cost per student approaches $20.

42. a. Let c be the average cost (in dollars) and m be the number

of months.

c = (cost per month)(number of months) + (initial cost) —————

number of months

= 59m + 100

— m


trace feature, the average cost falls to $69 per month after

about 10 months.

b. Because the horizontal asymptote is c = 59, the average

cost approaches $59.

43. The vertical asymptote is B, x = −4.

44. The x-intercept is B, 5.

45. a. Use a graphing calculator to graph the function. Using

the trace feature, the temperature is about 23°C for the

thunder to be heard 2.9 seconds after the lightning strike.

b. Compute the average rate of change.

t(10) − t(0)

— 10 − 0

≈ 2.97 − 3.02

— 10

≈ −0.005 sec/°C


Chapter 7

46. a.

2 3 41 x

y

60

30

90

−30

−60

−90

A reasonable domain is 0 ≤ x ≤ 1 and the corresponding

range is 0 ≤ y ≤ 150.

b. When x = 0.2, y = 15(0.2)

— 1.1 − 0.2

≈ $3333. When

x = 0.4, y = 15(0.4)

— 1.1 − 0.4

≈ $8571. When x = 0.8,

y = 15(0.8)

— 1.1 − 0.8

= $40,000. No; doubling the percentage

does not double the cost because this is not a linear

function.

47.

8

−2

10

−8

Because the graph is symmetric

with respect to the y-axis, the

function is even.

48.

10

−10

10

−10


with respect to the y-axis, the

function is even.

49.

10

−1

1

−10


with respect to the origin, the

function is odd.

50.

6

−6

6

−6


with respect to the origin, the

function is odd.

51. Your friend is correct; a rational function with a denominator

that has two zeros will have a graph that has two vertical

asymptotes. For example, y = 3 ——

(x + 1)(x − 1) has vertical

asymptotes x = 1 and x = −1.

52. The graph of the function of f has asymptotes x = −3 and

y = 2. The graph appears to approach the lines y = 2 and

x = −3.

53. The graph of y = 1 —

x is symmetric in the lines y = x and

y = −x. This says that the function f(x) = 1 —

x is its own inverse.

54.

55. Another point on the graph is (4, 3). The point (2, 1) is 1 unit

to the left and 1 unit down from (3, 2), so a point on the other

branch is 1 unit to the right and 1 unit up from (3, 2).

56. a. The graph is decreasing on (−∞, 0) and (0, ∞) because the

graph falls from left to right.

b. The graph is increasing on (−∞, 0) and (0, ∞) because the

graph rises from left to right.

57. The average cost for the Internet provider can be modeled

by c = 43x + 50

— x . The competitor is a better choice for

12 months of service or less because the average cost is less

for x = 6 and x = 12 and greater for x = 18 and x = 24.

58. a. The model for the frequency heard when the ambulance

is approaching is fℓ = 740(2000)

— 740 − r

and when the

ambulance is moving away is fℓ = 740(2000)

— 740 + r

.

b.

10 20 30 40 500 r

f

1700

1800

1900

2000

2100

2200

2300

2400

16000

Approaching

Moving away

c. The frequency heard is greater for an approaching sound

source.


59. 4x 2 − 4x − 80 = 4(x 2 − x − 20)

= 4(x − 5)(x + 4)

60. 3x 2 − 3x − 6 = 3(x 2 − x − 2)

= 3(x − 2)(x + 1)

61. 2x 2 − 2x − 12 = 2(x 2 − x − 6)

= 2(x − 3)(x + 2)

62. 10x 2 + 31x − 14 = (5x − 2)(2x + 7)


Chapter 7

63. 32 ⋅ 3 4 = 3 2+4 = 3 6

64. 2 1/2 ⋅ 2 3/5 = 2 1/2 + 3/5 = 2 11/10

65. 6 5 6 —

6 1/6 = 6 5/6−1/6 = 6 4/6 = 6 2/3

66. 6 8 —

6 10 = 6 8−10 = 6 −2 =

1 —

6 2

7.1–7.2 What Did You Learn? (p. 373)

1. The amount of storage on an MP3 player is fi nite, so as

the fi le size of the songs increases, the number of songs

decreases.

2. The inverse variation equation is true for both situations,

so the logic is correct.

3. You can use the table feature of a graphing calculator to

verify that your answer approaches the horizontal asymptote.

4. Sample answer: The value of x is a percent as a decimal, so

the domain for x is greater than zero and less than one. The

value of y must be positive, so the range must be greater than

zero.

7.1–7.2 Quiz (p. 374)


x + y = 7 y = −x + 7 neither


2 —

5 x = y y =

2 —

5 x direct


xy = 0.45 y = 0.45

— x inverse


x .



x .



x .

xy 144 144 108 72

y — x 36 9 3

9 —

8

So, x and y show neither direct nor inverse variation.

7. y = a —

x

10 = a —

5

50 = a


— x . When x = −2,

y = 50

— −2

= −25.

8. The graph is C because the asymptotes are x = 0 and y = 2.

9. The graph is A because the asymptotes are x = −3 and

y = −2.

10. The graph is B because the asymptotes are x = − 1 —

3 and y =

2 —

3 .


2

x + 8 ) ‾ 2x + 9

2x + 16

−7


— x + 8

+ 2.

−6 −2−4

6

4

8

x

−2

−4

−6

y10

−10−12−14

The graph of g is a translation 8 units to the left and 2 units up.

12. r = a —

t

70 = a —

20

1400 = a

So, the model for the pump is r = 1400

— t .

Pumping rate(gal/min)

Time(min)

50 28

60 23.3

65 21.5

70 20

xy 27 108 243 432

y — x 3 3 3 3

xy −24 −24 −24 −24

y — x −24 −12 −

8 —

3

−6 —

4 = −

3 —

2


Chapter 7

13. The model is f (x) = x + 16

— x + 38

.

10 20 30 40 500 x

y

.2

.3

.4

.5

.6

.7

.8

.9

1.0

.10

Consecutive strikes

Stri

ke p

erce

nta

ge

Using the graph, the strike percentage is 0.60, when the

number of consecutive strikes is 17.


1. a. 1 —

x − 1 ⋅

x − 2 —

x + 1 =

x − 2 ——

(x − 1)(x + 1) . This matches F because

the expressions x − 1 and x + 1 cannot be equal to zero.

b. 1 —

x − 1 ⋅

−1 —

x − 1 =

−1 —

(x − 1)2 . This matches H because the

expression x − 1 cannot be equal to zero.

c. 1 —

x − 2 ⋅

x − 2 —

x + 1 =

1 —

x + 1 , x ≠ 2. This matches D because


d. x + 2

— x − 1

⋅ −x

— x + 2

= −x

— x − 1

, x ≠ −2. This matches B because


e. x —

x + 2 ÷

x + 1 —

x + 2 =

x —

x + 1 , x ≠ −2. This matches G because

the expressions x + 2 and x + 1 cannot be equal to zero.

f. x —

x − 2 ÷

x + 1 —

x =

x 2 ——

(x − 2) (x + 1) , x ≠ 0. This matches A

because the expressions x, x − 2, and x + 1 cannot be

equal to zero.

g. x —

x + 2 ÷

x —

x − 1 =

x − 1 —

x + 2 , x ≠ 0. This matches C because

the expressions x, x + 2, and x − 1 cannot be equal to zero.

h. x + 2

— x ÷

x + 1 —

x − 1 =

(x + 2)(x − 1) ——

x(x + 1) , x ≠ 1. This

matches E because the expressions x, x − 1, and x + 1

cannot be equal to zero.

2. a. Sample answer: 1 —

x + 1 ⋅

x —

x + 1 . The expression x + 1 is

equal to zero when x = −1.

b. Sample answer: x —

x + 1 ⋅

x + 1 —

x − 3 . The expressions

x + 1 and x − 3 are equal to zero when x = −1 and x = 3, respectively.

c. Sample answer: 1 —

x + 1 ÷

x —

x − 3 . The expressions x, x + 1,

and x − 3 are equal to zero when x = 0, x = −1, and x = 3, respectively.

3. The excluded values are the values that make any expression

in the denominator of each product or dividend zero.

4. It is possible for the product or quotient of two rational

expressions to have no excluded values if each expression in

each denominator does not have any values for x that make

the expression zero. Sample answer: 1 —

x 2 + 1 ⋅

x 2 + 1 —

x 2 + 5 .


1. 2(x + 1) ——

(x + 1)(x + 3) =

2 —

x + 3 , x ≠ −1

2. x + 4 —

x 2 − 16 =

x + 4 ——

(x − 4)(x + 4)

= x + 4 ——

(x − 4)(x + 4)

= 1 —

x − 4 , x ≠ −4

3. 4 —

x(x + 2) is in simplifi ed form.

4. x 2 − 2x − 3 ——

x 2 − x − 6 =

(x − 3)(x + 1) ——

(x − 3)(x + 2)

= (x − 3)(x + 1)

—— (x − 3)(x + 2)

= x + 1

— x + 2

, x ≠ 3

5. 3x 5y 2 —

8xy ⋅

6xy 2 —

9x 3y =

18x 6y 4 —

72x 4y 2

= x 2y 2

— 4 , x ≠ 0, y ≠ 0

6. 2x 2 − 10x — x 2 − 25

⋅ x + 3

— 2x 2

= 2x(x − 5)

—— (x − 5)(x + 5)

⋅ x + 3

— 2x 2

= 2x(x − 5)(x + 3)

—— 2x 2(x − 5)(x + 5)

= x + 3

— x(x + 5)

, x ≠ 5


Chapter 7

7. x + 5 —

x 3 − 1 ⋅ ( x 2 + x + 1 ) =

x + 5 ——

(x − 1) ( x 2 + x + 1 ) ⋅ ( x 2 + x + 1 )

= (x + 5) ( x 2 + x + 1 )

—— (x − 1) ( x 2 + x + 1 )

= x + 5

— x − 1

8. 4x — 5x − 20

÷ x 2 − 2x

—— x 2 − 6x + 8

= 4x —

5x − 20 ⋅

x 2 − 6x + 8 ——

x 2 − 2x

= 4x —

5(x − 4) ⋅

(x − 4)(x − 2) ——

x(x − 2)

= 4x(x − 4)(x − 2)

—— 5x(x − 4)(x − 2)

= 4 —

5 , x ≠ 0, x ≠ 2, x ≠ 4

9. 2x 2 + 3x − 5 ——

6x ÷ ( 2x 2 + 5x ) =

2x 2 + 3x − 5 ——

6x ⋅

1 —

2x 2 + 5x

= (x − 1)(2x + 5)

—— 6x

⋅ 1 —

x(2x + 5)

= (x − 1)(2x + 5)

—— 6x 2(2x + 5)

= x − 1

— 6x 2

, x ≠ − 5 —

2



1. To multiply rational expressions, multiply numerators,

then multiply denominators, and write the new fraction

in simplifi ed form. To divide one rational expression

by another, multiply the fi rst rational expression by the

reciprocal of the second rational expression.

2. The expression x 2 + 4x − 12

—— x 2 + 6x

does not match the others

because it is the only expression that is not in simplifi ed

form.


3. 2x 2 —

3x 2 − 4x =

2x 2 —

x(3x − 4)

= 2x —

3x − 4 , x ≠ 0

4. 7x 3 − x 2 —

2x 3 =

x 2(7x − 1) —

2x 3

= 7x − 1

— 2x

5. x 2 − 3x − 18 ——

x 2 − 7x + 6 =

(x − 6)(x + 3) ——

(x − 6)(x − 1)

= x + 3

— x − 1

, x ≠ 6

6. x 2 + 13x + 36 ——

x 2 − 7x + 10 is in simplifi ed form.

