chapter 2 · 2016-09-30 · x ≥ 1; sample answer: the inequality is true for all values of x that...

Chapter 2 Maintaining Mathematical Profi ciency (p. 51)

1.

−4−8 0 4 8

6

2. ∣ 2 ∣ = 2

−2−4 0 2 4

3. ∣ −1 ∣ = 1

−2−4 0 2 4

1

4. 2 + ∣ −2 ∣ = 2 + 2 = 4

−2−4 0 2 4

5. 1 − ∣ −4 ∣ = 1 − 4 = −3

−2−4 0 2 4

−3

6. −5 + ∣ 3 ∣ = −5 + 3 = −2

−2−4 0 2 4

7. 9

2 4 6 8 10 120

2 < 9

8. 5

−4 −2 0 2 4 6−6

−6 < 5

9. −12 −10−14 −6 −4 −2−8

−12 < −4

10. −7−13

−12 −10−14−16 −6 −4−8

−7 > −13

11. 0 2 4 6 8 10−2

∣ −8 ∣ = ∣ 8 ∣

12.

−8 −4 0 4 8 12 16 20−12

−10 18

−10 < ∣ −18 ∣ 13. Because a < b and −a is the opposite of a and −b is the

opposite of b, −b < −a.

Chapter 2 Mathematical Practices (p. 52)

1. 2x + 3 < x − 1 2. −x − 1 > −2x + 2

6

−4.5

−6

4.5

4

−3

−4

3

The solution is x < −4. The solution is x > 3.

3. 1 — 2 x + 1 ≤

3 —

2 x + 3

4

−3

−4

3

The solution is x ≥ −2.

2.1 Explorations (p. 53)

1. a. Statement: The temperature t in Sweden is at least −10°C.

Inequality: t ≥ −10°C Graph:

−20−40 0 20 40

−10

b. Statement: The elevation e of Alabama is at most 2407 feet.

Inequality: e ≤ 2407 ft

Graph:

−2000 0 2000

2407

2. a. x ≥ 1; Sample answer: The inequality is true for all values

of x that are greater than or equal to 1.

b. x > 1; Sample answer: The inequality is true for all values

of x that are greater than 1.

c. x ≤ 1; Sample answer: The inequality is true for all values

of x that are less than or equal to 1.

d. x < 1; Sample answer: The inequality is true for all values

of x that are less than 1.

3. Sample answer: An inequality can be used to describe the

relationship between two expressions or quantities. More

specifi cally, it tells which quantity is greater and which is

less and whether the quantities may or may not be equal.

4. a. Sample answer: A tube has to pass through a hole with a

diameter of 3.5 millimeters. So, the diameter of the tube

must be less than 3.5 millimeters.

b. Sample answer: Your mom gives you $6 for dinner out

with your friends. So, you have to order something that

will cost at most $6 with tax.

c. Sample answer: The temperature must stay above −2°C,

or the school will cancel the ski bus.

d. Sample answer: In some school districts, each student

must do at least 10 hours of community service each

semester in order to graduate.

Chapter 2

Copyright © Big Ideas Learning, LLC Algebra 1 61All rights reserved. Worked-Out Solutions

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

Chapter 2

2.1 Monitoring Progress (pp. 54 –57)

1. A number b is fewer than 30.4.

b < 30.4

An inequality is b < 30.4.

2. − 7 — 10 is at least twice a number k minus 4.

− 7 — 10 ≥ 2 ⋅ k − 4

An inequality is − 7 — 10 ≥ 2k − 4.

3. c + 4 < −1

−6 + 4 <?

−1

−2 < −1 ✓

So, −6 is a solution of the inequality.

4. 10 ≤ 3 − m

10 ≤?

3 − (−6)

10 ≤?

3 + 6

10 ≤ 9 ✗

So, −6 is not a solution of the inequality.

5. 21 ÷ x ≥ −3.5

21 ÷ (−6) ≥?

−3.5

−3.5 ≥ −3.5 ✓

So, −6 is a solution of the inequality.

6. 4x − 25 > −2

4(−6) − 25 >?

−2

−24 − 25 >?

−2

−49 > −2 ✗

So, −6 is not a solution of the inequality.

7. b > −8 8. 1.4 ≥ g

40−4−8−12

0 0.4 0.8 1.2 1.6

1.4

9. r < 1 —

2 10. v ≥ √

— 36

0 1 2−1−2

12

0 2 4 6 8 10

11. An inequality that represents the graph is x ≥ −6.

2.1 Exercises (pp. 58 –60)

Vocabulary and Core Concept Check

1. A mathematical sentence using the symbols < , > , ≤ , or ≥ is

called an inequality.

2. x + 3 > 8

5 + 3 >?

8

8 > 8

no; Because 8 is not greater than itself, 5 is not a value of x

that makes the inequality true. So, 5 is not in the solution set.

3. Draw an open circle when a number is not part of the

solution. Draw a closed circle when a number is part of the

solution. Draw an arrow to the left or right to show that the

graph continues in that direction.

4. The sentence, “w is no more than −7” is different, because

an inequality that represents this sentence is w ≤ −7, but the

other three sentences can be represented by the inequality

w ≥ −7.

Monitoring Progress and Modeling with Mathematics

5. A number x is greater than 3.

x > 3

An inequality is x > 3.

6. A number n plus 7 is less than or equal to 9.

n + 7 ≤ 9

An inequality is n + 7 ≤ 9.

7. Fifteen is no more than a number t divided by 5.

15 ≥ t —

5

An inequality is 15 ≥ t —

5 .

8. Three times a number w is less than 18.

3w < 18

An inequality is 3w < 18.

9. One-half of a number y is more than 22.

1 —

2 ⋅ y > 22

An inequality is 1 —

2 y > 22.

10. Three is less than the sum of a number s and 4.

3 < s + 4

An inequality is 3 < s + 4.

11. Thirteen is at least the difference of a number v and 1.

13 ≥ v − 1

An inequality is 13 ≥ v − 1.


Chapter 2

12. Four is no less than the quotient of a number x and 2.1.

4 ≥ x —

2.1

An inequality is 4 ≥ x —

2.1 .

13. Let w be the weight of the second fi sh.

The second fi sh weighs at least 0.5 lb more than the fi rst fi sh.

w ≥ 0.5 + 1.2

An inequality is w ≥ 0.5 + 1.2, or w ≥ 1.7.

14. Let x be the additional people that can enter the pool.

Additional

people are at

most maximum

capacity minus people in

pool.

x ≤ 600 − 430

An inequality is x ≤ 600 − 430, or x ≤ 170.

15. r + 4 > 8

2 + 4 >?

8

6 > 8 ✗

So, r = 2 is not a solution of the inequality.

16. 5 − x < 8

5 − (−3) <?

8

5 + 3 <?

8

8 < 8 ✗

So, x = −3 is not a solution of the inequality.

17. 3s ≤ 19

3(−6) ≤?

19

−18 ≤ 19 ✓

So, s = −6 is a solution of the inequality.

18. 17 ≥ 2y

17 ≥?

2(7)

17 ≥ 14 ✓

So, y = 7 is a solution of the inequality.

19. −1 > − x —

2

−1 >?

− 3 —

2

−1 > −1 1 —

2 ✓

So, x = 3 is a solution of the inequality.

20. 4 — z ≥ 3

4 —

2 ≥?

3

2 ≥ 3 ✗

So, z = 2 is not a solution of the inequality.

21. 14 ≥ −2n + 4

14 ≥?

−2(−5) + 4

14 ≥?

10 + 4

14 ≥ 14 ✓

So, n = −5 is a solution of the inequality.

22. −5 ÷ (2s) < −1

−5 ÷ (2 ⋅ 10) <?

−1

−5 ÷ 20 <?

−1

−0.25 < −1 ✗

So, s = 10 is not a solution of the inequality.

23. 20 ≤ 10

— 2z

+ 20

20 ≤?

10

— 2(5)

+ 20

20 ≤?

10

— 10

+ 20

20 ≤?

1 + 20

20 ≤ 21 ✓

So, z = 5 is a solution of the inequality.

24. 3m — 6 − 2 > 3

3(8)

— 6 − 2 >

? 3

24

— 6 − 2 >

? 3

4 − 2 >?

3

2 > 3

So, m = 8 is not a solution of the inequality.

25. a. 8 feet 11 inches = 8(12) inches + 11 inches = 107 inches

Let h be the heights (in inches) of every other person that

has ever lived.

An inequality is h < 107.

b. 9 feet = 9(12) inches = 108 inches

h < 107

108 < 107

Because 108 < 107 is not true, 9 feet is not a solution of

the inequality.



Chapter 2

26. a. Words: Other

competitor

bench

press

weights

are

at

most

Winner’s

bench

press

weight

−

Least

amount

bench

press

weight

Variable: Let x be the weights that the other

competitors bench pressed.

Inequality: x ≤ 400 − 23

An inequality is x ≤ 400 − 23, or x ≤ 377.

b. x ≤ 400 − 23

379 ≤?

400 − 23

379 ≤ 377 ✗

Because x = 379 pounds does not make the inequality

true, it is not a solution of the inequality.

27. Because −1 is to the right of −4 on the number line, it is

greater than −4. So, y = 8 does not make the inequality true

and therefore is not a solution of the inequality.

−y + 7 < −4

−8 + 7 <?

−4

−1 < −4 ✗

So, 8 is not in the solution set.

28. The statement 6 ≤ 6 is true because 6 is less than or equal to

itself. So, x = 8 is a solution of the inequality.

1 — 2 x + 2 ≤ 6

1 — 2 (8) + 2 ≤

? 6

4 + 2 ≤?

6

6 ≤ 6 ✓

So, 8 is in the solution set.

29. x ≥ 2 30. z ≤ 5

0 2 4−2

0 2 4 6

5

31. −1 > t 32. −2 < w

0 2−2−4

−1 0 2−2−4

33. v ≤ −4 34. s < 1

0−2−4−6

0 2−2

1

35. 1 — 4 < p 36. r ≥ − ∣ 5 ∣

0 1− 12 1 1

212

14

0−2−4−6

−5

37. x < 7

7

0 2 4 6 8 10

38. n ≥ −2 39. z ≥ 1.3

−2−4 0 2

1.3

1.0 1.2 1.4 1.6

40. w < 5.2

4.6 4.8 5.0 5.2

41. An inequality is x ≤ 4. 42. An inequality is x ≥ −2.

43. An inequality is x > 3. 44. An inequality is x < −1.

45. C; The temperature must increase by no less than

76°F − 74°F = 2°F in order for the temperature to be no less

than 76°F. So, the increase must be greater than or equal

to 2°F and is represented by x ≥ 2.

46. For a truck with 2 axles, an inequality that represents the

total possible weights w (in pounds) of the vehicle and its

contents is w ≤ 40,000.

20,000 40,000 60,000 80,000

For a truck with 3 axles, an inequality is w ≤ 60,000.

20,000 40,000 60,000 80,000

For a truck with 4 axles, an inequality is w ≤ 80,000.

20,000 40,000 60,000 80,000

47. If the longest natural arch is 400(12) = 4800 inches, an

inequality that represents the lengthsℓ (in inches) of all other

natural arches isℓ < 4800.

4600 4800 5000

48. Sample answer: If a student works no more than 25 hours

each week and the hours are divided evenly over a 5-day

work week, then an inequality that represents how many

hours h the student can work each day is h ≤ 25

— 5 , or h ≤ 5.

49. Sample answer: If today is the 23rd day of the month, and

you let x represent the number of days left in the month, then

an inequality to represent this situation is 23 + x ≤ 31.

50. a. Let T be the known melting points (in degrees Celsius) of

all metallic elements.

An inequality is T ≥ −38.87.

b. T ≥ −38.87

−38.87 ≥ −38.87

Because −38.87°C is greater than or equal to itself, it is

possible for a metallic element to have a melting point of

−38.87°C.


Chapter 2

51. Words:Price per

one-way

ride⋅

Number

of

one-way

rides

is less

than or

equal to

Cost of a

monthly

pass

Variable: Let x be how many one-way rides you buy.

Inequality: $0.90 ⋅ x ≤ 24.00

0.9x ≤ 24

0.9(25) ≤?

24

22.5 ≤ 24 ✓

Because x = 25 makes the inequality true, it is cheaper to

pay the one-way fare for 25 rides than to buy a monthly pass.

52. Your cousin is correct because 1324 is less than or equal to

itself. So, x = 1324 is a solution of the inequality.

53. Sample answer: Some highways have a minimum speed

of 55 miles per hour and a maximum speed of 70 miles

per hour. Let s be the speed you can legally travel on the

highway. Then the situation can be represented by the

inequalities s ≥ 55 and s ≤ 70.

54. Let b be the amounts of all the losing bids. An inequality is

b < 17,000.

12,000 14,000 16,000 18,000

17,000

55. The area of the triangle is 1 — 2 bh =

1 —

2 (x)(6) = 3x, which equals

42 when x = 14. So, the area is less than 42 square meters

when x < 14 meters.

56. The area of the triangle is 1 — 2 bh =

1 —

2 (x)(10) = 5x, which

equals 8 when x = 1.6. So, the area is greater than or equal

to 8 square feet when x ≥ 1.6 feet.

57. The area of the trapezoid is 1 — 2 ( b1 + b2 ) h =

1 —

2 (8 + 4)(x) = 6x,

which equals 18 when x = 3. So, the area is less than

18 square centimeters when x < 3 centimeters.

58. The area of the rectangle isℓw = 2x, which equals 12 when

x = 6. So, the area is greater than 12 square inches when

x > 6 inches.

59. a. 200 meters —

35 seconds =

40 —

7 , or about 5.71 meters per second

So, an inequality is r > 40

— 7 or about 5.71.

5.71

5.68 5.70 5.72 5.74

b. no; The human body is only able to run so fast. Eventually

the graph will reach speeds at which it is impossible for

the human body to run.

Maintaining Mathematical Profi ciency

60. x + 2 = 3 Check: x + 2 = 3

− 2 − 2 1 + 2 =?

3

x = 1 3 = 3 ✓

The solution is x = 1.

61. y − 9 = 5 Check: y − 9 = 5

+ 9 + 9 14 − 9 =?

5

y = 14 5 = 5 ✓

The solution is y = 14.

62. 6 = 4 + y Check: 6 = 4 + y

− 4 − 4 6 =?

4 + 2

2 = y 6 = 6 ✓


63. −12 = y − 11 Check: −12 = y − 11

+ 11 + 11 −12 =?

−1 − 11

−1 = y −12 = −12 ✓

The solution is y = −1.

64. v = x ⋅ y ⋅ z

v — yz

= xyz

— yz

v — yz

= x

The rewritten literal equation is x = v —

yz .

65. s = 2r + 3x

s − 2r = 2r − 2r + 3x

s − 2r = 3x

s − 2r —

3 =

3x —

3

s − 2r —

3 = x

The rewritten literal equation is x = s − 2r

— 3 .

66. w = 5 + 3(x − 1)

w = 5 + 3(x) − 3(1)

w = 5 + 3x − 3

w = 3x + 2

w − 2 = 3x + 2 − 2

w − 2

— 3 =

3x —

3

w − 2

— 3 = x

The rewritten literal equation is x = w − 2

— 3 .


Chapter 2

67. n = 2x + 1

— 2

2 ⋅ n = 2 ⋅ 2x + 1 —

2

2n = 2x + 1

2n − 1 = 2x + 1 − 1

2n − 1 = 2x

2n − 1

— 2 = x

n − 1 — 2 = x

The rewritten literal equation is x = n − 1 —

2 .


1. a. no; It is possible, for T and C to be equal if every one of a

quarterback’s completed passes results in a touchdown.

b. yes; The sum of the number of completed passes and the

number of intercepted passes is always less than or equal

to the number of attempted passes because a quarterback

must attempt a pass in order for it to be either completed

or intercepted.

c. no; It is possible, for N and A to be equal if every one of

the quarterback’s attempted passes are intercepted.

d. yes; The difference of the number of attempted passes

and the number of completed passes is always greater

than or equal to the number of incomplete passes, because

all attempted passes must be completed, intercepted, or

incomplete.

2. a. Sample answer:

Attempts Completions Yards Touchdowns Interceptions

10 5 50 0 5

P = 8.4Y + 100C + 330T − 200N

——— A

= 8.4(50) + 100(5) + 330(0) − 200(5)

——— 10

= −80

— 10

= −8

Whenever a quarterback earns a negative ranking, the

inequality P < 0 will be true. Because −8 < 0 is a true

statement, P = −8 is a solution of the inequality. One

way a quarterback can earn a negative ranking after only

10 attempted passes is by completing only half of them

for only 50 yards and no touchdowns while having the

other half intercepted.

b. Sample answer:


100 72 653 11 6

P = 8.4Y + 100C + 330T − 200N

——— A

= 8.4(653) + 100(72) + 330(11) − 200(6)

———— 100

= 15,115.2

— 100

≈ 151

P + 100 ≥ 250 Check: P + 100 ≥ 250

− 100 − 100 151 + 100 ≥?

250

P ≥ 100 251 ≥ 250 ✓

All values of P that are greater than or equal to 150 will

make the inequality true. So, P = 151 is a solution of the

inequality. One way a quarterback can earn a ranking of

151 is by completing 72 out of 100 attempted passes for

653 yards and 11 touchdowns while only throwing

6 interceptions.

c. Sample answer:


100 83 712 13 5

P = 8.4Y + 100C + 330T − 200N

——— A

= 8.4(712) + 100(83) + 330(13) − 200(5)

———— 100

= 17,570.8

— 100

≈ 176

P − 250 > −80 Check: P − 250 > −80

+ 250 + 250 176 − 250 >?

−80

P > 170 −74 > −80 ✓

All values of P that are greater than or equal to 170 will

make the inequality true. So, P = 176 is a solution of the

inequality. One way a quarterback can earn a ranking of

176 is by completing 83 out of 100 attempted passes for

712 yards and 13 touchdowns while only throwing

5 interceptions.

3. In order to isolate a variable, you can undo addition by

subtracting the same value from each side of an inequality,

or you can undo subtraction by adding the same value to

each side of an inequality.


Chapter 2

4. a. x + 3 < 4 b. x − 3 ≥ 5

+ 3 + 3

x ≥ 8

The solution is x ≥ 8.

− 3 − 3

x < 1

The solution is x < 1.

c. 4 > x − 2 d. −2 ≤ x + 1

− 1 − 1

−3 ≤ x


+ 2 + 2

6 > x


2.2 Monitoring Progress (pp. 62 – 64)

1. b − 2 > −9

+ 2 + 2

b > −7

The solution is b > −7.

−7

−8 −6 −4 −2 0

2. m − 3 ≤ 5

+ 3 + 3

m ≤ 8

The solution is m ≤ 8.