7. x 2 + 11x + 18 ——

x 3 + 8 =

(x + 2)(x + 9) ——

( x + 2 ) ( x 2 − 2x + 4 )

= x + 9 ——

x 2 − 2x + 4 , x ≠ − 2

8. x 2 − 7x + 12 ——

x 3 − 27 =

(x − 4)(x − 3) ——

( x − 3 ) ( x 2 + 3x + 9 )

= x − 4 ——

x 2 + 3x + 9 , x ≠ 3

9. 32x 4 − 50 ——

4x 3 − 12x 2 − 5x + 15 =

2 ( 4x 2 − 5 ) ( 4x 2 + 5 ) ——

( x − 3 ) ( 4x 2 − 5 )

= 2 ( 4x 2 + 5 )

— x − 3

, x ≠ ± √—

5 —

4

10. 3x 3 − 3x 2 + 7x − 7 ——

27x 4 − 147 =

( x − 1 ) ( 3x 2 + 7 ) ——

3 ( 3x 2 − 7 ) ( 3x 2 + 7 )

= x − 1 —

3 ( 3x 2 − 7 )

11. 4xy 3 —

x 2y ⋅

y —

8x =

4x y 4 —

8x 3y

= y 3

— 2x 2

, y ≠ 0

12. 48x 5y 3 —

y 4 ⋅

x 2y —

6x 3y 2 =

48x 7y 4 —

6x 3y 6

= 8x 4

— y 2

, x ≠ 0

13. x 2(x − 4) —

x − 3 ⋅

(x − 3)(x + 6) ——

x 3 =

x 2(x − 4)(x − 3)(x + 6) ——

x 3(x − 3)

= (x − 4)(x + 6)

—— x , x ≠ 3

14. x 3(x + 5) —

x − 9 ⋅

(x − 9)(x + 8) ——

3x 3 =

x 3(x − 9)(x + 5)(x + 8) ——

3x 3(x − 9)

= (x + 5)(x + 8)

—— 3 , x ≠ 0, x ≠ 9

15. x 2 − 3x — x − 2

⋅ x 2 + x − 6

— x =

x(x − 3) —

x − 2 ⋅

(x + 3)(x − 2) ——

x

= x(x − 3)(x − 2)(x + 3)

—— x(x − 2)

= (x − 3)(x + 3), x ≠ 0, x ≠ 2

16. x 2 − 4x — x − 1

⋅ x 2 + 3x − 4

—— 2x

= x(x − 4)

— x − 1

⋅ (x + 4)(x − 1)

—— 2x

= x(x − 4)(x + 4)(x − 1)

—— 2x(x − 1)

= (x − 4)(x + 4)

—— 2 , x ≠ 0, x ≠ 1

17. x 2 + 3x − 4 —

x 2 + 4x + 4 ⋅

2x 2 + 4x —

x 2 − 4x + 3 =

(x + 4)(x − 1) ——

(x + 2)(x + 2) ⋅

2x(x + 2) ——

(x − 1)(x − 3)

= 2x(x + 4)(x − 1)(x + 2)

——— (x + 2)(x + 2)(x − 1)(x − 3)

= 2x(x + 4)

—— (x + 2)(x − 3)

, x ≠ 1


Chapter 7

18. x 2 − x − 6 —

4x 3 ⋅

2x 2 + 2x —

x 2 + 5x + 6 =

(x − 3)(x + 2) ——

4x 3 ⋅

2x(x + 1) ——

(x + 2)(x + 3)

= 2x(x − 3)(x + 2)(x + 1)

—— 4x 3(x + 2)(x + 3)

= (x − 3)(x + 1)

—— 2x 2(x + 3)

, x ≠ −2

19. x 2 + 5x − 36

—— x 2 − 49

⋅ ( x 2 − 11x + 28 )

= (x + 9)(x − 4)

—— (x − 7)(x + 7)

⋅ (x − 4)(x − 7)

—— 1

= (x + 9)(x − 4)(x − 4)(x − 7)

——— (x − 7)(x + 7)

= (x + 9)(x − 4) 2

—— (x + 7)

, x ≠ 7

20. x 2 − x − 12

—— x 2 − 16

⋅ ( x 2 + 2x − 8 )

= (x − 4)(x + 3)

—— (x − 4)(x + 4)

⋅ (x + 4)(x − 2)

—— 1

= (x − 4)(x + 3)(x + 4)(x − 2)

——— (x − 4)(x + 4)

= (x + 3)(x − 2), x ≠ −4, x ≠ 4

21. The error is that the expressions needed to be factored fi rst,

and then the common factors are divided out.

x 2 + 16x + 48

—— x 2 + 8x + 16

= (x + 4)(x + 12)

—— (x + 4)(x + 4)

= x + 12

— x + 4

22. The error is that the expression 3 − x should be factored into

−(x − 3) before dividing out common factors.

x2 − 25

— 3 − x

⋅ x − 3

— x + 5

= (x − 5)(x + 5)

—— −(x − 3)

⋅ x − 3

— x + 5

= (x − 5)(x + 5)(x − 3)

—— −(x − 3)(x + 5)

= −(x − 5)

= 5 − x, x ≠ 3, x ≠ −5

23. The rational expression in simplifi ed form is B.

24. 4x 2y — 2x 3

⋅ 12y 4

— 24x 2

= 48x 2y 5

— 48x 5

4x 2y

— 2x 3

⋅ 12y 4

— 24x 2

= 2y

— x ⋅

y 4 —

2x 2

= y 5

— x 3

= 2y 5

— 2x 3

= y 5

— x 3

Sample answer: Multiplying the numerators and

denominators before simplifying is preferred because you

only have to simplify once.

25. The expressions have the same simplifi ed form, but the

domain of f (x) is all real numbers except 7 —

3 , and the domain

of g(x) is all real numbers.

26. The total area of the base of the new building is

x 2 − 7x + 10

—— 6x − 12

⋅ 3x 2 − 12

— x 2 − x − 20

= (x + 2)(x − 2)

—— 2(x + 4)

, x ≠ 2, x ≠ 5.

27. 32x 3y — y 8

÷ y 7 —

8x 4 =

32x 3y —

y 8 ⋅

8x 4 —

y 7

= 256x 7y

— y 15

= 256x 7

— y 14

, x ≠ 0

28. 2xyz — x 3z 3

÷ 6y 4

— 2x 2z 2

= 2xyz

— x 3z 3

⋅ 2x 2z 2

— 6y 4

= 4x 3yz 3

— 6x 3y 4z 3

= 2 —

3y 3 , x ≠ 0, z ≠ 0

29. x 2 − x − 6 —

2x 4 − 6x 3 ÷

x + 2 —

4x 3 =

x 2 − x − 6 —

2x 4 − 6x 3 ⋅

4x 3 —

x + 2

= (x − 3)(x + 2)

—— 2x 3(x − 3)

⋅ 4x 3

— x + 2

= 4x 3(x − 3)(x + 2)

—— 2x 3(x − 3)(x + 2)

= 2, x ≠ −2, x ≠ 0, x ≠ 3

30. 2x 2 − 12x —— x 2 − 7x + 6

÷ 2x —

3x − 3 =

2x 2 − 12x ——

x 2 − 7x + 6 ⋅

3x − 3 —

2x

= 2x(x − 6)

—— (x − 1)(x − 6)

⋅ 3(x − 1)

— 2x

= 6x(x − 6)(x − 1)

—— 2x(x − 6)(x − 1)

= 3, x ≠ 0, x ≠ 1, x ≠ 6

31. x 2 − x − 6 —

x + 4 ÷(x 2 − 6x + 9) =

x 2 − x − 6 —

x + 4 ⋅

1 ——

x 2 − 6x + 9

= (x − 3)(x + 2)

—— x + 4

⋅ 1 ——

(x − 3)(x − 3)

= (x − 3)(x + 2)

—— (x + 4)(x − 3)(x − 3)

= (x + 2) ——

(x + 4)(x − 3)

32. x 2 − 5x − 36

—— x + 2

÷ (x 2 − 18x + 81)

= x 2 − 5x − 36

—— x + 2

⋅ 1 ——

x 2 − 18x + 81

= (x − 9)(x + 4)

—— x + 2

⋅ 1 ——

(x − 9)(x − 9)

= (x − 9)(x + 4)

—— (x + 2)(x − 9)(x − 9)

= x + 4 ——

(x + 2)(x − 9)


Chapter 7

33. x 2 + 9x + 18

—— x 2 + 6x + 8

÷ x 2 − 3x − 18 ——

x 2 + 2x − 8

= x 2 + 9x + 18

—— x 2 + 6x + 8

⋅ x 2 + 2x − 8

—— x 2 − 3x − 18

= (x + 3)(x + 6)

—— (x + 2)(x + 4)

⋅ (x − 2)(x + 4)

—— (x − 6)(x + 3)

= (x + 3)(x + 6)(x − 2)(x + 4)

——— (x + 2)(x + 4)(x − 6)(x + 3)

= (x + 6)(x − 2)

—— (x + 2)(x − 6)

, x ≠ −4, x ≠ −3

34. x 2 − 3x − 40

—— x 2 + 8x − 20

÷ x 2 + 13x + 40

—— x 2 + 12x + 20

= x 2 − 3x − 40

—— x 2 + 8x − 20

⋅ x 2 + 12x + 20

—— x 2 + 13x + 40

= (x − 8)(x + 5)

—— (x + 10)(x − 2)

⋅ (x + 10)(x + 2)

—— (x + 8)(x + 5)

= (x − 8)(x + 5)(x + 10)(x + 2)

——— (x + 10)(x − 2)(x + 8)(x + 5)

= (x − 8)(x + 2)

—— (x − 2)(x + 8)

, x ≠ −10, x ≠ −5, x ≠ −2

35. a. For a cylindrical can, the surface area is S = 2πr 2 + 2πrh

and the volume is V = πr 2h. So, the efficiency ratio is

S — V

= 2πr(r + h)

— πr 2h

= 2(r + h)

— rh

b. Soup: 2(3.4 + 10.2)

—— (3.4)(10.2)

≈ 0.784

Coffee: 2(7.8 + 15.9)

—— (7.8)(15.9)

≈ 0.382

Paint: 2(8.4 + 19.4)

—— (8.4)(19.4)

≈ 0.341

Because 0.341 < 0.382 < 0.784, the paint can is most

effi cient, followed by the coffee can, and then the soup can.

36. a. and b. The surface area and volume of the original

tin are S = 2s 2 + 4sh and V = s 2h. The surface area

and volume of the new tin are S = 2s 2 + 8sh and

V = 2s 2h. So, the effi ciency ratio for the original tin is

S — V

= 2s 2 + 4sh

— s 2h

= 2s + 4h

— sh

and the effi ciency ratio for

the new tin is S —

V =

2s 2 + 8sh —

2s 2h =

s + 4h —

sh

c. The new tin is more effi cient because s + 4h < 2s + 4h

when s and h are positive and the denominators are the

same in both ratios.

37. To fi nd a model M for the annual healthcare expenditure per

resident, divide the total amount I by the population P.

M = 171t + 1361

—— 1 + 0.018t

÷ (2960t + 278,649)

= 171t + 1361

—— 1 + 0.018t

⋅ 1 ——

2960t + 278,649

= 171t + 1361

——— (1 + 0.018t)(2960t + 278,649)

Because I is in billions of dollars and P is in thousands, M is

in millions of dollars. To estimate the annual healthcare

expenditure per resident in 2010, let t = 10 in the model.

M = 171 ⋅ 10 + 1361

———— (1 + 0.018 ⋅ 10)(2960 ⋅ 10 + 278,649)

≈ 0.008443

In 2010, the annual health care expenditure per resident was

about $8443.

38. To fi nd a model M for the annual education expenditure per

student, divide the total amount I by the number of students P.

M = 17,913t + 709,569

—— 1 − 0.028t

÷ (590.6t + 70,219)

= 17,913t + 709,569

—— 1 − 0.028t

⋅ 1 ——

590.6t + 70,219

= 17,913t + 709,569

——— (1 − 0.028t)(590.6t + 70,219)

Because I is in millions of dollars and P is in thousands, M is

in thousands of dollars. To estimate the annual education

expenditure per student in 2009, let t = 8 in the model.

M = 17,913 ⋅ 8 + 709,569

——— (1 − 0.028 ⋅ 8)(590.6 ⋅ 8 + 70,219)

≈ 14.665

In 2009, the annual education expenditure per student was

about $14,665.

39. a. This means the population increases by 2,960,000 people

per year.

b. This means the population was 278,649,000 in 2000.

40. For h(x) the excluded value is x = 1 and for k(x) the

excluded values are x = −4 and x = 1. The value x = 1 is

not in the domain of f and y = 0 when x = −4 in the graph

of g, so x = −4 is not in the domain of 1 —

g .

41. x y

−3.5 −0.1333

−3.8 −0.1282

−3.9 −0.1266

−4.1 −0.1235

−4.2 −0.1220

The graph does not have a value for y when x = −4 and

approaches y = −0.125.


Chapter 7

42. You are correct. The values –9 and 5 make the denominator

equal to zero in the original expression, so both values are

not in the domain.

43. By using the Pythagorean Theorem the unlabeled sides of the

triangle are

√——

(6x) 2 + (8x) 2 = √——

36x 2 + 64x 2 = √—

100x 2 = 10x

and

√——

(8x)2 + (15x) 2 = √——

64x 2 + 225x 2

= √—

289x 2

= 17x, when x > 0.

The perimeter of the triangle is P = 6x + 15x + 17x + 10x =

48x and the area is A = 1 —

2 (8x)(6x + 15x) = 84x 2. So, the

ratio of the perimeter to the area is 48x

— 84x 2

= 4 —

7x .

44. x − 5 ——

x 2 + 2x − 35 ÷ ——

x 2 − 3x − 10

= x − 5 ——

(x + 7)(x − 5) ⋅

(x − 5)(x + 2) ——

= (x − 5)(x − 5)(x + 2)

—— (x + 7)(x − 5)

= x + 2

— x + 7

The missing factor is x − 5.