0 2 4 6 8

3. 1 — 4 > y −

1 —

4

+ 1 —

4 +

1 —

4 ( 1 —

4 +

1 —

4 =

2 —

4 =

1 —

2 )

1 — 2 > y

The solution is y < 1 —

2 .

−2 −1 0 1 2

12

4. k + 5 ≤ −3

− 5 − 5 k ≤ −8

The solution is k ≤ −8.

−12 −10 −8 −6 −4

5. 5 —

6 ≤ z +

1 —

6

− 1 —

6 −

1 —

6 ( 5 —

6 −

1 —

6 =

4 —

6 =

2 —

3 )

2 —

3 ≤ z

The solution is z ≥ 2 —

3 .

0 2

3 113 22

32

6. p + 0.7 > −2.3

− 0.7 − 0.7

p > −3

The solution is p > −3.

−6 −4 −2 0 2

−3

7. Words: Watts used by

microwave

oven+

Watts used

by the

toaster<?

Overload

wattage

1000 + 800 <?

1800

1800 < 1800 ✗

no; If both the microwave oven and the toaster are plugged

in at the same time, then 1800 watts are being used by the

circuit, which is the point at which the circuit overloads. So,

you cannot have both the microwave oven and the toaster

plugged into the circuit at the same time.

2.2 Exercises (pp. 65 – 66)


1. Because 5 is subtracted from each side of the inequality, then

by the Subtraction Property of Inequality, the inequalities are

equivalent.

2. Sample answer: When you are solving both equations and

inequalities with addition, in order to isolate the variable on

one side, you undo subtraction by adding the same number

to each side, which produces an equivalent equation or

inequality, respectively.


3. k + 11 < −3

You would subtract 11 from each side.

4. v − 2 > 14

You would add 2 to each side.

5. −1 ≥ b − 9

You would add 9 to each side.

6. −6 ≤ 17 + p

You would subtract 17 from each side.

7. x − 4 < −5

+ 4 + 4

x < −1

The solution is x < −1.

0 2 4−2−4

−1


Chapter 2

8. 1 ≤ s − 8

+ 8 + 8

9 ≤ s

The solution is s ≥ 9.

4 6 8 10 12 14

9

9. 6 ≥ m − 1

+ 1 + 1

7 ≥ m


0 2 4 6 8 10

7

10. c − 12 > −4

+ 12 + 12

c > 8

The solution is c > 8.

4 6 8 10 12 14

11. r + 4 < 5

− 4 − 4

r < 1

The solution is r < 1.

0 2 4−2−4

1

12. −8 ≤ 8 + y

− 8 − 8

−16 ≤ y

The solution is y ≥ −16.

−20 −18 −16 −14 −12 −10

13. 9 + w > 7

− 9 − 9

w > −2

The solution is w > −2.

0 2 4−2−4

14. 15 ≥ q + 3

− 3 − 3

12 ≥ q

The solution is q ≤ 12.

7 9 11 13 15 17

12

15. h − (−2) ≥ 10

h + 2 ≥ 10

− 2 − 2

h ≥ 8

The solution is h ≥ 8.

3 5 7 9 11 13

8

16. −6 > t − (−13)

−6 > t + 13

− 13 − 13

−19 > t

The solution is t < −19.

−25 −23 −21 −19 −17 −15

17. j + 9 − 3 < 8

j + 6 < 8

− 6 − 6

j < 2

The solution is j < 2.

0 2 4−2−4

18. 1 − 12 + y ≥ −5

−11 + y ≥ −5

+ 11 + 11

y ≥ 6

The solution is y ≥ 6.

0 2 4 6 8 10

19. 10 ≥ 3p − 2p − 7

10 ≥ p − 7

+ 7 + 7

17 ≥ p

The solution is p ≤ 17.

11 13 15 17 19 21

20. 18 − 5z + 6z > 3 + 6

18 + z > 9

− 18 − 18

z > −9

The solution is z > −9.

−12 −10 −8 −6 −4 −2

−9


Chapter 2

21. A number plus 8 is greater than 11.

n + 8 > 11

n + 8 > 11

− 8 − 8

n > 3

An inequality is n + 8 > 11, and the solution is n > 3.

22. A number minus 3 is at least −5.

n − 3 ≥ −5

n − 3 ≥ −5

+ 3 + 3 n ≥ −2

An inequality is n − 3 ≥ −5, and the solution is n ≥ −2.

23. The difference of a number and 9 is fewer than 4.

n − 9 < 4

n − 9 < 4

+ 9 + 9 n < 13

An inequality is n − 9 < 4, and the solution is n < 13.

24. Six is less than or equal to the sum of a number and 15.

6 ≤ n + 15

6 ≤ n + 15

− 15 − 15

−9 ≤ n

An inequality is 6 ≤ n + 15, and the solution is n ≥ −9.

25. a. Words: Current

weight of

your bag+

Additional

weight ≤ Maximum

weight

Variable: Let w be how much weight (in pounds) you

can add to your bag.

Inequality: 38 + w ≤ 50

38 + w ≤ 50

− 38 − 38

w ≤ 12

So, you can add no more than 12 pounds to your bag.

b. w ≤ 12

9 + 5 ≤?

12

14 ≤ 12

no; Because 9 + 5 = 14 is not less than or equal to 12,

you cannot add both a laptop and a pair of boots to your

bag without going over the weight limit.

26. Words: Price of

book + Additional

money spent≥ Amount for

free shipping

Variable: Let x be how much more you must spend.

Equation: 19.76 + x ≥ 25

19.76 + x ≥ 25

− 19.76 − 19.76

x ≥ 5.24

In addition to the book, you must spend at least $5.24 in

order to get free shipping.

27. The values that are to the right of −3 should have been

shaded, not the values to the left of −3.

−17 < x − 14

+ 14 + 14

−3 < x

So, the solution is x > −3.

−6 −4 −2 0

−3

28. A 10 should have been added to each side of the inequality,

not just the left.

−10 + x ≥ −9

−10 + 10 + x ≥ −9 + 10

x ≥ 1 So, the solution is x ≥ 1.

0 2 4−2

1

29. Words: Number

of goals

so far+

Additional

goals

needed≥

Record

number

of goals

Variable: Let g be the possible numbers of

additional goals the player can score.

Inequality: 59 + g ≥ 92

59 + g ≥ 92

− 59 − 59

g ≥ 33

In order to match or break the NHL record, the player must

score 33 or more goals.


Chapter 2

30. a.

Words: Your

1st

jump

score

+

Your

2nd

jump

score

> Competitor’s

1st jump

score+

Competitor’s

2nd jump

score

Variable: Let s be your score on your second jump.

Inequality: 119.5 + s > 117.1 + 119.8

119.5 + s > 117.1 + 119.8

119.5 + s > 236.9

− 119.5 − 119.5

s > 117.4

So, you must earn a score greater than 117.4 points on

your second jump.

b. Both are correct because 118.4 > 117.4 and 117.5 > 117.4.

31. A; By the Subtraction Property of Equality you can subtract

3 from each side of the inequality. x − b < 3, and the result

when simplifi ed is the equivalent inequality x − b − 3 < 0.

D; By the Addition Property of Equality, you can add

(b − x − 3) to each side of the inequality, and the result

when simplifi ed is −3 < b − x.

32. Perimeter < 51.3

x + 14.2 + 15.5 < 51.3

x + 29.7 < 51.3

− 29.7 − 29.7

x < 21.6

So, the side labeled x is less than 21.6 inches in length.

33. Perimeter ≤ 18.7

x + 6.4 + 4.9 + 4.1 ≤ 18.7

x + 15.4 ≤ 18.7

− 15.4 − 15.4

x ≤ 3.3

So, the side labeled x is at most 3.3 feet long.

34. Sample answer: You received $50 for your birthday, and you

have already decided to buy a shirt that costs $34. What is

the most you can spend on some new songs and application

downloads? An inequality to represent this situation is

34 + x ≤ 50, and the solution is x ≤ 16, which matches

the graph.

35. no; You cannot check all of the numbers in the solution set

of an inequality, because the solution set contains infi nitely

many numbers.

x − 11 ≥ −3

+ 11 + 11

x ≥ 8

For the equation x − 11 ≥ −3, the solution set includes

all numbers greater than or equal to 8. So, you can check

numbers such as 8, 9.5, 55

— 3 , or 2,576. All will make the

inequality true as well as all the numbers in between

and greater.

36. a. no; Sample answer: The set of students who have brown

hair and the set of students who have brown eyes are

separate sets with some overlap. You cannot determine

from the diagram whether there are more students with

brown hair or more with brown eyes.

b. no; Sample answer: Because you cannot determine

whether there are more students with brown hair or more

with brown eyes, adding 10 to the number of students

with brown hair does not guarantee that the sum will be at

least as big as the number of students with brown eyes.

c. yes; Sample answer: The number of students with brown

hair is at least as large as the number of students with both

brown hair and brown eyes because the set of students

with both brown hair and brown eyes is contained

completely within the set of students with brown hair.

d. yes; Sample answer: Because the inequality H ≥ X from

part (c) must be true, if you add 10 to the number of

students with brown hair, the sum is still at least as large

as the number of students with both brown hair and

brown eyes.

e. no; Sample answer: It is possible for all of the students

with brown hair also to have brown eyes, in which case H

would be equal to X, and the inequality H > X would not

be true.

f. yes; Sample answer: Because the inequality H ≥ X from

part (c) must be true if you add 10 to the number of

students with brown hair, the sum is always going to be

greater than the number of students with brown eyes.

37. a. x + 8 ≥ 14

− 8 − 8

x ≥ 6

The numbers that are not solutions of x + 8 < 14 are x ≥ 6.