45. 2x 2 + x − 15

—— 2x 2 − 11x − 21

⋅ (6x + 9) ÷ 2x − 5 —

3x − 21

= 2x 2 + x − 15

—— 2x 2 − 11x − 21

⋅ 6x + 9

— 1 ⋅

3x − 21 —

2x − 5

= (2x − 5)(x + 3)

—— (2x + 3)(x − 7)

⋅ 3(2x + 3)

— 1 ⋅

3(x − 7) —

2x − 5

= 9(2x − 5)(x + 3)(2x + 3)(x − 7)

——— (2x + 3)(2x − 5)(x − 7)

= 9(x + 3), x ≠ − 3 —

2 , x ≠

5 —

2 , x ≠ 7

46. (x 3 + 8) ⋅ x − 2 ——

x 2 − 2x + 4 ÷

x 2 − 4 —

x − 6

= x 3 + 8

— 1 ⋅

x − 2 ——

x 2 − 2x + 4 ⋅

x − 6 —

x 2 − 4

= (x + 2)(x 2 − 2x + 4)

—— 1 ⋅

x − 2 ——

x 2 − 2x + 4 ⋅

x − 6 ——

(x + 2)(x − 2)

= (x + 2)(x 2 − 2x + 4)(x − 2)(x − 6)

——— (x 2 − 2x + 4)(x + 2)(x − 2)

= x − 6, x ≠ −2, x ≠ 2, x ≠ 6

47. The surface area to volume is 2πr(r + h)

— πr 2h

= 2(h + r)

— rh

.

Galapagos Penguin: 2(53 + 6)

— (6)(53)

≈ 0.371

King Penguin: 2(94 + 11)

— (11)(94)

≈ 0.203

The King Penguin has a smaller surface area to volume ratio,

so it is better equipped to live in a colder environment.

48. no; Sample answer: When f (x) = √—

x and g(x) = √—

x 3 ,

(fg)(x) = x 2, x ≥ 0, and ( f — g ) (x) =

1 —

x , x > 0, but both graphs

exclude negative values of x.

49. Let f (x) = x(x − 1)

— x + 2

and g(x) = x(x + 2)

— x − 1

. Then

(fg)(x) = x(x − 1)

— (x + 2)

⋅ x(x + 2)

— (x − 1)

= x 2(x − 1)(x + 2)

—— (x + 2)(x − 1)

= x 2

and

( f — g ) (x) =

x(x − 1) —

(x + 2) ÷

x(x + 2) —

(x − 1)

= x(x − 1)

— (x + 2)

⋅ (x − 1)

— x(x + 2)

= x(x − 1)(x − 1)

—— x(x + 2)(x + 2)

= (x − 1) 2

— (x + 2) 2

.


50. 1 — 2 x + 4 =

3 —

2 x + 5

x + 8 = 3x + 10

−2 = 2x

−1 = x

51. 1 — 3 x − 2 =

3 —

4 x

4x − 24 = 9x

−24 = 5x

− 24

— 5 = x

52. 1 — 4 x −

3 —

5 =

9 —

2 x −

4 —

5

5x − 12 = 90x − 16

4 = 85x

4 — 85

= x

53. 1 — 2 x +

1 —

3 =

3 —

4 x −

1 —

5

30x + 20 = 45x − 12

32 = 15x

32 —

15 = x

54. 42 = 2 ⋅ 3 ⋅ 7 55. 91 = 7 ⋅ 13

56. 72 = 23 ⋅ 32 57. 79 is prime.


Chapter 7


1. a. 1 —

x − 1 +

3 —

x − 1 =

1 + 3 —

x − 1 =

4 —

x − 1

The domain is C because when x = 1, the denominator of

each expression is zero.

b. 1 —

x − 1 +

1 —

x =

x —

x(x − 1) +

x − 1 —

x(x − 1)

= x + (x − 1)

— x(x − 1)

= 2x − 1

— x(x − 1)

The domain is F because when x = 1, the denominator

of the fi rst expression is zero, and when x = 0, the

denominator of the second expression is zero.

c. 1 —

x − 2 +

1 —

2 − x =

1 —

x − 2 −

1 —

x − 2 = 0

The domain is G because when x = 2, the denominator of


d. 1 —

x − 1 +

−1 —

x + 1 =

x + 1 ——

(x − 1 )(x + 1) −

x − 1 ——

(x − 1)(x + 1)

= (x + 1) − (x − 1)

—— (x + 1)(x − 1)

= 2 ——

(x − 1)(x + 1)

The domain is B because when x = 1, the denominator

of the fi rst expression is zero, and when x = −1, the


e. x —

x + 2 −

x + 1 —

2 + x =

x —

x + 2 −

x + 1 —

x + 2 =

x − (x + 1) —

x + 2 =

−1 —

x + 2

The domain is A because when x = −2, the denominator

of each expression is zero.

f. x — x − 2

− x + 1

— x =

x 2 —

x(x − 2) −

(x + 1)(x − 2) ——

x(x − 2)

= x 2 − (x 2 − x − 2)

—— x(x − 2)

= x + 2

— x(x − 2)

The domain is H because when x = 2, the denominator



g. x — x + 2

− x —

x − 1 =

x(x − 1) ——

(x + 2)(x − 1) −

x(x + 2) ——

(x + 2)(x − 1)

= (x 2 − x) − (x 2 + 2x)

—— (x + 2)(x − 1)

= −3x ——

(x + 2)(x − 1)

The domain is E because when x = −2, the denominator



h. x + 2 —

x −

x + 1 —

x =

(x + 2) − (x + 1) ——

x =

1 —

x

The domain is D because when x = 0, the denominator of


2. a. Sample answer: x —

x + 1 −

x + 2 —

x + 1

The denominator of each expression must have a factor of

x + 1 because the domain is all real numbers except −1.

b. Sample answer: x + 3

— x + 1

+ x —

x − 3

The denominators must have the factors x + 1 and x − 3

because the domain is all real numbers except −1 and 3.

c. Sample answer: x − 2

— x(x + 1)

+ x —

x − 3

The denominators must have the factors x + 1, x, and

x − 3 because the domain is all real numbers except −1,

0, and 3.

3. The domain is all real numbers except any value of x that

makes the denominator of any term equal to zero.

4. The numerators and denominators were added. The terms

must be rewritten using a common denominator before

adding the numerators.

x —

x + 4 +

3 —

x − 4 =

x(x − 4) ——

(x + 4)(x − 4) +

3(x + 4) ——

(x + 4)(x − 4)

= x(x − 4) + 3(x + 4)

—— (x + 4)(x − 4)

= x 2 − x + 12

—— (x + 4)(x − 4)


1. 8 —

12x −

5 —

12x =

8 − 5 —

12x =

3 —

12x =

1 —

4x

2. 2 —

3x 2 +

1 —

3x 2 =

2 + 1 —

3x 2 =

3 —

3x 2 =

1 —

x 2

3. 4x — x − 2

− x —

x − 2 =

4x − x —

x − 2 =

3x —

x − 2

4. 2x 2 —

x 2 + 1 +

2 —

x 2 + 1 =

2x 2 + 2 —

x 2 + 1 =

2(x 2 + 1) —

x 2 + 1 = 2

5. Step 1 Factor each polynomial.

5x 3

10x 2 − 15x = 5x(2x − 3)

Step 2 The LCM is the product of the highest power of each

factor that appears in either polynomial.

LCM = 5x 3(2x − 3)

6. 3 — 4x

− 1 —

7 =

21 —

28x −

4x —

28x =

21 − 4x —

28x


Chapter 7

7. 1 —

3x 2 +

x —

9x 2 − 12 =

1 —

3x 2 +

x —

3(3x 2 − 4)

= 3x 2 − 4

—— 3x 2(3x 2 − 4)

+ x3

—— 3x 2(3x 2 − 4)

= (3x 2 − 4) + x 3

—— 3x 2(3x 2 − 4)

= x 3 + 3x 2 − 4

—— 3x 2(3x 2 − 4)

= (x − 1)(x + 2) 2

—— 3x 2(3x 2 − 4)

8. x —— x 2 − x − 12

+ 5 —

12x − 48

= x ——

(x − 4)(x + 3) +

5 —

12(x − 4)

= 12x ——

12(x − 4)(x + 3) +

5(x + 3) ——

12(x − 4)(x + 3)

= 12x + 5(x + 3)

—— 12(x − 4)(x + 3)

= 17x + 15

—— 12(x − 4)(x + 3)

9. Rewrite by inspection:

2x − 4

— x − 3

= 2x − 6 + 2

— x − 3

= 2x − 6

— x − 3

+ 2 —

x − 3 =

2 —

x − 3 + 2

y

4 6 8−2

4

6

x

−2

g


up of the graph of f (x) = 2 —

x .

10. x —

6 −

x —

3

— x —

5 −

7 —

10

=

2x

— 12

− 4x

— 12

— 2x

— 10

− 7 —

10

=

−2x

— 12

— 2x − 7

— 10

= −2x

— 12

⋅ 10 —

2x − 7

= −20x

— 12(2x − 7)

= −5x —

3(2x − 7)

11. 2 —

x − 4

— 2 —

x + 3

=

2 —

x −

4x —

x —

2 —

x +

3x —

x

=

2 − 4x

— x

— 2 + 3x

— x

= 2 − 4x

— x ⋅

x —

2 + 3x

= 2 − 4x

— 2 + 3x

= 2(1 − 2x)

— 2 + 3x

, x ≠ 0

12.

3 —

x + 5

——

2 —

x − 3 +

1 —

x + 5

=

3 —

x + 5

———

2(x + 5) ——

(x − 3)(x + 5) +

x − 3 ——

(x − 3)(x + 5)

=

3 —

x + 5

—— 2(x + 5) + (x − 3)

—— (x − 3)(x + 5)

=

3 —

x + 5

——

3x + 7 ——

(x − 3)(x + 5)

= 3 —

x + 5 ⋅

(x − 3)(x + 5) ——

3x + 7

= 3(x − 3)(x + 5)

—— (x + 5)(3x + 7)

= 3(x − 3)

— 3x + 7

, x ≠ −5, x ≠ 3



1. A fraction that contains a fraction in its numerator or

denominator is called a complex fraction.

2. Both expressions require a common denominator before

adding or subtracting the numerators.


3. 15 —

4x +

5 —

4x =

15 + 5 —

4x =

20 —

4x =

5 —

x

4. x — 16x 2

− 4 —

16x 2 =

x − 4 —

16x 2

5. 9 —

x + 1 −

2x —

x + 1 =

9 − 2x —

x + 1

6. 3x 2 —

x − 8 +

6x —

x − 8 =

3x 2 + 6x —

x − 8 =

3x(x + 2) —

x − 8

7. 5x — x + 3

+ 15 —

x + 3 =

5x + 15 —

x + 3 =

5(x + 3) —

x + 3 = 5, x ≠ −3


Chapter 7

8. 4x 2 —

2x − 1 −

1 —

2x − 1 =

4x 2 − 1 —

2x − 1 =

(2x − 1)(2x + 1) ——

2x − 1

= 2x + 1, x ≠ 1 —

2

9. The LCM is the product of the highest power of each factor

that appears in either polynomial.

LCM = 3x(x − 2)

10. Step 1 Factor each polynomial. Write numerical factors as

products of primes.

2x 2

4x + 12 = 2 2(x + 3)



LCM = 2 2x 2(x + 3) = 4x 2(x + 3)

11. The LCM is the product of the highest power of each factor

that appears in either polynomial.

LCM = 2x(x − 5)

12. Step 1 Factor each polynomial. Write numerical factors as

products of primes.

24x 2 = 2 3 ⋅ 3x 2

8x 2 − 16x = 2 3x(x − 2)



LCM = 2 3 ⋅ 3x 2(x − 2) = 24x 2(x − 2)


x 2 − 25 = (x + 5)(x − 5)

x − 5



LCM = (x + 5)(x − 5)


9x 2 − 16 = (3x + 4)(3x − 4)

3x 2 + x − 4 = (x − 1)(3x + 4)



LCM = (3x + 4)(3x − 4)(x − 1)


x 2 + 3x − 40 = (x + 8)(x − 5)

x − 8



LCM = (x − 5)(x − 8)(x + 8)


x 2 − 2x − 63 = (x − 9)(x + 7)

x + 7



LCM = (x − 9)(x + 7)

17. The error is that the LCM of 5x and x 2 is 5x 2, so multiply the

fi rst term by x —

x and the second term by

5 —

5 before adding the

numerators.

2 —

5x +

4 —

x 2 =

2x —

5x 2 +

20 —

5x 2 =

2x + 20 —

5x 2 =

2(x + 10) —

5x2

18. The error is that the LCM is (x + 2)(x − 5), so multiply x by

x − 5 and 4 by x + 2.

x —

x + 2 +

4 —

x − 5 =

x(x − 5) ——

(x + 2)(x − 5) +

4(x + 2) ——

(x + 2)(x − 5)

= x(x − 5) + 4(x + 2)

—— (x + 2)(x − 5)

= x 2 − x + 8

—— (x + 2)(x − 5)

19. 12 —

5x −

7 —

6x =

72 —

30x −

35 —

30x

= 37

— 30x

20. 8 —

3x2 +

5 —

4x =

32 —

12x2 +

15x —

12x2

= 15x + 32

— 12x2

21. 3 —

x + 4 −

1 —

x + 6 =

3(x + 6) ——

(x + 4)(x + 6) −

x + 4 ——

(x + 4)(x + 6)

= 3(x + 6) − (x + 4)

—— (x + 4)(x + 6)

= 2(x + 7)

—— (x + 4)(x + 6)

22. 9 —

x − 3 +

2x —

x + 1 =

9(x + 1) ——

(x − 3)(x + 1) +

2x(x − 3) ——

(x − 3)(x + 1)

= 9(x + 1) + 2x(x − 3)

—— (x − 3)(x + 1)

= 2x2 + 3x + 9

—— (x − 3)(x + 1)

23. 12 ——

x2 + 5x − 24 +

3 —

x − 3 =

12 ——

(x + 8)(x − 3) +

3 —

x − 3

= 12 ——

(x + 8)(x − 3) +

3(x + 8) ——

(x + 8)(x − 3)

= 12 + 3(x + 8)

—— (x + 8)(x − 3)

= 3(x + 12)

—— (x + 8)(x − 3)


Chapter 7

24. x2 − 5 ——

x2 + 5x − 14 −

x + 3 —

x + 7 =

x2 − 5 ——

(x − 2)(x + 7) −

x + 3 —

x + 7

= x2 − 5 ——

(x − 2)(x + 7) −

(x + 3)(x − 2) ——

(x − 2)(x + 7)

= (x2 − 5) − (x + 3)(x − 2)

——— (x − 2)(x + 7)

= −x + 1

—— (x − 2)(x + 7)

25. x + 2 —

x − 4 +

2 —

x +

5x —

3x − 1 =

x(3x − 1)(x + 2) ——

x(3x − 1)(x − 4)

+ 2(3x − 1)(x − 4)

—— x(3x − 1)(x − 4)

+ 5x2(x − 4)

—— x(3x − 1)(x − 4)

= x(3x − 1)(x + 2) + 2(3x − 1)(x − 4) + 5x2(x − 4)

————— x(3x − 1)(x − 4)

= 3x3 + 5x2 − 2x + 6x2 − 26x + 8 + 5x3 − 20x2

———— x(3x − 1)(x − 4)

= 8x3 − 9x2 − 28x + 8

—— x(3x − 1)(x − 4)

26. x + 3 —

x2 − 25 −

x − 1 —

x − 5 +

3 —

x + 3 =

x + 3 ——

(x − 5)(x + 5) −

x − 1 —

x − 5 +

3 —

x + 3

= (x + 3)(x + 3)

—— (x − 5)(x + 5)(x + 3)

− (x − 1)(x + 3)(x + 5)

—— (x − 5)(x + 5)(x + 3)

+ 3(x − 5)(x + 5)

—— (x − 5)(x + 5)(x + 3)

= (x + 3)(x + 3) − (x − 1)(x + 3)(x + 5) + 3(x − 5)(x + 5)

————— (x − 5)(x + 5)(x + 3)

= x2 + 6x + 9 − (x3 + 7x2 + 7x − 15) + 3x2 − 75

———— (x − 5)(x + 5)(x + 3)

= −x3 − 3x2 − x − 51

—— (x − 5)(x + 5)(x + 3)

27. The statement is sometimes true. When the denominators

have no common factors, the product of the denominators

is the LCD. When the denominators have common factors,

use the LCM to fi nd the LCD.

28. The statement is always true. The LCD is the product of

the highest power of each factor that appears in any of the

denominators.

29. You would begin with A.

30. You would begin with B.


5x − 7

— x − 1

= 5x − 5 − 2

— x − 1

= 5(x − 1) − 2

—— x − 1

= 5(x − 1)

— x − 1

+ −2

— x − 1

= −2

— x − 1

+ 5

y

42 6−2−4

4

2

8

6

x

10

g

The graph of g is a translation 1 unit to the right and 5 units

up of the graph of f (x) = −2

— x .


6x + 4

— x + 5

= 6x + 30 − 26

—— x + 5

= 6(x + 5) − 26

—— x + 5

= 6(x + 5)

— x + 5

+ −26

— x + 5

= −26

— x + 5

+ 6

y

84−4−8

8

4

x

12

16

−12−16

−4

g



— x .


Chapter 7


12x —

x − 5 =

12x − 60 + 60 ——

x − 5 =

12(x − 5) + 60 ——

x − 5

= 12(x − 5)

— x − 5

+ 60 —

x − 5 =

60 —

x − 5 + 12

y

8 12 16 204−4−8

8

4

x

16

20

24

−4

g


up of the graph of f (x) = 60

— x .


8x —

x + 13 =

8x + 104 − 104 ——

x + 13 =

8(x + 13) − 104 ——

x + 13

= 8(x + 13)

— x + 13

+ −104

— x + 13

= −104

— x + 13

+ 8

8−8 x

−8

y

16

−16−24−32

24

g



— x .


2x + 3

— x =

2x —

x +

3 —

x =

3 —

x + 2

y

84−4−8

8

4

x

−4

g

The graph of g is a translation 2 units up of the graph of

f (x) = 3 —

x .


4x − 6

— x =

4x —

x +

−6 —

x =

−6 —

x + 4

y

84−4−8

8

x

−4

g

The graph of g is a translation 4 units up of the graph of

f (x) = −6

— x .


3x + 11

— x − 3

= 3x − 9 + 20

—— x − 3

= 3(x − 3) + 20

—— x − 3

= 3(x − 3)

— x − 3

+ 20 —

x − 3 =

20 —

x − 3 + 3

y

8 124−4−8

8

4

x

−4

−8

12

g



— x .


Chapter 7


7x − 9

— x + 10

= 7x + 70 − 79

—— x + 10

= 7(x + 10) − 79

—— x + 10

= 7(x + 10)

— x + 10

+ −79

— x + 10

= −79

— x + 10

+ 7

8−8 x

−8

y

16

−16−24

24

8

g



— x .

39. x — 3 − 6

— 10 +

4 —

x =

x —

3 −

18 —

3

— 10x

— x +

4 —

x

=

x − 18

— 3

— 10x + 4

— x

= x − 18

— 3 ⋅

x —

10x + 4

= x(x − 18)

— 6(5x + 2)

, x ≠ 0

40. 15 −

2 —

x —

x —

5 + 4

=

15x

— x −

2 —

x —

x —

5 +

20 —

5

=

15x − 2

— x

— x + 20

— 5

= 15x − 2

— x ⋅

5 —

x + 20

= 5(15x − 2)

— x(x + 20)

41.

1 —

2x − 5 −

7 —

8x − 20

——

x —

2x − 5 =

4 —

4(2x − 5) −

7 —

4(2x − 5)

——

x —

2x − 5

=

4 − 7

— 4(2x − 5)

—

x —

2x − 5

=

−3 —

4(2x − 5)

—

x —

2x − 5

= −3 —

4(2x − 5) ⋅

2x − 5 —

x

= −3(2x − 5)

— 4x(2x − 5)

= −3

— 4x

, x ≠ 5 —

2

42.

16 —

x − 2

—

4 —

x + 1 +

6 —

x =

16 —

x − 2

——

4x —

x(x + 1) +

6(x + 1) —

x(x + 1)

=

16 —

x − 2

—— 4x + 6(x + 1)

—— x(x + 1)

=

16 —

x − 2

— 10x + 6

— x(x + 1)

= 16 —

x − 2 ⋅

x(x + 1) —

10x + 6

= 16x(x + 1)

—— 2(x − 2)(5x + 3)

= 8x(x + 1)

—— (x − 2)(5x + 3)

, x ≠ −1, x ≠ 0


Chapter 7

43.

1 —

3x2 − 3

——

5 —

x + 1 −

x + 4 —

x2 − 3x − 4

=

1 ——

3(x − 1)(x + 1)

——

5 —

x + 1 −

x + 4 ——

(x + 1)(x − 4)

=

1 ——

3(x − 1)(x + 1)

———

5(x − 4) ——

(x + 1)(x − 4) −

x + 4 ——

(x + 1)(x − 4)

=

1 ——

3(x − 1)(x + 1)

—— 5(x − 4) − (x + 4)

—— (x + 1)(x − 4)

=

1 ——

3(x − 1)(x + 1)

——

4x − 24 ——

(x + 1)(x − 4)

= 1 ——

3(x − 1)(x + 1) ⋅

(x + 1)(x − 4) ——

4(x − 6)

= (x + 1)(x − 4)

—— 12(x − 1)(x + 1)(x − 6)

= x − 4 ——

12(x − 1)(x − 6) , x ≠ −1, x ≠ 4

44.

3 —

x − 2 −

6 —

x2 − 4

——

3 —

x + 2 +

1 —

x − 2

=

3(x + 2)

—— (x − 2)(x + 2)

− 6 ——

(x − 2)(x + 2)

———

3(x − 2) ——

(x − 2)(x + 2) +

x + 2 ——

(x − 2)(x + 2)

=

3(x + 2) − 6

—— (x − 2)(x + 2)

—— 3(x − 2) + (x + 2)

—— (x − 2)(x + 2)

=

3x ——

(x − 2)(x + 2)

——

4x − 4 ——

(x − 2)(x + 2)

= 3x ——

(x − 2)(x + 2) ⋅ (x − 2)(x + 2)

—— 4(x − 1)

= 3x(x − 2)(x + 2)

—— 4(x − 1)(x − 2)(x + 2)

= 3x —

4(x − 1) , x ≠ 2, x ≠ −2

45. T = d —

a − j +

d —

a + j

= d(a + j)

—— (a − j)(a + j)

+ d(a − j)

—— (a − j)(a + j)

= d(a + j) + d(a − j)

—— (a − j)(a + j)

= 2ad ——

(a − j)(a + j)

When d = 2468, a = 510 and j = 115,

T = 2(510)(2468)

——— (510 − 115)(510 + 115)

≈ 10.2.

So, the travel time is about 10.2 hours.

46. Rt = 1 —

1 —

R1

+ 1 —

R2

=

1 —

1

——

R2 — R1R2

+ R1 —

R1R2

=

1 —

1

— R2 + R1 —

R1R2

= 1 —

1 ⋅

R1R2 — R1 + R2

= R1R2 —

R1 + R2

, R1 ≠ 0, R2 ≠ 0

When R1 = 2000 and R2 = 5600,

Rt = (2000)(5600)

—— 2000 + 5600

≈ 1474. So, the resistance is about

1474 ohms.

47. y = y1 + y2

= 40

— x +

100 —

x + 30

= 40(x + 30)

— x(x + 30)

+ 100x —

x(x + 30)

= 40(x + 30) + 100x

—— x(x + 30)

= 140x + 1200

—— x(x + 30)

= 20(7x + 60)

—— x(x + 30)

48. a. t = 0.5

— r +

22 —

15r +

6 —

r + 5

= 0.5(15)(r + 5)

—— 15r(r + 5)

+ 22(r + 5)

— 15r(r + 5)

+ 6(15r)

— 15r(r + 5)

= 119.5r + 147.5

—— 15r(r + 5)

b. When r = 2, t = 119.5(2) + 147.5

—— 15(2)(2 + 5)

≈ 1.84. So, the

amount of time is about 1 hour and 50 minutes.


Chapter 7

49. You friend is incorrect because the LCM of 2 and 4 is 4,

which is greater than one number and equal to the other

number.

50. The asymptotes are y = −4 and x = −1, so h = −1 and

k = −4.

51. a. M = Pi ——

1 − ( 1 —

1 + i )

12t

= Pi ——

1 − 1 —

(1 + i)12t ⋅

(1 + i)12t —

(1 + i)12t

= Pi(1 + i)12t

—— (1 + i)12t − 1

b. When P = 15,500, i = 0.005 and t = 4,

M = (15,500)(0.005)12 ⋅ 4

—— (1 + 0.005)12 ⋅ 4 − 1

≈ 364.02.

So, the monthly payment is $364.02.

52. It is possible to write two rational functions whose sum is a

quadratic function. Sample answer: x3 − 1

— x +

1 —

x = x2, x ≠ 0.

53. g(x) = (97.6)(0.024) + x(0.003)

——— 12.2 + x

= 2.3058

— 12.2 + x

+ 0.003

The graph of g is a translation 12.2 units to the left and

0.003 unit up of the graph of f (x) = 2.3058

— x .