0 2 4 6 8 10

b. x − 12 < 5.7

+ 12 + 12

x < 17.7

The numbers that are not solutions of x − 12 ≥ 5.7 are

x < 17.7.

17.0 17.2 17.4 17.6 17.8 18.0

17.7


Chapter 2

38. c − 3 ≥ d b + 4 < a + 1 a − 2 ≤ d − 7

+ 3 + 3 − 1 − 1 + 7 + 7 c ≥ d + 3 b + 3 < a a + 5 ≤ d

Because b + 3 < a, it must be true that b < a. Similarly,

because a + 5 ≤ d, you can conclude that a < d. Finally,

because c ≥ d + 3, you know that c > d, or d < c. So,

b < a < d < c.


39. 7 ⋅ (−9) = −63 40. −11 ⋅ (−12) = 132

41. −27 ÷ (−3) = 9 42. 20 ÷ (−5) = −4

43. 6x = 24 Check: 6x = 24

6x — 6 =

24 —

6 6(4) =

? 24

x = 4 24 = 24 ✓


44. −3y = −18 Check: −3y = −18

−3y — −3

= −18

— −3

−3(6) =?

−18

y = 6 −18 = −18 ✓


45. s —

−8 = 13 Check:

s —

−8 = 13

−8 ⋅ s —

−8 = −8(13)

−104 —

−8 =

? 13

s = −104 13 = 13 ✓

The solution is s = −104.

46. n —

4 = −7.3 Check:

n —

4 = −7.3

4 ⋅ n —

4 = 4 ⋅ (−7.3)

−29.2 —

4 =

? −7.3

n = −29.2 −7.3 = −7.3 ✓

The solution is n = −29.2.


1. a. x −1 0 1 2 3 4 5

3x −3 0 3 6 9 12 15

6 <?

3x No No No No Yes Yes Yes

The graph on the right represents the solution set of the

inequality 6 < 3x. So, the solution is x > 2.

b. i. 2x < 4

x −1 0 1 2 3 4 5

2x −2 0 2 4 6 8 10

2x <?

4 Yes Yes Yes No No No No


ii. 3 ≥ 3x

x −1 0 1 2 3 4 5

3x −3 0 3 6 9 12 15

3 ≥?

3x Yes Yes Yes No No No No

The solution is x ≤ 1.

iii. 2x < 8

x −1 0 1 2 3 4 5

2x −2 0 2 4 6 8 10

2x <?

8 Yes Yes Yes Yes Yes No No


iv. 6 ≥ 3x

x −1 0 1 2 3 4 5

3x −3 0 3 6 9 12 15

6 ≥?

3x Yes Yes Yes Yes No No No


Dividing each side of an inequality by the same positive

number produces an equivalent inequality.


Chapter 2

2. a. x −5 −4 −3 −2 −1 0 1

−3x 15 12 9 6 3 0 −3

6 <?

−3x Yes Yes Yes No No No No

The graph on the left represents the solution set of the

inequality 6 < −3x. So, the solution is x < −2.

b. i. −2x < 4

x −5 −4 −3 −2 −1 0 1

−2x 10 8 6 4 2 0 −2

−2x <?

4 No No No No Yes Yes Yes

The solution is x > −2.

ii. 3 ≥ −3x

x −5 −4 −3 −2 −1 0 1

−3x 15 12 9 6 3 0 −3

3 ≥?

−3x No No No No Yes Yes Yes


iii. −2x < 8

x −5 −4 −3 −2 −1 0 1

−2x 10 8 6 4 2 0 −2

−2x <?

8 No No Yes Yes Yes Yes Yes


iv. 6 ≥ −3x

x −5 −4 −3 −2 −1 0 1

−3x 15 12 9 6 3 0 −3

6 ≥?

−3x No No No Yes Yes Yes Yes


When dividing each side of an inequality by the same

negative number, the direction of the inequality symbol

must be reversed to produce an equivalent inequality.

3. Divide each side of the inequality by the same number. If the

number is positive, this produces an equivalent inequality. If

the number is negative, the inequality must be reversed to

be equivalent.

4. a. 7x < −21 b. 12 ≤ 4x

12

— 4 ≤

4x —

4

3 ≤ x


7x — 7 <

−21 —

7

x < −3


c. 10 < −5x d. −3x ≤ 0

−3x — −3

≥ 0 —

−3

x ≥ 0


10

— −5

> −5x

— −5

−2 > x



1. n — 7 ≥ −1 2. −6.4 ≥

1 —

5 w

5 ⋅ (−6.4) ≥ 5 ⋅ 1 —

5 w

−32 ≥ w The solution is w ≤ −32.

−34 −32 −30

7 ⋅ n — 7 ≥ 7 ⋅ (−1)

n ≥ −7 The solution is n ≥ −7.

−9 −7 −5

3. 4b ≥ 36 4. −18 > 1.5q

−18

— 1.5

> 1.5q

— 1.5

−12 > q

The solution is q < −12.

−14 −12 −10

4b — 4 ≥

36 —

4

b ≥ 9

The solution is b ≥ 9.

7 9 11

5. p —

−4 < 7 6.

x — −5

≤ −5

−5 ⋅ x — −5

≥ −5 ⋅ (−5)

x ≥ 25


21 23 25 27 29

−4 ⋅ p — −4

> −4 ⋅ 7

p > −28

The solution is p > −28.

−32 −30 −28 −26 −24

7. 1 ≥ − 1 — 10 z

−10 ⋅ 1 ≤ −10 ⋅ ( − 1 — 10 z )

−10 ≤ z

The solution is z ≥ −10.

−6−8−10 −4 −2 0

8. −9m > 63

−9m — −9

< 63 — −9

m < −7

The solution is m < −7.

−6−8−10 −4 −2 0

−7


Chapter 2

9. −2r ≥ −22 10. −0.4y ≥ −12

−0.4y — −0.4

≤ −12

— −0.4

y ≤ 30

The solution is y ≤ 30.

26 28 30 32 34

−2r — −2

≤ −22

— −2

r ≤ 11

The solution is r ≤ 11.

7 9 11 13 15

11. Words: Cost per

copy ⋅ Number

of copies ≤ 3.65

Variable: Let c be the numbers of copies you can make.

Inequality: 0.25 ⋅ c ≤ 3.65

0.25c ≤ 3.65

0.25c — 0.25

≤ 3.65 — 0.25

c ≤ 14.6 Because you cannot make a partial copy, you can make at

most 14 copies.

12. Consider the formula for distance d = rt.

Words: Distance ≤ Maximum

speed limit ⋅ Time

Variable: Let t be the time (in hours) it takes the school bus

to travel 165 miles.

Inequality: 165 ≤ 55 ⋅ t

165 ≤ 55t

165

— 55

≤ 55t —

55

3 ≤ t

If the maximum speed limit for a school bus is 55 miles per

hour, it will take the school bus at least 3 hours to travel

165 miles.



1. Sample answer: When you solve 2x < −8, you divide each

side by positive 2 in order to undo multiplication, which

produces an equivalent inequality. When you solve

−2x < 8, you divide each side by negative 2. So, you must

reverse the direction of the inequality symbol in order to

produce an equivalent inequality.

2. Sample answer: −5x ≥ 30


3. 4x < 8 4. 3y ≤ −9

3y — 3 ≤

−9 —

3

y ≤ −3

The solution is y ≤ −3.

0−2−4−6

−3

4x — 4 <

8 —

4

x < 2


0 2 4 6−2

5. −20 ≤ 10n 6. 35 < 7t

35

— 7 <

7t —

7

5 < t

The solution is t > 5.

0 2 4 6 8

5

−20

— 10

≤ 10n

— 10

−2 ≤ n

The solution is n ≥ −2.

0 2−2−4−6

7. x — 2 > −2 8.

a — 4 < 10.2

4 ⋅ a —

4 < 4 ⋅ 10.2

a < 40.8

The solution is a < 40.8.

40.4 40.6 40.8 41.0 41.2

2 ⋅ x —

2 > 2 ⋅ (−2)

x > −4


0 2−2−4−6

9. 20 ≥ 4 —

5 w 10. −16 ≤

8 —

3 t

3 — 8 ⋅ (−16) ≤ 3 —

8 ⋅

8 —

3 t

−6 ≤ t The solution is t ≥ −6.

−6−8 −4 −2 0

5 — 4 ⋅ 20 ≥ 5 —

4 ⋅

4 —

5 w

25 ≥ w The solution is w ≤ 25.

20 22 24 26 28

25

11. −6t < 12 12. −9y > 9

−9y — −9

< 9 —

−9

y < −1 The solution is y < −1.

0 2 4−2−4

−1

−6t — −6

> 12

— −6

t > −2 The solution is t > −2.

0 2 4−2−4

13. −10 ≥ −2z 14. −15 ≤ −3c

−15

— −3

≥ −3c

— −3

5 ≥ c

The solution is c ≤ 5.

0 2 4 6 8

5

−10 — −2

≤ −2z

— −2

5 ≤ z The solution is z ≥ 5.

0 2 4 6 8

5


Chapter 2

15. n —

−3 ≥ 1 16.

w — −5

≤ 16

−5 ⋅ w —

−5 ≥ −5 ⋅ 16

w ≥ −80

The solution is w ≥ −80.

−84 −82 −80 −78 −76

−3 ⋅ n —

−3 ≤ −3 ⋅ 1

n ≤ −3

The solution is n ≤ −3.

−6−8 −4 −2 0

−3

17. −8 < − 1 — 4 m

−4 ⋅ (−8) > −4 ⋅ ( − 1 — 4 m )

32 > m

The solution is m < 32.