54. A = 2 ( x + 5 —

x ⋅

x —

x + 1 ) + 2 ( x + 5

— x ⋅

x + 1 —

3 )

+ 2 ( x —

x + 1 ⋅

x + 1 —

3 )

= 2 ( x + 5 —

x + 1 ) + 2 ( (x + 5)(x + 1)

—— 3x

) + 2 ( x — 3 )

= 2 ( x + 5 —

x + 1 +

(x + 5)(x + 1) ——

3x +

x —

3 )

= 2 ( 3x(x + 5) —

3x(x + 1) +

(x + 5)(x + 1)2

—— 3x(x + 1)

+ x2(x + 1)

— 3x(x + 1)

) = 2 ( 3x(x + 5) + (x + 5)(x + 1)2 + x2(x + 1)

———— 3x(x + 1)

) =

2(2x3 + 11x2 + 26x + 5) ———

3x(x + 1)

55. a. Your rate is R1 = 1 —

40 , the second employee’s rate is R2 =

1 —

x ,

and the third employee’s rate is R3 = 1 —

x + 10 .

b. The combined rate is

R = R1 + R2 + R3

= 1 —

40 +

1 —

x +

1 —

x + 10

= x(x + 10)

— 40x(x + 10)

+ 40(x + 10)

— 40x(x + 10)

+ 40x —

40x(x + 10)

= x2 + 90x + 400

—— 40x(x + 10)

c. When x = 35, R = 352 + 90(35) + 400

—— 40(35)(35 + 10)

≈ 0.0758.

So, the combined rate is about 0.0758 cars per minute, or

about 4.5 cars per hour. This was obtained by multiplying

the number of cars per minute by the rate 60 minutes per

hour to obtain an answer in cars per hour.

56. a. The model for the second dose is

A = 391(t − 1)2 + 0.112

——— 0.218(t − 1)4 + 0.991(t − 1)2 + 1

.

b. The total amount of aspirin in the bloodstream after the

second dose is taken can be modeled by

A = 391t2 + 0.112

—— 0.218t4 + 0.991t2 + 1

+ 391(t − 1)2 + 0.112

——— 0.218(t − 1)4 + 0.991(t − 1)2 + 1

57. The next two expressions are

1 + 1 ——

2 + 1 ——

2 + 1 —

2 + 1 —

2 + 1 —

2

and

1 + 1 ———

2 + 1 ——

2 + 1 ——

2 + 1 —

2 + 1 —

2 + 1 —

2

The values of the expressions are 1.4, about 1.4167,

about 1.4138, about 1.4143, and about 1.4142, and they

approach √—

2 .


58.

x

y

4

8

10

−4 −2 42

The graphs intersect at (1, 7) and (2, 10). So, the solutions

are (1, 7) and (2, 10).


Chapter 7

59. Solve for y in each equation.

y = 2x2 − 3x

y = 5 —

2 x −

9 —

4

x

y4

−2

−4 −2 42

The two graphs intersect at ( 1 — 2 , −1 ) and ( 9 —

4 ,

27 —

8 ) .

So, the solutions are ( 1 — 2 , −1 ) and ( 9 —

4 ,

27 —

8 ) .

60. Solve for y in each equation

y = x2 + x + 3

y = −x2 − 3x − 5

x

y4

2

−2

−4 −2 42

The graphs do not intersect. So, the system has no solution.

61.

x

y

2

−2−4−6

The graphs do not intersect. So, the system has no solution.


1. a. The graph is F. The graphed equations are y = 2 —

x − 1 and

y = 1. The solution is x = 3.

b. The graph is D. The graphed equations are y = 2 —

x − 2 and

y = 2. The solution is x = 3.

c. The graph is A. The graphed equations are y = −x − 1

— x − 3

and y = x + 1. The solutions are x = −1 and x = 2.

d. The graph is C. The graphed equations are y = 2 —

x − 1 and

y = x. The solutions are x = −1 and x = 2.

e. The graph is B. The graphed equations are y = 1 —

x and

y = −1

— x − 2

. The solution is x = 1.

f. The graph is E. The graphed equations are y = 1 —

x and

y = x2. The solution is x = 1.

2. a. x 2 —

x − 1 x

−2 −0.6667 −2

−1 −1 −1

0 −2 0

1 undef. 1

2 2 2

3 1 3

When x = −1 and x = 2, the expressions are equal to

each other. So, the solutions are x = −1 and x = 2.

x 1 —

x −1 —

x − 2

−2 −0.5 0.25

−1 −1 0.33333

0 undef. 0.5

1 1 1

2 0.5 undef.

3 0.33333 −1

When x = 1, the expressions are equal to each other. So,

the solution is x = 1.

b. 2 —

x − 1 = x

2 = x(x − 1)

0 = x 2 − x − 2

0 = (x − 2)(x + 1)

x = 2 or x = −1

So, the solutions are x = −1 and x = 2.

1 —

x =

−1 —

x − 2

x − 2 = x( −1)

2x = 2

x = 1

So, the solution is x = 1.

3. To solve a rational equation when each side of the equation

is a single rational expression, use cross multiplication.

To solve any rational equation, multiply each side of the

equation by the least common denominator of the rational

expressions.


Chapter 7

4. a. x + 1

— x − 1

= x − 1

— x + 1

(x + 1)2 = (x − 1)2

x 2 + 2x + 1 = x 2 − 2x + 1

2x = −2x

4x = 0

x = 0

So, the solution is x = 0.

b. 1 —

x + 1 =

1 —

x 2 + 1

x 2 + 1 = x + 1

x 2 = x

x 2 − x = 0

x(x − 1) = 0

x = 0 or x = 1

So, the solutions are x = 0 and x = 1.

c. 1 —

x 2 − 1 =

1 —

x − 1

x − 1 = x2 − 1

x = x2

0 = x2 − x

0 = x(x − 1)

x = 0 or x = 1

x = 1 is not in the domain of either expression. So, the

solution is x = 0.


1. 3 —

5x =

2 —

x − 7 Check

3(x − 7) = 2(5x) 3 —

5(−3) =

?

2 —

(−3) − 7

3x − 21 = 10x − 3 —

15 =

? −

2 —

10

−21 = 7x − 1 —

5 = −

1 —

5 ✓

−3 = x

The solution is x = −3.

2. −4

— x + 3

= 5 —

x − 3 Check

−4(x − 3) = 5(x + 3) −4 —

( − 1 —

3 ) + 3

=?

5 —

( − 1 —

3 ) − 3

−4x + 12 = 5x + 15

−9x = 3 −4

— 8 —

3

=?

5 —

− 10 —

3

x = − 1 —

3

− 3 —

2 = − 3 —

2 ✓

The solution is x = − 1 —

3 .

3. 1 —

2x + 5 =

x —

11x + 8 Check

11x + 8 = x(2x + 5) 1 —

2(4) + 5 =

?

4 —

11(4) + 8

11x + 8 = 2x2 + 5x 1 —

13 =

?

4 —

52

0 = 2x2 − 6x − 8 1 —

13 =

1 —

13 ✓

0 = 2(x − 4)(x + 1) 1 —

2(−1) + 5 =

?

−1 —

11(−1) + 8

x = 4 or x = −1 1 —

3 =

? −1

— −3

1 — 3 = 1 —

3 ✓

The solutions are x = 4 and x = −1.

4. 15

— x +

4 —

5 =

7 —

x Check

5x ( 15 —

x +

4 —

5 ) = 5x ( 7 —

x )

15 —

−10 +

4 —

5 =

?

7 —

−10

75 + 4x = 35 15

— −10

+ −8

— −10

=?

7 —

−10

4x = −40 7 —

−10 =

7 —

−10 ✓

x = −10

The solution is x = −10.

5. 3x —

x + 1 −

5 —

2x =

3 —

2x

2x(x + 1) ( 3x —

x + 1 −

5 —

2x ) = 2x(x + 1) ( 3 —

2x )

6x 2 − 5(x + 1) = 3(x + 1)

6x 2 − 5x − 5 = 3x + 3

6x 2 − 8x − 8 = 0

2(x − 2)(3x + 2) = 0

x = 2 or x = − 2 —

3

Check

3(2)

— 2 + 1

− 5 —

2(2) =

?

3 —

2(2)

2 − 5 —

4 =

? 3 —

4

3 —

4 =

3 —

4 ✓

3 ( −

2 —

3 ) —

− 2 —

3 + 1

− 5 —

2 ( − 2 —

3 ) =

?

3 —

2 ( − 2 —

3 )

−6 + 15

— 4 =

? −

9 —

4

− 9 —

4 = −

9 —

4 ✓

The solutions are x = 2 or x = − 2 —

3 .


Chapter 7

6. 4x + 1

— x + 1

= 12 —

x 2 − 1 + 3

(x 2 − 1) ( 4x + 1 —

x + 1 ) = (x 2 − 1) ( 12

— x 2 − 1

+ 3 ) (4x + 1)(x − 1) = 12 + 3(x 2 − 1)

4x2 − 3x − 1 = 3x 2 + 9

x 2 − 3x − 10 = 0

(x + 2)(x − 5) = 0

x = −2 or x = 5

Check

4(−2) + 1

— −2 + 1

=?

12 —

(−2)2 − 1 + 3

7 =?

4 + 3

7 = 7 ✓

4(5) + 1

— 5 + 1

=?

12 —

5 2 − 1 + 3

7 —

2 =

? 1 —

2 + 3

7 —

2 =

7 —

2 ✓

The solutions are x = −2 and x = 5.

7. 9 —

x − 2 +

6x —

x + 2 =

9x 2 —

x 2 − 4

( x 2 − 4 ) ( 9 —

x − 2 +

6x —

x + 2 ) = ( x 2 − 4 )

9x 2 —

x 2 − 4

9(x + 2) + 6x(x − 2) = 9x 2

6x 2 − 3x + 18 = 9x 2

0 = 3x 2 + 3x − 18

0 = 3 ( x 2 + x − 6 )

0 = 3(x + 3)(x − 2)

x = −3 or x = 2

Check

9 —

−3 − 2 +

6(−3) —

−3 + 2 =

?

9(−3)2

— (−3)2 − 4

− 9 —

5 + 18 =

? 81

— 5

81

— 5 =

81 —

5 ✓

4 —

2 − 2 +

6(2) —

2 + 2 =

?

9(2)2

— 2 2 − 4

4 —

0 + 3 =

? 36

— 0

Division by zero is undefi ned.

The apparent solution x = 2 is extraneous. So, the only

solution is x = −3.

8. 7 —

x − 1 − 5 =

6 —

x 2 − 1

(x 2 − 1) ( 7 —

x − 1 − 5 ) = ( x 2 − 1 ) ( 6

— x 2 − 1

) 7(x + 1) − 5 ( x2 − 1 ) = 6

−5x 2 + 7x + 12 = 6

0 = 5x 2 − 7x − 6

0 = (x − 2)(5x + 3)

x = 2 or x = − 3 —

5

Check

7 —

2 − 1 − 5 =

?

6 —

2 2 − 1

7 − 5 =?

2

2 = 2 ✓

7 —

( − 3 —

5 ) − 1

− 5 =?

6 —

( − 3 —

5 )

2

− 1

− 35

— 8 − 5 =

? − 75 —

8

− 75

— 8 = −

75 —

8 ✓

The solutions are x = 2 and x = − 3 —

5 .

9. Graph the function f.

x

y4

2

−2

−4 −2 42

Notice that no horizontal line intersects the graph more than

once. So, the inverse of f is a function. Find the inverse.

y = 1 —

x − 2

x = 1 —

y − 2

x + 2 = 1 —

y

y(x + 2) = 1

y = 1 —

x + 2

So, the inverse of f is g(x) = 1 —

x + 2 .


Chapter 7

10. Substitute 90 for c and solve by cross multiplying.

90 = 50m + 800

— m

90m = 50m + 800

40m = 800

m = 20

The average cost falls to $90 per model after 20 models

are printed.



1. You can solve a rational equation by cross multiplying when

each side of the equation is a single rational expression. Each

side of the equation must contain one single polynomial

fraction for cross multiplication to be used.

2. The apparent solution x = 3 is extraneous because

when x = 3 is substituted into the original equation, the

denominators are equal to zero.


3. 4 —

2x =

5 —

x + 6 Check

4(x + 6) = 5(2x) 4 —

2(4) =

?

5 —

4 + 6

4x + 24 = 10x 4 —

8 =

?

5 —

10

24 = 6x 1 —

2 =

1 —

2 ✓

4 = x

The solution is x = 4.

4. 9 —

3x =

4 —

x + 2 Check

9(x + 2) = 4(3x) 9 —

3(6) =

?

4 —

6 + 2

9x + 18 = 12x 9 —

18 =

? 4 —

8

18 = 3x 1 —

2 =

1 —

2 ✓

6 = x


5. 6 —

x − 1 =

9 —

x + 1 Check

6(x + 1) = 9(x − 1) 6 —

5 − 1 =

?

9 —

5 + 1

6x + 6 = 9x − 9 6 —

4 =

? 9 —

6

15 = 3x 3 —

2 =

? 3 —

2 ✓

5 = x


6. 8 —

3x − 2 =

2 —

x − 1 Check

8(x − 1) = 2(3x − 2) 8 —

3(2)−2 =

?