28 30 32 34 36

18. −6 > − 2 — 3 y

− 3 — 2 ⋅ (−6) < −

3 — 2 ⋅ ( −

2 — 3 y )

9 < y

The solution is y > 9.

5 7 9 11 13

19. Words: Cost per

fi sh ⋅ Number

of fi sh ≤ Amount you

have to spend

Variable: Let p be the prices (in dollars) you can pay per

fi sh.

Inequality: p ⋅ 5 ≤ 12

5p ≤ 12

5p — 5 ≥

12 —

5

p ≤ 2.4

You can spend no more than $2.40 per fi sh.

20. Words: Change in

temperature

per hour⋅

Number

of

hours≤

Total

change in

temperature

Variable: Let t be the time (in hours) the temperature

is dropping.

Inequality: −8 ⋅ t ≤ −36

−8t ≤ −36

−8t — −8

≥ −36 — −8

t ≥ 4.5

It will take at least 4.5 hours for the temperature to drop at

least 36° F.

21. 36 < 3y

16

−6

0

6

36

— 3 <

3y —

3

12 < y

The solution is y > 12.

22. 17v ≥ 51

4

−3

−4

3

17v — 17

≥ 51

— 17

v ≥ 3

The solution is v ≥ 3.

23. 2 ≤ − 2 — 9 x

0

−6

−16

6

− 9 — 2 ⋅ 2 ≥ −

9 — 2 ⋅ ( −

2 — 9 x )

−9 ≥ x

The solution is x ≤ −9.

24. 4 > n —

−4

0

−6

−16

6

−4 ⋅ 4 < −4 ⋅ n —

−4

−16 < n

The solution is n > −16.

25. 2x > 3 —

4

2

−1.5

−2

1.5

1 — 2 ⋅ 2x >

1 —

2 ⋅

3 —

4

x > 3 —

8

The solution is x > 3 —

8 .

26. 1.1y < 4.4

6

−3

−2

3

1.1y — 1.1

< 4.4

— 1.1

y < 4

The solution is y < 4.

27. The direction of the inequality symbol should be reversed only

if each side is multiplied or divided by a negative number.

−6 > 2 —

3 x

3 — 2 ⋅ (−6) >

3 —

2 ⋅

2 —

3 x

−9 > x


28. Because each side of the inequality is divided by negative

4, the direction of the inequality symbol must be reversed to

produce an equivalent inequality.

−4y ≤ −32

−4y — −4

≥ −32

— −4

y ≥ 8 The solution is y ≥ 8.


Chapter 2

29. Words: Cost

per square

foot⋅

Area of

carpet≤

Total

amount you

can spend

Variable: Let c be the costs per square foot that you

can pay.

Inequality: $ c —

ft2 ⋅ (14 ft)(14 ft) ≤ $700

$c ⋅ 196 ft2

— ft2

≤ $700

$c ⋅ 196 ≤ $700

$c ⋅ 196

— 196

≤ $700 —

196

$c ≤ $25 — 7 ≈ $3.57

So, you can spend no more than $3.57 per square foot on

carpet for your bedroom.

30. a. C; Multiplying each side by m results in x < −m.

b. A; Multiplying each side by m results in x > m.

c. B; Multiplying each side by m results in x < m.

d. D; Multiplying each side by −m and reversing the

inequality symbol results in x > −m.

31. a. Consider the formula for distance d = rt.

Words: Distance

run≤

Maximum

speed ⋅ Number

of hours

Variable: Let d be the distance (in miles) you run.

Inequality: d ≤ 6.3 ⋅ 2

d ≤ 6.3 ⋅ 2

d ≤ 12.6

You run at most 12.6 miles.

b. d ≤ 6.3 ⋅ 4

d ≤ 25.2

Your friend is correct. In 4 hours, the farthest you could

run is 25.2 miles.

32. x — 4 ≤ 5

4 ⋅ x —

4 ≤ 4 ⋅ 5

x ≤ 20

Sample answer: 5x ≤ 100

33. Words: Production

cost per

penny⋅

Number

of pennies

produced>

Total spent

in production

costs

Variable: Let n be how many pennies are produced.

Inequality: $0.02 ⋅ n > 6,000,000

0.02n > 6,000,000

0.02n — 0.02

> 6,000,000

— 0.02

n > 300,000,000

Over 300 million pennies are produced when the U.S. Mint

pays more than $6 million in production costs.

34. −3x ≤ −2

−3x — −3

≥ −2

— −3

x ≥ 2 — 3

no; By the Division Property of Inequality for c < 0,

−3x ≤ −2 is equivalent to x ≥ 2 —

3 as shown. So, −3x ≤ −2 is

not equivalent to x ≤ 2 — 3 .

35. a. A > B or B < A

b. −A < −B or −B > −A

c. As numbers move farther from zero on a number line,

their absolute values become larger; A > B and ∣ A ∣ > ∣ B ∣ ; −A < −B and ∣ A ∣ > ∣ B ∣ .

36. Sample answer: If x > 0, then when you multiply each side

by x, 4 ≥ 2x would be equivalent to 4 —

x ≥ 2. However, if

x < 0, then when you multiply each side by x, the direction of

the inequality symbol must be reversed, and 4 ≤ 2x is

equivalent to 4 —

x ≥ 2.

37. r > 5

C

— 2π

> 5

2π ⋅ C

— 2π

> 2π ⋅ 5

C > 10π So the circumference of the circle is greater than 10π units

in length.


Chapter 2

38. Consider the formula for distance d = rt.

Words: Distance

Time < Speed

Variable: Let d be the possible distances a beginner can

travel.

Inequality: d —

0.75 < 18 ( Note: 45 min = 0.75 h )

d —

0.75 < 18

0.75 ⋅ d —

0.75 < 0.75 ⋅ 18

d < 13.5

A beginner will travel less than 13.5 miles in 45 minutes of

practice time.

39. Words: Total

number of

employees⋅

Fraction

of the

employees

who work

full-time

≥

Maximum

number of

employees

who work

part-time

Variable: Let p be the fraction of the employees who work

part-time.

Inequality: 36 ⋅ p ≥ 12

36p ≥ 12

36p — 36

≥ 12

— 36

p ≥ 1 —

3

At least 1 —

3 of the employees work part-time.

0 1 2

13


40. 5x + 3 = 13 Check: 5x + 3 = 13

− 3 − 3 5(2) + 3 =?

13

5x = 10 10 + 3 =?

13

5x — 5 =

10 —

5 13 = 13 ✓

x = 2


41. 1 — 2 y − 8 = −10 Check:

1 —

2 y − 8 = −10

+ 8 + 8 1 —

2 (−4) − 8 =

? −10

1 —

2 y = −2 −2 − 8 =

? −10

2 ⋅ 1 —

2 y = 2 ⋅ (−2) −10 = −10 ✓

y = −4

The solution is y = −4.

42. −3n + 2 = 2n − 3 Check: −3n + 2 = 2n − 3

+ 3n + 3n −3(1) + 2 =?

2(1)−3

2 = 5n − 3 −3 + 2 =?

2 − 3

+ 3 + 3 −1 = −1 ✓

5 = 5n

5 —

5 =

5n —

5

1 = n

The solution is n = 1.

43. 1 —

2 z + 4 =

5 —

2 z − 8

− 1 —

2 z −

1 —

2 z

4 = 2z − 8 ( 5 — 2 −

1 —

2 =

4 —

2 = 2 )

+ 8 + 8

12 = 2z

12

— 2 =

2z —

2

6 = z

The solution is z = 6.

Check: 1 — 2 z + 4 =

5 —

2 z − 8

1 —

2 (6) + 4 =

? 5 —

2 (6) − 8

3 + 4 =?

15 − 8

7 = 7 ✓

44. 85% = 0.85 and 0.8 = 0.80

So, 85% > 0.8.

45. 50% = 50

— 100

= 1 —

2 =

15 —

30

So, 16

— 30

> 15

— 30

.

46. 120% = 1.2

So, 120% > 0.12.

47. 60% = 60

— 100

= 3 —

5 =

9 —

15 and

2 —

3 =

10 —

15

So, 2 —

3 > 60%.


Chapter 2


1. a. 2x + 3 ≤ x + 5 Write the inequality.

− x − x Subtract x from each side.

x + 3 ≤ 5 Simplify.

− 3 − 3 Subtract 3 from each side.

x ≤ 2 Simplify.

The solution is x ≤ 2, which is represented by graph B.

b. −2x + 3 > x + 9 Write the inequality.

+ 2x + 2x Add 2x to each side.

3 > 3x + 9 Simplify.

− 9 − 9 Subtract 9 from each side.

−6 > 3x Simplify.

−6

— 3 >

3x —

3 Divide each side by 3.

−2 > x Simplify.

The solution is x < −2, which is represented by graph A.

c. 27 ≥ 5x + 4x Write the inequality.

27 ≥ 9x Simplify.

27

— 9 ≥

9x —


3 ≥ x Simplify.

The solution is x ≤ 3, which is represented by graph E.

d. −8x + 2x − 16 < −5x + 7x Write the inequality.

−6x − 16 < 2x Simplify.


−16 < 8x Simplify.

−16

— 8 <

8x —


−2 < x Simplify.

The solution is x > −2, which is represented by graph C.

e. 3(x − 3) − 5x > −3x − 6 Write the inequality.

3(x) − 3(3) − 5x > −3x − 6 Distributive Property

3x − 9 − 5x > −3x − 6 Simplify.

−2x − 9 > −3x − 6 Simplify.


x − 9 > −6 Simplify.

+ 9 + 9 Add 9 to each side.

x > 3 Simplify.

The solution is x > 3, which is represented by graph D.

f. −5x − 6x ≤ 8 − 8x − x Write the inequality.

−11x ≤ 8 − 9x Simplify.


−2x ≤ 8 Simplify.

−2x

— −2

≥ 8 —

−2 Divide each side by −2

and reverse the direction

of the inequality symbol.

x ≥ −4 Simplify.

The solution is x ≥ −4, which is represented by graph F.

2. To solve a multi-step inequality, simplify each side of the

inequality, if possible. Then use inverse operations to isolate

the variable. Be sure to reverse the inequality symbol when

multiplying or dividing by a negative number.

3. Sample answer: 5x − 7 ≥ 12x and 4(x + 3) ≤ x + 9

Check: 5x − 7 ≥ 12x 4(x + 3) ≤ x + 9

4(x) + 4(3) ≤ x + 9

4x + 12 ≤ x + 9

− x − x

3x + 12 ≤ 9

− 12 − 12

3x ≤ −3

3x

— 3 ≤

−3 —

3

x ≤ −1

− 5x − 5x

−7 ≥ 7x

−7

— 7 ≥

7x —

7

−1 ≥ x

The solution of both inequalities is x ≤ 1.


1. 4b − 1 < 7

+ 1 + 1

4b < 8

4b — 4 <

8 —

4

b < 2

The solution is b < 2.

0 2 4−2−4

2. 8 − 9c ≥ −28

− 8 − 8

−9c ≥ −36

−9c — −9

≤ −36

— −9

c ≤ 4

The solution is c ≤ 4.

0 2 4 6 8

3. n — −2

+ 11 > 12

− 11 − 11

n —

−2 > 1

−2 ⋅ n —

−2 < −2 ⋅ 1

n < −2

The solution is n < −2.

0 2−2−4−6


Chapter 2

4. 6 ≥ 5 − v —

3

− 5 − 5

1 ≥ − v —

3

−3 ⋅ 1 ≤ −3 ⋅ ( − v —

3 )

−3 ≤ v

The solution is v ≥ −3.

0 2−2−4−6

−3

5. 5x − 12 ≤ 3x − 4

− 3x − 3x

2x − 12 ≤ −4

+ 12 + 12

2x ≤ 8

2x — 2 ≤

8 —

2

x ≤ 4


6. 2(k − 5) < 2k + 5

2(k) − 2(5) < 2k + 5

2k − 10 < 2k + 5

− 2k − 2k

−10 < 5

The inequality −10 < 5 is true. So, all real numbers

are solutions.

7. −4(3n − 1) > −12n + 5.2

−4(3n) − 4(−1) > −12n + 5.2

−12n + 4 > −12n + 5.2

+ 12n + 12n

4 > 5.2

The inequality 4 > 5.2 is false. So, there is no solution.

8. 3(2a − 1) ≥ 10a − 11

3(2a) − 3(1) ≥ 10a − 11

6a − 3 ≥ 10a − 11

− 6a − 6a

−3 ≥ 4a − 11

+ 11 + 11

8 ≥ 4a

8 —

4 ≥

4a —

4

2 ≥ a

The solution is a ≤ 2.

9. 95 + 91 + 77 + 89 + x —— 5 ≥ 85

352 + x —

5 ≥ 85

5 ⋅ 352 + x

— 5 ≥ 5 ⋅ 85

352 + x ≥ 425

x ≥ 73

A score of at least 73 points will allow you to advance.

2.4 Exercises (pp. 77 –78)


1. Sample answer: For solving both multi-step inequalities and

multi-step equations, you begin by simplifying each side,

if necessary. Then, you use inverse operations to isolate the

variable. The processes are different, however, because when

solving multi-step inequalities only, you have to remember to

reverse the inequality symbol when multiplying or dividing

by a negative number.

2. Sample answer: Because the terms with the variable are the

same, they will cancel. Then the resulting inequality, 8 ≤ −3

is false.


3. 7b − 4 ≤ 10

+ 4 + 4

7b ≤ 14

7b — 7 ≤

14 —

7

b ≤ 2

The solution is b ≤ 2, which is represented by graph B.

4. 4p + 4 ≥ 12

− 4 − 4

4p ≥ 8

4p — 4 ≥

8 —

4

p ≥ 2

The solution is p ≥ 2, which is represented by graph A.

5. −6g + 2 ≥ 20

− 2 − 2

−6g ≥ 18

−6g — −6

≤ 18

— −6

g ≤ −3

The solution is g ≤ −3, which is represented by graph C.


Chapter 2

6. 3(2 − f ) ≤ 15

3(2 − f )

— 3 ≤

15 —

3

2 − f ≤ 5

−2 −2

−f ≤ 3

(−1)(−f ) ≥ (−1)(3)

The solution is f ≥ −3, which is represented by graph D.

7. 2x − 3 > 7

+ 3 + 3

2x > 10

2x

— 2 >

10 —

2

x > 5

The solution is x > 5.

0 2 4 6 8

5

8. 5y + 9 ≤ 4

− 9 − 9

5y ≤ −5

5y — 5 ≤

−5 —

5

y ≤ −1

The solution is y ≤ −1.

0 2 4−2−4

−1

9. −9 ≤ 7 − 8v

− 7 − 7

−16 ≤ −8v

−16

— −8

≥ −8v

— −8

2 ≥ v

The solution is v ≤ 2.

0 2 4−2−4

10. 2 > −3t − 10

+ 10 + 10

12 > −3t

12

— −3

< −3t

— −3

−4 < t

The solution is t > −4.

−6−8 −4 −2 0

11. w — 2 + 4 > 5

− 4 − 4

w — 2 > 1

2 ⋅ w

— 2 > 2 ⋅ 1

w > 2

The solution is w > 2.

0 2 4−2−4

12. 1 + m

— 3 ≤ 6

− 1 − 1

m

— 3 ≤ 5

3 ⋅ m

— 3 ≤ 3 ⋅ 5

m ≤ 15


0 4 8 12 16 20

15

13. p —

−8 + 9 > 13

− 9 − 9

p —

−8 > 4

−8 ⋅ p —

−8 < −8 ⋅ 4

p < −32

The solution is p < −32.

−36 −34 −32 −30 −28

14. 3 + r —

−4 ≤ 6

− 3 − 3

r —

−4 ≤ 3

−4 ⋅ r —

−4 ≥ −4 ⋅ 3

r ≥ −12

The solution is r ≥ −12.

−16 −12 −8 −4 0

15. 6 ≥ −6(a + 2)

6 —

−6 ≤

−6(a + 2) —

−6

−1 ≤ a + 2

−2 −2

−3 ≤ a

The solution is a ≥ −3.

−6−8 −4 −2 0

−3


Chapter 2

16. 18 ≤ 3(b − 4)

18

— 3

≤ 3(b − 4)

— 3

6 ≤ b − 4

+ 4 + 4 10 ≤ b The solution is b ≥ 10.

0 4 8 12 16

10

17. 4 − 2m > 7 − 3m

+ 3m + 3m

4 + m > 7

− 4 − 4

m > 3

The solution is m > 3.

18. 8n + 2 ≤ 8n − 9

− 8n − 8n

2 ≤ −9

The inequality 2 ≤ −9 is false. So, there is no solution.

19. −2d − 2 < 3d + 8

+ 2d + 2d

−2 < 5d + 8

− 8 − 8

−10 < 5d

−10

— 5 <

5d —

5

−2 < d

The solution is d > −2.

20. 8 + 10f > 14 − 2f

+ 2f + 2f

8 + 12f > 14

− 8 − 8

12f > 6

12f

— 12

> 6 —

12

f > 1 —

2

The solution is f > 1 —

2 .

21. 8g − 5g − 4 ≤ −3 + 3g

3g − 4 ≤ −3 + 3g

− 3g − 3g

−4 ≤ −3

The inequality −4 ≤ −3 is true. So, all real numbers are

solutions.

22. 3w − 5 > 2w + w − 7

3w − 5 > 3w − 7

− 3w − 3w

−5 > −7

The inequality −5 > −7 is true. So, all real numbers

are solutions.

23. 6(ℓ+ 3) < 3(2ℓ+ 6)

6(ℓ) + 6(3) < 3(2ℓ) + 3(6)

6ℓ + 18 < 6ℓ + 18

− 6ℓ − 6ℓ

18 < 18

The inequality 18 < 18 is false. So, there is no solution.

24. 2(5c − 7) ≥ 10(c − 3)

2(5c) − 2(7) ≥ 10(c) − 10(3)

10c − 14 ≥ 10c − 30

− 10c − 10c

−14 ≥ −30

The inequality −14 ≥ −30 is true. So, all real numbers are

solutions.

25. 4 ( 1 — 2 t − 2 ) > 2(t − 3)

4 ( 1 — 2 t ) − 4(2) > 2(t) − 2(3)

2t − 8 > 2t − 6

− 2t − 2t

−8 > −6

The inequality −8 > −6 is false. So, there is no solution.

26. 15 ( 1 — 3 b + 3 ) ≤ 6(b + 9)

15 ( 1 — 3 b ) + 15(3) ≤ 6(b) + 6(9)

5b + 45 ≤ 6b + 54

− 5b − 5b

45 ≤ b + 54

− 54 − 54

−9 ≤ b

The solution is b ≥ −9.