2 —

2−1

8x − 8 = 6x − 4 8 —

4 =

? 2

— 1

2x = 4 2 = 2 ✓

x = 2

The solution is x = 2

7. x —

2x + 7 =

x − 5 —

x − 1 Check

x (x − 1) = (2x + 7)(x − 5) 7 —

2(7) + 7 =

? 7 − 5

— 7 − 1

x 2 − x = 2x 2 − 3x − 35 7 —

21 =

? 2 —

6

0 = x 2 − 2x − 35 1 —

3 =

1 —

3 ✓

0 = (x − 7)(x + 5) −5 —

2(−5) + 7 =

? −5 − 5

— −5 − 1

x = 7 or x = −5 −5

— −3

=?

−10

— −6

5 —

3 =

5 —

3 ✓


8. −2

— x − 1

= x − 8

— x + 1

Check

−2(x + 1) = (x − 1)(x − 8) −2

— 5 − 1

=?

5 − 8

— 5 + 1

−2x − 2 = x 2 − 9x + 8 −2

— 4 =

? −3

— −6

0 = x 2 − 7x + 10 − 1 —

2 = −

1 —

2 ✓

0 = (x − 5)(x − 2) −2

— 2 − 1

=?

2 − 8

— 2 + 1

x =5 or x = 2 −2

— 1 =

? −6

— 3

−2 = −2 ✓

The solutions are x = 5 or x = 2.

9. x2 − 3

— x + 2

= x − 3

— 2 Check

2 ( x 2 − 3 ) = (x + 2)(x − 3) 02 − 3

— 0 + 2

=?

0 − 3

— 2

2x 2 − 6 = x 2 − x − 6 −3

— 2 =

−3 —

2 ✓

x 2 + x = 0 (−1)2 − 3

— −1 + 2

=?

−1 − 3

— 2

x(x + 1) = 0 −2

— 1 =

? −4

— 2

x = 0 or x = −1 −2 = −2 ✓



Chapter 7

10. −1

— x − 3

= x − 4

— x2 − 27

− ( x2 − 27 ) = (x − 3)(x − 4)

−x2 + 27 = x2 − 7x + 12

0 = 2x2 − 7x − 15

0 = (x − 5)(2x + 3)

x = 5 or x = − 3 —

2

Check

−1

— 5 − 3

=?

5 − 4

— 52 − 27

− 1 —

2 =−

1 —

2 ✓

−1 —

( − 3 —

2 ) − 3

=?

− 3 —

2 − 4

——

( − 3 —

2 )

2

− 27

−1

—

− 7 —

2

=?

−

11 —

2 —

− 99

— 4

2 —

9 =

2 —

9 ✓


2 .

11. 90

— 100

= 37 + x

— 44 + x

90(44 + x) = 100(37 + x)

3960 + 90x = 3700 + 100x

260 = 10x

26 = x

You need to put 26 consecutive serves into play.

12. 0.360 = 12 + x

— 60 + x

0.360(60 + x) = 12 + x

21.5 + 0.360x = 12 + x

9.6 = 0.64x

15 = x

You need 15 consecutive hits.

13. percent of copper in mixture = weight of copper in mixture

——— total weight of mixture

55

— 100

= 25 —

25 + x

55(25 + x) = 100 ⋅ 25

1375 + 55x = 2500

55x = 1125

x ≈ 20.5

You need 20.5 ounces of zinc.

14. 12 = 16 ⋅ 0.2

— 0.2 + x

12(0.2 + x) = 3.2

2.4 + 12x = 3.2

12x = 0.8

x ≈ 0.07

You should add about 0.07 liters of water.

15. The LCD is x(x + 3).

16. The LCD is x(x − 1).

17. The LCD is 2(x + 1)(x + 4).

18. The LCD is 3(x + 9)(2x − 1).

19. 3 —

2 +

1 —

x = 2 Check

2x ( 3 — 2 +

1 —

x ) = 2x(2)

3 —

2 +

1 —

2 =

? 2

3x + 2 = 4x 2 = 2 ✓

2 = x


20. 2 —

3x +

1 —

6 =

4 —

3x Check

6x ( 2 — 3x

+ 1 —

6 ) = 6x ( 4 —

3x )

2 —

3(4) +

1 —

6 =

?

4 —

3(4)

4 + x = 8 1 —

6 +

1 —

6 =

? 1 —

3

x = 4 1 —

3 =

1 —

3 ✓


21. x − 3

— x − 4

+ 4 = 3x

— x

x − 3

— x − 4

+ 4 = 3

(x − 4) ( x − 3 —

x − 4 + 4 ) = (x − 4)(3)

(x − 3) + 4(x − 4) = 3(x − 4)

5x − 19 = 3x − 12

2x = 7

x = 7 —

2

Check

7 —

2 − 3

— 7 —

2 − 4

+ 4 =?

3 ( 7 — 2 ) —

7 —

2

−1 + 4 =?

3

3 = 3 ✓

The solution is x = 7 —

2 .


Chapter 7

22. 2 —

x − 3 +

1 —

x =

x − 1 —

x − 3

x(x − 3) ( 2 —

x − 3 +

1 —

x ) = x(x − 3) ( x − 1

— x − 3

) 2x + (x − 3) = x(x − 1)

3x − 3 = x2 − x

0 = x2 − 4x + 3

0 = (x − 3)(x − 1)

x = 3 or x = 1

Check

2 —

3 − 3 +

1 —

3 =

? 3 − 1

— 3 − 3

2 —

0 +

1 —

3 =

? 2 —

0


2 —

1 − 3 +

1 —

1 =

? 1 − 1

— 1 − 3

−1 + 1 =?

0

0 = 0 ✓

The apparent solution x = 3 is extraneous. So, the only

solution is x = 1.

23. 6x —

x + 4 + 4 =

2x + 2 —

x − 1

(x + 4)(x − 1) ( 6x —

x + 4 + 4 ) = (x + 4)(x − 1) ( 2x + 2

— x − 1

) 6x(x − 1) + 4(x + 4)(x − 1) = (x + 4)(2x + 2)

10x2 + 6x − 16 = 2x2 + 10x + 8

8x2 − 4x − 24 = 0

4(x − 2)(2x + 3) = 0

x = 2 or x = − 3 —

2

Check

6(2)

— 2 + 4

+ 4 =?

2(2) + 2

— 2 − 1

2 + 4 =?

6

6 = 6 ✓

6

( − 3 —

2 ) —

− 3 —

2 + 4

+ 4 =?

2 ( − 3 —

2 ) + 2

—

− 3 —

2 − 1

−18

— 5 + 4 =

? 2 —

5

2 —

5 =

2 —

5 ✓


2 .

24. 10

— x + 3 =

x + 9 —

x − 4

x(x − 4) ( 10 —

x + 3 ) = x(x − 4) ( x + 9

— x − 4

) 10(x − 4) + 3x(x − 4) = x(x + 9)

3x 2 − 2x − 40 = x 2 + 9x

2x 2 − 11x − 40 = 0

(x − 8)(2x + 5) = 0

x = 8 or x = − 5 —

2

Check

10

— 8 + 3 =

? 8 + 9

— 8 − 4

17

— 4 =

17 —

4 ✓

10

—

− 5 —

2

+ 3 =?

− 5 —

2 + 9

—

− 5 —

2 − 4

−4+3 =?

−1

−1 = −1 ✓


2 .

25. 18 —

x2 − 3x −

6 —

x − 3 =

5 —

x

x(x − 3) ( 18 —

x 2 − 3x −

6 —

x − 3 ) = x(x − 3) ( 5 —

x )

18 − 6x = 5(x − 3)

18 − 6x = 5x − 15

33 = 11x

3 = x

Check

18 —

32 − 3(3) −

6 —

3 − 3 =

? 5 —

3

18

— 0 −

6 —

0 =

? 5 —

3


The apparent solution x = 3 is extraneous. So, the equation

has no solution.


Chapter 7

26. 10 —

x2 − 2x +

4 —

x =

5 —

x − 2

x(x − 2) ( 10 —

x2 − 2x +

4 —

x ) = x(x − 2) ( 5

— x − 2

) 10 + 4(x − 2) = 5x

4x + 2 = 5x

2 = x

Check

10 —

22 − 2(2) +

4 —

2 =

?

5 —

2 − 2

10

— 0 +

4 —

2 =

? 5 —

0


The apparent solution x = 2 is extraneous. So, the equation

has no solution.

27. x + 1

— x + 6

+ 1 —

x =

2x + 1 —

x + 6

x(x + 6) ( x + 1 —

x + 6 +

1 —

x ) = x(x + 6) ( 2x + 1

— x + 6

) x(x + 1) + (x + 6) = x(2x + 1)

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

0 = x 2 − x − 6

0 = (x − 3)(x + 2)

x = 3 or x = −2

Check

3 + 1

— 3 + 6

+ 1 —

3 =

? 2(3) + 1

— 3 + 6

4 —

9 +

1 —

3 =

? 7 —

9

7 —

9 =

7 —

9 ✓

−2 + 1

— −2 + 6

+ 1 —

−2 =

? 2(−2)+1

— −2 + 6

− 1 —

4 + −

1 —

2 =

? −

3 —

4

− 3 —

4 = −

3 —

4 ✓

The solutions are x = 3 and x = −2.

28. x + 3

— x − 3

+ x —

x − 5 =

x + 5 —

x − 5

(x − 3)(x − 5) ( x + 3 —

x − 3 +

x —

x − 5 ) = (x − 3)(x − 5) ( x + 5

— x − 5

) (x − 5)(x + 3) + x(x − 3) = (x − 3)(x + 5)

2x2 − 5x − 15 = x2 + 2x − 15

x2 − 7x = 0

x(x − 7) = 0

x = 0 or x = 7

Check

0 + 3

— 0 − 3

+ 0 —

0 − 5 =

? 0 + 5

— 0 − 5

−1 + 0 =?

−1

−1 = −1 ✓

7 + 3

— 7 − 3

+ 7 —

7 − 5 =

? 7 + 5

— 7 − 5

5 —

2 +

7 —

2 =

? 6

6 = 6 ✓

The solutions are x = 0 and x = 7.

29. 5 —

x − 2 =

2 —

x + 3

x(x + 3) ( 5 — x − 2 ) = x(x + 3) ( 2

— x + 3

) 5(x + 3) − 2x(x + 3) = 2x

−2x2 − x + 15 = 2x

0 = 2x2 + 3x − 15

x = −3 ± √

— 129 —

4

Check

5 —

−3.59 − 2 =

?

2 —

−3.59 + 3

−1.393 − 2 =?

2 —

−0.59

−3.393 ≈ −3.390 ✓

5 —

2.09 − 2 =

?

2 —

2.09 + 3

2.392 − 2 =?

2 —

5.09

0.392 ≈ 0.393 ✓

The solutions are x = −3 − √

— 129 —

4 ≈ −3.59 and

x = −3+ √

— 129 —

4 ≈ 2.09.


Chapter 7

30. 5 —

x2 + x − 6 = 2 +

x − 3 —

x − 2

(x − 2)(x + 3) ( 5 ——

(x − 2)(x + 3) )

= (x − 2)(x + 3) ( 2 + x − 3

— x − 2

) 5 = 2(x − 2)(x + 3) + (x + 3)(x − 3)

5 = 3x 2 + 2x − 21

0 = 3x 2 + 2x − 26

x = −1 ± √

— 79 —

3

Check

5 ———

(−3.30)2 + (−3.30) − 6 =

? 2 +

−3.30 − 3 —

−3.30 − 2

5 —

1.59 =

? 2 +

−6.30 —

−5.30

3.145 ≈ 3.189 ✓

5 ——

(2.63)2 + 2.63 − 6 =

? 2 +

2.63 − 3 —

2.63 − 2

5 —

3.5469 =

? 2 +

−0.37 —

0.63

1.450 ≈ 1.413 ✓

The solutions are x = −1 − √

— 79 —

3 ≈ −3.30 and

x = −1+ √

— 79 —

3 ≈ 2.63.

31. The error is that both sides of the equation should be

multiplied by the same expression.

5 —

3x +

2 —

x2 = 1

3x3 ⋅ 5 —

3x + 3x3 ⋅

2 —

x2 = 3x3 ⋅ 1

32. The error is that each term of the equation should be

multiplied by the LCD.

7x + 1

— 2x + 5

+ 4 = 10x − 3

— 3x

(2x + 5)(3x) ⋅ 7x + 1

— 2x + 5

+ (2x + 5)(3x) ⋅ 4

= (2x + 5)(3x) ( 10x − 3 —

3x )

33. a. Work rate Time Work done

You 1 room

— 8 hours

5 hours 5 —

8 room

Friend 1 room

— t hours

5 hours 5 —

t room

b. The sum is the amount of the time it would take you and

your friend to paint the room together: 5 —

8 +

5 —

t = 1.

5 —

8 +

5 —

t = 1

5t + 40 = 8t

40 = 3t

13. — 3 = t

So, it would take 13. — 3 hours, or 13 hours and 20 minutes,

to paint the room.