27. 9j − 6 + 6j ≥ 3(5j − 2)

15j − 6 ≥ 3(5j) − 3(2)

15j − 6 ≥ 15j − 6

− 15j − 15j

−6 ≥ −6

The inequality −6 ≥ −6 is true So, all real numbers are

solutions.


Chapter 2

28. 6h − 6 + 2h < 2(4h − 3)

8h − 6 < 2(4h) − 2(3)

8h − 6 < 8h − 6

− 8h − 8h

−6 < −6

The inequality −6 < −6 is false. So, there is no solution.

29. In the fi rst step, you need to use the Distributive Property on

the left side.

x — 4 + 6 ≥ 3

4 ( x — 4 + 6 ) ≥ 4 ⋅ 3

4 ( x — 4 ) + 4(6) ≥ 12

x + 24 ≥ 12

− 24 − 24

x ≥ −12


30. The inequality −2 ≤ −7 is false because −2 is farther to the

right on the number line than −7. So, the original inequality

has no solution.

−2(1 − x) ≤ 2x − 7

−2(1) − 2(−x) ≤ 2x − 7

−2 + 2x ≤ 2x − 7

− 2x − 2x

−2 ≤ −7

Because −2 ≤ −7 is false, there is no solution.

31. Words: Value

per bill ⋅ Number

of bills +Minimum

balance≤ Current

balance

Variable: Let n be the number of $20 bills you can

withdraw.

Inequality: 20 ⋅ n + 100 ≤ 320

20n + 100 ≤ 320

− 100 − 100

20n — 20

≤ 220

— 20

n ≤ 11

You can withdraw no more than 11 $20 bills.

32. Words: Minimum

earnings

per hour ⋅

Number

of

hours+

Cost

of

materials≤

Selling

price

Variable: Let h be how long (in hours) the woodworker

can spend building the cabinet.

Inequality: 25 ⋅ h + 125 ≤ 500

25h + 125 ≤ 500

25h ≤ 375

h ≤ 15

In order to earn at least $25 an hour, the woodworker can

spend no more than 15 hours building the cabinet.

33. Area > 60 ft2

12 ft ⋅ (2x − 3)ft > 60 ft2

12(2x − 3) > 60

12(2x) − 12(3) > 60

24x − 36 > 60

+ 36 + 36

24x > 96

24x — 24

> 96

— 24

x > 4

So, the value of x must be greater than 4.

34. Words: Forest Park

membership

fee

+ Forest Park

nightly fee ⋅ Number

of nights

≥

Woodland

membership

fee

+

Woodland

nightly fee

⋅ Number

of nights

Variable: Let n be the number of nights you plan to camp.

Inequality: 100 + 35n ≥ 20 + 55n

100 + 35n ≥ 20 + 55n

− 35n − 35n

100 ≥ 20 + 20n

− 20 − 20

80 ≥ 20n

80

— 20

≥ 20n

— 20

4 ≥ n

no; If you plan to camp for less than 4 nights, Woodland

Campgrounds would charge less and therefore be the better

choice. If you plan to camp for more than 4 nights, Forest

Park Campgrounds charges less and would therefore be the

better choice.


Chapter 2

35.

8 ft8 ft

74 ft

24 ft

h

Use the Pythagorean Theorem to fi nd the height the ladder

will reach.

h2 + 242 = 742

h2 = 742 − 242

h2 = 4900

h = 70

So, the ladder and the truck can reach a height of

70 + 8 = 78 feet. Solve the inequality 10n ≤ 78 to fi nd how

many stories the ladder can reach.

10n ≤ 78

10n

— 10

≤ 78

— 10

n ≤ 7.8

The ladder can reach 7 stories.

36. a. no more than $40

b. Gasoline is $3.55 per gallon, and a car wash is $8.

c. 3.55x + 8 ≤ 40

d. The graph of y = 3.55x + 8 intersects y = 40 when

x = 9. So, you can buy no more than 9 gallons of gas.

37. Area ≥ 9(π − 2)

π r2 − 4 ( 1 — 2 r ⋅ r ) ≥ 9(π − 2)

π r2 − 2r2 ≥ 9(π − 2)

r2(π − 2) ≥ 9(π − 2)

r2(π − 2)

— (π − 2)

≥ 9(π − 2)

— (π − 2)

r2 ≥ 9

√—

r2 ≥ √—

9

r ≥ 3

So, when r is at least 3 units long, the area of the shaded

region is greater than or equal to 9(π − 2).

38. Sample answer: Let x be the runner’s time (in minutes) in his

last race.

25.5 + 24.3 + 24.8 + 23.5 + x ———

5 < 24

98.1 + x —

5 < 24

5 ⋅ 98.1 + x

— 5 < 5 ⋅ 24

98.1 + x < 120

− 98.1 − 98.1

x < 21.9

In order to achieve his goal, the runner must fi nish his last

race in under 21.9 minutes.

39. a(x + 3) < 5x + 15 − x

a(x) + a(3) < 4x + 15

ax + 3a < 4x + 15

If a = 4, then an equivalent inequality is 4x + 12 < 4x + 15,

and when you subtract 4x from each side, you get 12 < 15,

which is always true. So, all real numbers are solutions of the

original inequality when a = 4.

40. 3x + 8 + 2ax ≥ 3ax − 4a

3x + 2ax + 8 ≥ 3ax − 4a

x(3 + 2a) + 8 ≥ 3ax − 4a

Let 3 + 2a = 3a.

− 2a − 2a

3 = a

Then x[3 + 2(3)] + 8 ≥ 3(3)x − 4(3)

x(3 + 6) + 8 ≥ 9x − 12

9x + 8 ≥ 9x − 12

− 9x − 9x

8 ≥ −12

If a = 3, then the inequality 8 ≥ −12 is equivalent to the

original inequality. Because 8 ≥ −12 is always true, all real

numbers are a solution of the original inequality when a = 3.


41. Six times a number y is less than or equal to 10.

6y ≤ 10

An inequality is 6y ≤ 10.

42. A number p plus 7 is greater than 24.

p + 7 > 24

An inequality is p + 7 > 24.

43. The quotient of a number r and 7 is no more than 18.

r — 7 ≤ 18

An inequality is r — 7 ≤ 18.


Chapter 2

2.1–2.4 What Did You Learn? (p. 79)

1. Sample answer: Because the Xianren Bridge is the world’s

longest natural arch, no other natural arch is the same length

or longer. So, the length of every other arch is less than

400 feet, the length of the Xianren Bridge.

2. Sample answer: If you make a mistake and write your

inequality incorrectly for part (a), then you may also get the

wrong answer when you answer part (b). So, checking its

reasonableness is important to be sure that the correct result

is used.

3. Sample answer: Because carpet is sold by the square foot,

you need to know the area of the room in square feet. To fi nd

the area of the room, you multiply the length and width of

the room, and because both were measured in feet, the units

of the product are square feet.

2.1–2.4 Quiz (p. 80)

1. A number z minus 6 is greater than or equal to 11.

z − 6 ≥ 11

An inequality is z − 6 ≥ 11.

2. Twelve is no more than the sum of −1.5 times a number w and 4.

12 ≤ −1.5w + 4

An inequality is 12 ≤ −1.5w + 4.

3. x < 0 4. x ≥ 8

5. 9 + q ≤ 15

− 9 − 9 q ≤ 6

The solution is q ≤ 6.

2 4 6 80−2

6. z − (−7) < 5

z + 7 < 5

− 7 − 7 z < −2

The solution is z < −2.

0 2−2−4

7. −3 < y − 4

+ 4 + 4

1 < y The solution is y > 1.

0 2 4

1

8. 3p ≥ 18

3p — 3 ≥

18 —

3

p ≥ 6

The solution is p ≥ 6.

0 2 4 6 8 10

9. 6 > w —

−2

−2 ⋅ 6 < −2 ⋅ w — −2

−12 < w

The solution is w > −12.

0−2−4−6−8−10−12

10. −20x > 5

−20x — −20

< 5 —

−20

x < − 1 —

4

The solution is x < − 1 —

4 .

−1

−

− 0

14

12

11. 3y − 7 ≥ 17

+ 7 + 7

3y

— 3 ≥

24 —

3

y ≥ 8

The solution is y ≥ 8.

12. 8(3g − 2) ≤ 12(2g + 1)

8(3g) − 8(2) ≤ 12(2g) + 12(1)

24g − 16 < 24g + 12

− 24g − 24g

−16 < 12

The inequality −16 < 12 is true. So, all real numbers are

solutions.

13. 6(2x − 1) ≥ 3(4x + 1)

6(2x) − 6(1) ≥ 3(4x) + 3(1)

12x − 6 ≥ 12x + 3

− 12x − 12x

−6 ≥ 3

The inequality −6 ≥ 3 is false. There is no solution.


Chapter 2

14. a. Let s be the distance (in yards) you can swim.

s ≥ 100

1000

Let t be how long (in minutes) you can tread water.

t ≥ 5

0 2 4 6 8−2

5

Let u be the length (in yards) you can swim under water

without taking a breath.

u ≥ 10

0 10 20

b. 250 ft ⋅ 1 yd

— 3 ft

= 250 yd

— 3 ≈ 83.3 yd

35 ft ⋅ 1 yd

— 3 ft

= 35 yd

— 3 ≈ 11.7 yd

no; Your swimming distance, 250 feet, is approximately

equal to 83.3 yards, which is less than the required

minimum of 100 yards. So, even though the other two

satisfy the respective requirements, this one does not.

15. Words: Current volume

of water + Additional

volume ≤ Maximum

volume

Variable: Let V be the additional volumes the pelican’s

bill can contain.