34. a. Work rate Time Work done

You 1 park

— 2 hours

1.2 hours 0.6 park

Friend 1 park

— t hours

1.2 hours 1.2

— t park

b. The sum is the amount of time it would take for you and

your friend to clean the park together: 0.6 + 1.2

— t = 1.

0.6 + 1.2

— t = 1

0.6t + 1.2 = t

1.2 = 0.4t

3 = t

So, it would take 3 hours to clean the park.

35. Sample answer: x + 1

— x + 2

= 3 —

x + 4 ; cross multiplication can be

used when each side of the equation is a single rational

expression.

Sample answer: x + 1

— x + 2

+ 3 —

x + 4 =

1 —

x + 3 ; Multiplying by the

LCD can be used when there is more than one rational

expression on one side of the equation.

36. Sample answer: A rational equation can be used to determine

how long it would take you and your friend working together

to clean a park.


Chapter 7


x

y

4

2

−2

4 62



y = 2 —

x − 4

x = 2 —

y − 4

x(y − 4) = 2

y − 4 = 2 —

x

y = 2 —

x + 4


x + 4.


x

y8

4

−4

−8

−8 −4



y = 7 —

x + 6

x = 7 —

y + 6

x(y + 6) = 7

y + 6 = 7 —

x

y = 7 —

x −6


x − 6.


x

y

2

−2

−4

−4 −2 4



y = 3 —

x −2

x = 3 —

y −2

x + 2 = 3 —

y

y(x + 2) = 3

y = 3 —

x + 2


x + 2 .


x

y

−4

−8

−4 −2 42



y = 5 —

x − 6

x = 5 —

y − 6

x + 6 = 5 —

y

y(x + 6) = 5

y = 5 —

x + 6


x + 6 .


Chapter 7


x

y4

2

−2

−4

4



y = 4 —

11 − 2x

x = 4 —

11 − 2y

x(11 − 2y) = 4

11 − 2y = 4 —

x

−2y = 4 —

x − 11

y = − 2 —

x +

11 —

2

So, the inverse of f is g(x) = − 2 —

x +

11 —

2 .


x

y

−2

4

−4

−2 2



y = 8 —

9 + 5x

x = 8 —

9 + 5y

x(9 + 5y) = 8

9 + 5y = 8 —

x

5y = 8 —

x − 9

y = 8 —

5x −

9 —

5


5x −

9 —

5 .


x

y

4

6

2

−4 −2 42

Notice that some horizontal lines intersect the graph

more than once. So, the inverse of f is not a function. Find

the inverse.

y = 1 —

x2 + 4

x = 1 —

y2 + 4

x − 4 = 1 —

y2

y2(x − 4) = 1

y2 = 1 —

x − 4

y = ± √—

1 —

x − 4

So, the inverse is y = ± √—

1 —

x − 4 .


x

y

−8

−4 −2 42

Notice that some horizontal lines intersect the graph more

than once. So, the inverse of f is not a function. Find the

inverse.

y = 1 —

x 4 − 7

x = 1 —

y 4 − 7

x + 7 = 1 —

y 4

y 4(x + 7) = 1

y4 = 1 —

x + 7

y = ± 4 √—

1 —

x + 7

So, the inverse is y = ± 4 √—

1 —

x + 7 .


Chapter 7

45. a. 1389 = 9000 ∙ 3.24

— e

1389e = 29,160

e ≈ 21

The fuel-effi ciency rate is about 21 miles per gallon.

b. c = m ∙ g

— e

ce = mg

e = mg

— c

e = 9000 ∙ 3.24

— 1389

≈ 21

The fuel-effi ciency rate is about 21 miles per gallon.

46. a. 0.47 = 105.07

— d + 33

0.47(d + 33) = 105.07

0.47d + 15.51 = 105.07

0.47d = 89.56

d ≈ 190.6

The depth is about 190.6 feet.

b. p = 105.07

— d + 33

p(d + 33) = 105.07

pd + 33p = 105.07

pd = −33p + 105.07

d = −33p + 105.07

—— p

d = −33(0.47) + 105.07

—— 0.47

d ≈ 190.6

The depth is about 190.6 feet.

47.

6

−4

−6

4

Graph the functions. Using the intersect feature, the graphs

intersect at x ≈ ± 0.8165. So, the solutions are x ≈ −0.8165

and x ≈ 0.8165.

48.

6

−4

−6

4


intersect at x ≈ ± 0.7746. So, the solutions are x ≈ −0.7746

and x ≈ 0.7746.

49.

6

−4

−6

4


intersect at x ≈ 1.3247. So, the solution is x ≈ 1.3247.

50.

6

−4

−6

4


intersect at x ≈ 1.2599. So, the solution is x ≈ 1.2599.

51. Use the equation 1 — ℓ =

ℓ — ℓ + 1 .

1 — ℓ =

ℓ — ℓ + 1

ℓ + 1 =ℓ2

0 =ℓ2 −ℓ − 1

ℓ = 1 ± √

— 5 —

2

Because length and width are positive, the ratio must be

positive. So, the golden ratio is 1 + √

— 5 —

2 .

52. The solution is x = −3 because the graphs intersect at x = −3.

53. Rewrite the expression.

y = 3x + 1

— x − 4

y = 3x − 12 + 13

—— x − 4

y = 3(x − 4) + 13

—— x − 4

y = 3 + 13 —

x − 4

Find the inverse.

y = 3 + 13 —

x − 4

x = 3 + 13 —

y − 4

x − 3 = 13 —

y − 4

(x − 3)(y − 4) = 13

y − 4 = 13 —

x − 3

y = 13 —

x − 3 + 4

y = 4x + 1

— x − 3

The inverse of f is g(x) = 4x + 1

— x − 3

.


Chapter 7

54. Rewrite the expression.

y = 4x − 7

— 2x + 3

y = 4x + 6 − 13

—— 2x + 3

y = 2(2x + 3) − 13

—— 2x + 3

y = 2 + −13

— 2x + 3

Find the inverse.

y = 2 + −13

— 2x + 3

x = 2 + −13

— 2y + 3

x − 2 = −13

— 2y + 3

(x − 2)(2y + 3) = −13

2y + 3 = −13

— x − 2

2y = −13

— x − 2

− 3

y = −13

— 2(x − 2)

− 3 —

2

y = −3x − 7

— 2x − 4

The inverse of f is g(x) = −3x − 7

— 2x − 4

.

55. Find the inverse.

y = ax + b

— cx + d

x = ay + b

— cy + d

(cy + d)x = ay + b

cxy + dx = ay + b

cxy − ay = b − dx

(cx − a)y = b − dx

y = b − dx

— cx − a

In Exercise 53, f(x) = 3x + 1

— x − 4

, so a = 3, b = 1, c = 1, and

d = −4. Substituting these values, the inverse is y = 1 + 4x

— x − 3

.

In Exercise 54, f(x) = 4x − 7

— 2x + 3

, so a = 4, b = −7, c = 2, and

d = 3. Substituting these values, the inverse is y = −7 − 3x

— 2x − 4

.

56. a. Yes, it is possible. Sample answer: 1 —

x =

1 —

x + 1 has no solution.

b. Yes, it is possible. Sample answer: 1 —

x =

1 —

2x + 1 has the

solution x = −1.

c. Yes, it is possible. Sample answer: 1 —

x =

x —

4x − 3 has the

solutions x = 1 and x = 3.

d. Yes, it is possible. Sample answer: 1 —

x + 1 =

3 —

3x + 3 has

infi nitely many solutions.

57. a. The statement is always true. When x = a, the denominator

of the fractions are both zero.

b. The statement is sometimes true. The equation will have

exactly one solution except when a = 3.

c. The statement is always true. x = a is an extraneous

solution, so the equation has no solution.

58. Your friend is correct. Extraneous solutions make the

denominator in one of the terms of the equation equal

to zero. The equation does not have variables in either

denominator, so there are no extraneous solutions.


59. The domain is discrete because the number of quarters in

your pocket is an integer.

2 4 6 80 x

y

2

1

0

Quarters

Mo

ney

(d

olla

rs)

60. The domain is continuous because the weight of the broccoli

can be any nonnegative number.

2 40 p

t

2

3

4

5

6

7

8

9

10

10

t = 2p

Broccoli (pounds)

Tota

l co

st (

do

llars

)

61. f ( −2 ) = ( −2 ) 3 − 2 ( −2 ) + 7

= −8 + 4 + 7

= 3

62. g ( 3 ) = −2 ( 3 ) 4 + 7 ( 3 ) 3 + ( 3 ) − 2 = −162 + 189 + 3 − 2 = 28


Chapter 7

63. h ( 3 ) = − ( 3 ) 3 + 3 ( 3 ) 2 + 5 ( 3 )

= −27 + 27 + 15

= 15

64. k ( −5 ) = −2 ( −5 ) 3 − 4 ( −5 ) 2 + 12 ( −5 ) − 5

= 250 − 100 − 60 − 5

= 85

7.3–7.5 What Did You Learn? (p. 399)

1. A rational equation is appropriate because you are fi nding

the annual healthcare expenditures per resident.

2. Sample answer: a —

ab ÷

▭ — ac

= c —

b

Replacing each factor with a variable shows how the

expressions are related.

3. The conjecture was that the expressions approach √—

2

because the numbers were getting closer to √—

2 ≈ 1.4142.

4. Both methods produce the same result. When solving the

equation, the constants are substituted fi rst. When using the

inverse, the constants are substituted last.

Chapter 7 Review (pp. 400–402)


xy = 5 y = 5 —

x inverse


5y = 6x y = 6 —

5 x direct


15 = x —

y y =

x —

15 direct


y − 3 = 2x y = 2x + 3 neither


x .

xy 245 605 1125 2000

y —

x 5 5 5 5



x .

xy 32 32 32 32

y —

x 1.28 0.5 0.32 0.08


7. y = a —

x

5 = a —

1

5 = a

The inverse equation is y = 5 —

x . When x = −3, y =

5 —

−3 = −

5 —

3 .

8. y = a —

x

−6 = a —

−4

24 = a

The inverse equation is y = 24

— x . When x = −3, y =

24 —

−3 = −8.

9. y = a —

x

18 = a —

5 —

2

45 = a


— x . When x = −3, y =

45 —

−3 = −15.

10. y = a —

x

2 —

3 =

a —

−12

−8 = a

The inverse equation is y = −8

— x . When x = −3, y =

−8 —

−3 =

8 —

3 .


Step 2 Plot points to the left of the vertical asymptote, such as

( 0, − 4 — 3 ) , (1, −2), and (2, −4). Plot points to the right

of the vertical asymptote, such as (4, 4), (5, 2), and

( 6, 4 —

3 ) .



asymptotes.

y

42 6 8 10−2

4

2

6

x

−4

−6

−2




Chapter 7

12. Step 1 Draw the asymptotes x = −5 and y = 2.


as ( −8, 5 —

3 ) , ( −7,

3 —

2 ) , and (−6, 1). Plot points to the

right of the vertical asymptote, such as (−4, 3),

( −3, 5 —

2 ) , and ( −2,

7 —

3 ) .



asymptotes.

y

2−2−4−6−8

4

6

8

x

−4

−2

−10−12

The domain is all real numbers except −5 and the range is

all real numbers except 2.

13. Step 1 Draw the asymptotes. Solve x − 4 = 0 for x to

fi nd the vertical asymptote x = 4. The horizontal


c =

3 —

1 = 3.


as ( 1, − 1 — 3 ) , (2, −2), and (3, −7). Plot points to the

right of the vertical asymptote, such as (5, 13),

(6, 8), and ( 7, 19

— 3 ) .



asymptotes.

y

2 6 8 10−2

4

2

6

8

x

−4

−2

10



14. 80x 4 —

y 3 ⋅

xy —

5x 2 =

80x 5y —

5x 2y 3

= 16x 3

— y 2

, x ≠ 0

15. x − 3 —

2x − 8 ⋅

6x 2 − 96 —

x 2 − 9 =

x − 3 —

2(x − 4) ⋅

6(x − 4)(x + 4) ——

(x + 3)(x − 3)

= 6(x − 3)(x − 4)(x + 4)

—— 2(x − 4)(x − 3)(x + 3)

= 3(x + 4)

— x + 3

, x ≠ 3, x ≠ 4

16. 16x 2 − 8x + 1 ——

x 3 − 7x 2 + 12x ÷

20x 2 − 5x —

15x 3 =

(4x − 1) 2 ——

x(x − 4)(x − 3) ⋅

15x 3 —

5x(4x − 1)

= 15x 3 (4x − 1) 2

——— 5x 2(x − 4)(x − 3)(4x − 1)

= 3x(4x − 1)

—— (x − 4)(x − 3)

, x ≠ 0, x ≠ 1 —

4

17. x 2 − 13x + 40 ——

x 2 − 2x − 15 ÷ (x 2 − 5x − 24)

= (x − 8)(x − 5)

—— (x − 5)(x + 3)

⋅ 1 ——

(x − 8)(x + 3)

= (x − 8)(x − 5)

——— (x − 5)(x + 3)(x − 8)(x + 3)

= 1 —

(x + 3) 2 , x ≠ 5, x ≠ 8

18. 5 —

6(x + 3) +

x + 4 —

2x =

5x —

6x(x + 3) +

3(x + 3)(x + 4) ——

6x(x + 3)

= 5x + 3(x + 3)(x + 4)

—— 6x(x + 3)

= 5x + (3x 2 + 21x + 36)

—— 6x(x + 3)

= 3x 2 + 26x + 36

—— 6x(x + 3)

19. 5x — x + 8

+ 4x − 9

—— x2 + 5x − 24

= 5x —

x + 8 +

4x − 9 ——

(x − 3)(x + 8)

= 5x(x − 3)

—— (x − 3)(x + 8)

+ 4x − 9 ——

(x − 3)(x + 8)

= 5x(x − 3) + (4x − 9)

—— (x − 3)(x + 8)

= 5x 2 − 15x + (4x − 9)

—— (x − 3)(x + 8)

= 5x 2 − 11x − 9

—— (x − 3)(x + 8)

20. x + 2 —

x2 + 4x + 3 −

5x —

x2 − 9 =

x + 2 ——

(x + 1)(x + 3) −

5x ——

(x − 3)(x + 3)

= (x + 2)(x − 3)

—— (x − 3)(x + 1)(x + 3)

−

5x(x + 1)

—— (x − 3)(x + 1)(x + 3)

= (x + 2)(x − 3) − 5x(x + 1)

——— (x − 3)(x + 1)(x + 3)

= (x 2 − x − 6) − (5x 2 − 8x)

——— (x − 3)(x + 1)(x + 3)

= −2(2x 2 + 3x + 3)

—— (x − 3)(x + 1)(x + 3)


Chapter 7


5x + 1

— x − 3

= 5x − 15 + 16

—— x − 3

= 5(x − 3) + 16

—— x − 3

= 5(x − 3)

— x − 3

+ 16 —

x − 3 = 5 +

16 —

x − 3


x − 3 + 5.

y

4 8 12−4−8

8

4

x

−4

12

16



— x .


4x + 2

— x + 7

= 4x + 28 − 26

—— x + 7

= 4(x + 7) − 26

—— x + 7

= 4(x + 7)

— x + 7

+ −26

— x + 7

= 4 + −26

— x + 7


— x + 7

+ 4.

y

4−4−8

8

x

−4

−8

12

−12−16



— x .


9x − 10

— x − 1

= 9x − 9 − 1

— x − 1

= 9(x − 1) − 1

—— x − 1

= 9(x − 1)

— x − 1

+ −1

— x − 1

= 9 + −1

— x − 1

y

4 8−4−8

8

4

x

12


— x − 1

+ 9. The graph of g

is a translation 1 unit to the right and 9 units up of the graph

of f (x) = −1

— x .

24. f = 1 —

1 —

p +

1 —

q

= 1 —

q —

pq +

p —

pq

=

1 —

1

— q + p

— pq

= 1 —

1 ⋅

pq —

q + p

= pq —

q + p , p ≠ 0, q ≠ 0

25. 5 —

x =

7 —

x + 2 Check

5 —

5 =

? 7

— 5 + 2

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

7 — 7

5x + 10 = 7x 1 = 1 ✓

10 = 2x 5 = x



Chapter 7

26. 8(x − 1)

— x 2 − 4

= 4 —

x + 2

8(x − 1)(x + 2) = 4(x 2 − 4)

8x 2 + 8x − 16 = 4x 2 − 16

4x 2 + 8x = 0

4x(x + 2) = 0

x = 0 or x = −2

Check 8(0 − 1) —

02 − 4 =

?

4 —

0 + 2

8(−2 − 1) —

(−2)2 − 4 =

?

4 —

−2 + 2

−8

— −4

=?

4 —

2

−24 —

0 =

? 4 —

0

2 = 2 ✓

The apparent solution x = −2 is extraneous. So, the only

solution is x = 0.

27. 2(x + 7)

— x + 4

− 2 = 2x + 20

— 2x + 8

2(x + 7)

— x + 4

− 2 = 2x + 20

— 2(x + 4)

2(x + 4) ( 2(x + 7) —

x + 4 − 2 ) = 2(x + 4) ( 2x + 20

— 2(x + 4)

) 4(x + 7) − 4(x + 4) = 2x + 20

12 = 2x + 20

−8 = 2x

−4 = x

Check 2(−4 + 7)

— −4 + 4

− 2 =?

2(−4) + 20 —

2(−4) + 8

−6

— 0 − 2 =

12 —

0


The apparent solution x = −4 is extraneous. So, there is

no solution.


x

y4

2

−2

−4

−8 −4



y = 3 —

x + 6

x = 3 —

y + 6

x(y + 6) = 3

y + 6 = 3 —

x

y = 3 —

x − 6


x − 6.


x

y4

2

8 12 164



y = 10 —

x − 7

x = 10 —

y − 7

x(y − 7) = 10

y − 7 = 10

— x

y = 10

— x + 7

So, the inverse of f is g(x) = 10

— x + 7.

Division by zero is

undefi ned.


Chapter 7


x

y

8

12

16

−4 −2 42



y = 1 —

x + 8

x = 1 —

y + 8

x − 8 = 1 —

y

y(x − 8) = 1

y = 1 —

x − 8


x − 8 .

31. a. 4.75 = 4n + 3

— n

4.75n = 4n + 3

0.75n = 3

n = 4

You need to bowl 4 games.

b. c = 4n + 3

— n

cn = 4n + 3

cn − 4n = 3

n(c − 4) = 3

n = 3 —

c − 4

n = 3 —

4.75 − 4

n = 4

You need to bowl 4 games.

Chapter 7 Test (p. 403)

1. y = a —

x

2 = a —

5

10 = a


— x . When x = 4, y =

10 —

4 =

5 —

2 .

2. y = a —

x

7 —

2 =

a —

–4

−14 = a

The inverse equation is y = –14

— x . When x = 4, y =

–14 —

4 = – 7 —

2 .

3. y = a —

x

5 —

8 =

a —

3 —

4

15

— 32

= a


— 32x

.

When x = 4, y = 15 —

32(4) =

15 —

128 .

4. The value of h is negative because the vertical asymptote is

to the left of the y-axis. The value of k is negative because

the horizontal asymptote is below the x-axis.

5. The value of h is positive because the vertical asymptote is to

the right of the y-axis. The value of k is positive because the

horizontal asymptote is above the x-axis.

6. The value of h is negative because the vertical asymptote is

to the left of the y-axis. The value of k is zero because the

horizontal asymptote is the x-axis.

7. 3x 2y — 4x 3y 5

÷ 6y 2

— 2xy 3

= 3x 2y

— 4x 3y 5

⋅ 2xy 3

— 6y 2

= 6x 3y 4

— 24x 3y 7

= 1 —

4y 3 , x ≠ 0

8. 3x —— x 2 + x − 12

− 6 —

x + 4 =

3x ——

(x − 3)(x + 4) −

6 —

x + 4

= 3x ——

(x − 3)(x + 4) −

6(x − 3) ——

(x − 3)(x + 4)

= 3x − 6(x − 3)

—— (x − 3)(x + 4)

= 3x − 6x + 18

—— (x − 3)(x + 4)

= −3x + 18

—— (x − 3)(x + 4)

= −3(x − 6)

—— (x − 3)(x + 4)

9. x2 − 3x − 4

—— x2 − 3x − 18

⋅ x − 6

— x + 1

= (x − 4)(x + 1)

—— (x − 6)(x + 3)

⋅ x − 6

— x + 1

= (x − 6)(x − 4)(x + 1)

—— (x − 6)(x + 1)(x + 3)

= x − 4

— x + 3

, x ≠ −1, x ≠ 6


Chapter 7

10. 4 —

x + 5 +

2x —

x2 − 25 =

4 —

x + 5 +

2x ——

(x − 5)(x + 5)

= 4(x − 5)

—— (x − 5)(x + 5)

+ 2x ——

(x − 5)(x + 5)

= 4(x − 5) + 2x

—— (x − 5)(x + 5)

= 4x − 20 + 2x

—— (x + 5)(x − 5)

= 6x − 20

—— (x + 5)(x − 5)

= 2(3x − 10)

—— (x + 5)(x − 5)

11. g(x) = (x + 3)(x − 2)

—— x + 3

= x − 2, x ≠ −3

The graphs of f and g are the same everywhere except at x = −3, where g(x) is undefi ned.

12. The model for the average cost is c = 1.25x + 500

—— x .

1.79 = 1.25x + 500

—— x

1.79x = 1.25x + 500

0.54x = 500

x ≈ 926

You have to produce about 926 pounds of honey before the

average cost is $1.79 per pound.

13. F = a —

d

60 = a —

10

600 = a

The model is F = 600

— d . So when the distance is 15 feet,

F = 600

— 15

= 40 foot-pounds.

14. a. The expression for the height is h = 6r. The formula for

the volume is V = πr 2h = πr 2(6r) = 6πr 3.

b. The total volume of the tennis balls is 3 ( 4 — 3 πr 3 ) , so the percent

of the can’s volume not occupied by the tennis balls is

6πr 3 − 4πr 3

— 6πr 3

= 2πr 3

— 6πr 3

= 2 —

6 = 0.33

— 3

So, 0.33 — 3 , or 33.

— 3 %, of the can’s volume is not occupied

by the tennis balls.

Chapter 7 Standards Assessment (pp. 404–405)

1. The graph shows A or C because the x-intercepts are x = 1

and x = 5 and the vertex is (3, 8).

2. 9 = 12

— s +

12 —

6 − s

s(6 − s)(9) = s(6 − s) ( 12 —

s +

12 —

6 − s )

9s(6 − s) = 12(6 − s) + 12s

54s − 9s2 = 72 − 12s + 12s

54s − 9s2 = 72

0 = 9s2 − 54s + 72

0 = 9(s − 4)(s − 2)

s = 4 or s = 2

So, two possible speeds of the escalator are 2 feet per second

and 4 feet per second.

3. Because the asymptotes intersect at (4, 3), the asymptotes

are x = 4 and y = 3. So, a —

c = 3, which means that a = 9

and c = 3. The function is undefi ned at x = 4, so the

denominator must be 3x + (−12). So, the function is

y = 9x + 6

— 3x + (−12)

.

8 12−4

8

4

x

12y

−8

−4

4. a. Checking Account

t 1 2 3 4

A 5000 5110 5220 5330

110 110 110

The fi rst differences are constant, so the function is linear.

Savings Account

t 1 2 3 4

A 5000 5100 5202 5306.04

× 1.02 × 1.02 × 1.02

The ratios are constant, so the function is exponential.

b. The amount of money in the checking account increases

by the same amount each year. The amount of money in

the savings account increases exponentially.

c. The value of the checking account can be modeled

by A = 110t + 4890 and the value of the savings

account can be modeled by A = 5000(1.02) t − 1.

After 10 years, the checking account has a value of

A = 110(10) + 4890 = $5990 and the savings account

has a value of A = 5000(1.02)10−1 = $5975.46. After

15 years, the checking account has a value of

A = 110(15) + 4890 = $6540 and the savings account

has a value of A = 5000(1.02)15−1 = $6597.39. So, the

checking account has a greater value after 10 years and

the savings account has a greater value after 15 years.


Chapter 7

5. ( 3 √—

125 ) 2 = 25, 251/2 = 5, ( √

— 25 )

2 = 25, 1253/2 ≈ 1398,

( √—

5 ) 3 ≈ 11.18

So, the order is

5 = 251/2 < ( √—

5 ) 3 < ( 3 √

— 125 ) 2 = ( √

— 25 )

2 < 1253/2

6. Because C is in billions of gallons and P is in thousands, the

quotient C ⋅ 109

— P ⋅ 103

can be used to fi nd a model for the daily

per capita water consumption (in gallons per person).

C ⋅ 109

— P ⋅ 10 3

= (93.14x − 234.2) ⋅ 109

—— x − 2.8

÷ [ (2200x + 182,000) ⋅ 103 ]

= (93.14x − 234.2) ⋅ 109

—— x − 2.8

⋅ 1 ———

(2200x + 182,000) ⋅ 103

= (93.14x − 234.2) ⋅ 106

——— (x − 2.8)(2200x + 182,000)

7. The exponential decay function is y = 37(1 − 0.30) x−1 =

37(0.70) x−1. The domain is x ≥ 1 and the range is y ≤ 37.

The value x = 1 represents the sales from the fi rst week,

which is the maximum value of the range.

8. The relationship is D because p = − 1 —

3 (q + r).

9. The model is the equation 90 = 95x + 5 ⋅ 83

—— x + 5

for the average

score on the quizzes. Solve for x.

90 = 95x + 5 ⋅ 83

—— x + 5

90(x + 5) = 95x + 415

90x + 450 = 95x + 415

35 = 5x

7 = x

You need to take 7 quizzes to raise your average score to 90.

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