Inequality: 100 + V ≤ 700

100 + V ≤ 700

− 100 − 100

V ≤ 600

The pelican’s bill can contain at most an additional

600 cubic inches.

16. a. Words: Amount saved

per week ⋅ Number

of weeks≥

Minimum

price

Variable: Let w be how long (in weeks) you need to

save to purchase a bike.

Inequality: 15 ⋅ w ≥ 120

15w ≥ 120

15w — 15

≥ 120

— 15

w ≥ 8

You need to save for at least 8 weeks in order to have

enough money to purchase a bike.

b. It will take less time to save up enough money to buy a

bike. The inequality would now be 65 + 15w ≥ 120.

65 + 15w ≥ 120

− 65 − 65

15w ≥ 55

15w — 15

≥ 55 —

15

w ≥ 11 —

3 , or 3

2 —

3

So, it will take at least 3 2 —

3 weeks to save up the rest of the

money needed to buy a bike.


1. a. x ≥ −6 and x < 3

b. x > −5 and x ≤ 4

c. x ≥ −4 and x ≤ 5

d. x > −3 and x < 6

e. Sample answer: You use “and” because the numbers in

the shaded interval make both inequalities true.

2. a. x ≤ −6 or x > 3

b. x < −5 or x ≥ 4

c. x ≤ −4 or x ≥ 5

d. x < −3 or x > 6

e. Sample answer: You use “or” because the numbers in the

shaded interval make either inequality true, but not both.

3. Sample answer: You can join two inequalities with the word

“and” or the word “or”. If you are describing all the numbers

between two endpoints, you use the word “and” because

the numbers in the interval make both inequalities true. If

you are describing the numbers either below or above two

endpoints, you use the word “or”, because the numbers in the

shaded interval make either inequality true, but not both.


1. A number d is more than 0 and less than 10.

d > 0 and d < 10

An inequality is 0 < d < 10.

0 4 8 12−4

10

2. A number a is fewer than −6 or no less than −3.

a < −6 or a ≥ −3

An inequality is a < −6 or a ≥ −3.

0−2−4−6−8

−3


Chapter 2

3. 5 ≤ m + 4 < 10

− 4 − 4 − 4 1 ≤ m < 6

The solution is 1 ≤ m < 6.

0 2 4 6 8

1

4. −3 < 2k − 5 < 7

+ 5 + 5 + 5 2 < 2k < 12

2 —

2 <

2k —

2 <

12 —

2

1 < k < 6

The solution is 1 < k < 6.

0 2 4 6 8

1

5. 4c + 3 ≤ −5 or c − 8 > −1

− 3 − 3 + 8 + 8 4c ≤ −8 c > 7

4c — 4 ≤

−8 —

4

c ≤ −2 or c > 7

The solution is c ≤ −2 or c > 7.

0 4 8−4

−2 7

6. 2p + 1 < −7 or 3 − 2p ≤ −1

− 1 − 1 − 3 − 3 2p < −8 −2p ≤ −4

2p — 2 <

−8 —

2

−2p —

−2 ≥

−4 —

−2

p < −4 or p ≥ 2

The solution is p < −4 or p ≥ 2.

0 4 8−4−8

2

7. −40 ≤ C ≤ 15

−40 ≤ 5 —

9 (F − 32) ≤ 15

9 —

5 ⋅ (−40) ≤

9 —

5 ⋅

5 —

9 (F − 32) ≤

9 —

5 ⋅ 15

−72 ≤ F − 32 ≤ 27

+ 32 + 32 + 32

−40 ≤ F ≤ 59

The solution is −40 ≤ F ≤ 59. So, the temperature rating of

the winter boots is −40°F to 59°F.



1. Sample answer: The graph of −6 ≤ x ≤ −4 and the graph of

x ≤ −6 or x ≥ −4 both have closed dots as endpoints on −6

and −4. The fi rst one, however, is shaded between −6 and

−4, and the second one is shaded below −6 and above −4.

2. Sample answer: The compound inequality that does not belong is a < 6 or a > −9. When you graph this compound

inequality, the whole number line will be shaded, because

the shaded regions of the individual inequalities overlap. So,

all real numbers are a solution of one part or the other of the

compound inequality.


3. −3 < x ≤ 2 (x > −3 and x ≤ 2)

4. 7 < x < 14 (x > 7 and x < 14)

5. x ≤ −7 or x ≥ −4 6. x ≤ 4 or x > 6

7. A number p is less than 6 and greater than 2.

p < 6 and p > 2

An inequality is 2 < p < 6.

0 2 4 6 8

8. A number n is less than or equal to −7 or greater than 12.

n ≤ −7 or n > 12

An inequality is n ≤ −7 or n > 12.

0 8 16−8

12−7

9. A number m is more than −7 2 —

3 or at most −10.

m > −7 2 —

3 or m ≤ −10

An inequality is m ≤ −10 or m > −7 2 —

3 .

−13 −11 −9 −7 −5

−10 −723

10. A number r is no less than −1.5 and fewer than 9.5.

r ≥ −1.5 and r < 9.5

An inequality is −1.5 ≤ r < 9.5.

0 2 4 6 8 10−2

9.5−1.5

11. Let e be the depths (in feet) of slitsnails.

−2500 ≤ e ≤ −100

0 1000−1000−2000−3000

−100−2500


Chapter 2

12. a. Let h be the elevations of trees in the low-elevation forest

zone.

1700 < h ≤ 2500

b. Let h be the elevations of fl owers in the subalpine and

alpine zones.

4000 < h ≤ 14,410

13. 6 < x + 5 ≤ 11

− 5 − 5 − 5 1 < x ≤ 6

The solution is 1 < x ≤ 6.

0 2 4 6 8

1

14. 24 > −3r ≥ −9

24

— −3

< −3r

— −3

≤ −9

— −3

−8 < r ≤ 3 The solution is −8 < r ≤ 3.

0 4−4−8

3

15. v + 8 < 3 or −8v < −40

− 8 − 8 −8v

— −8

> −40

— −8

v < −5 or v > 5

The solution is v < −5 or v > 5.

0 4 8−4−8

−5 5

16. −14 > w + 3 or 3w ≥ −27

− 3 − 3 3w

— 3 ≥

−27 —

3

−17 > w or w ≥ −9

The solution is w < −17 or w ≥ −9.

0−6−12−18−24

−9−17

17. 2r + 3 < 7 or −r + 9 ≤ 2

− 3 − 3 − 9 − 9

2r < 4 −r ≤ −7

2r — 2 <

4 —

2

−r —

−1 ≥

−7 —

−1

r < 2 or r ≥ 7 The solution is r < 2 or r ≥ 7.

0 4 8 12

72

18. −6 < 3n + 9 < 21

− 9 − 9 − 9 −15 < 3n < 12

−15

— 3 <

3n —

3 <

12 —

3

−5 < n < 4

The solution is −5 < n < 4.

0 4 8−4−8

−5

19. −12 < 1 —

2 (4x + 16) < 18

2 ⋅ (−12) < 2 ⋅ 1 —

2 (4x + 16) < 2 ⋅ 18

−24 < 4x + 16 < 36

− 16 − 16 − 16

−40 < 4x < 20

−40

— 4 <

4x —

4 <

20 —

4

−10 < x < 5

The solution is −10 < x < 5.

0 6−6−12

−10 5

20. 35 < 7(2 − b) or 1 —

3 (15b − 12) ≥ 21

1 —

3 (15b) −

1 —

3 (12) ≥ 21

5b − 4 ≥ 21

+ 4 + 4

5b ≥ 25

5b — 5 ≥

25 —

5

b ≥ 5

35 < 7(2) − 7(b)

35 < 14 − 7b

− 14 − 14

21 < − 7b

21

— −7

> −7b

— −7

−3 > b or

The solution is b < −3 or b ≥ 5.

0 4 8−4

−3 5

21. In the second step, 3 should have been subtracted from the 4

as well as the 3 and the 9.

4 < −2x + 3 < 9

− 3 − 3 − 3

1 < −2x < 6

1 —

−2 >

−2x —

−2 >

6 —

−2

− 1 —

2 > x > −3

The solution is −3 < x < − 1 —

2 .

0−2−4

−12


Chapter 2

22. The graph should be shaded to the left of −10 and to the

right of 5, but not between them.

x − 2 > 3 or x + 8 < −2

+ 2 + 2 − 8 − 8 x > 5 or x < −10

The solution is x < −10 or x > 5.

0 10−10

5

23. −20 ≤ C ≤ −15

−20 ≤ 5 —

9 (F − 32) ≤ −15

9 —

5 ⋅ (−20) ≤

9 —

5 ⋅

5 —

9 (F − 32) ≤

9 —

5 ⋅ (−15)

−36 ≤ F − 32 ≤ −27

+ 32 + 32 + 32

−4 ≤ F ≤ 5

The solution is −4 ≤ F ≤ 5. So, the possible temperatures of

the interior of the iceberg are −4°F to 5°F.

24. Let h be the heights of skiers the shop does not provide

skis for.

1.16h < 150 or 1.16h > 220

1.16h — 1.16

< 150

— 1.16

1.16h

— 1.16

> 220

— 1.16

h < 150

— 1.16

or h > 220

— 1.16

The shop does not provide skis for skiers of heights less than

150

— 1.16

or about 129.3 centimeters tall or greater than 220

— 1.16

or

about 189.6 centimeters tall.

25. 22 < −3c + 4 < 14

− 4 − 4 − 4 18 < −3c < 10

18

— −3

> −3c

— −3

> 10

— −3

−6 > c > −3 1 —

3

Because there are no numbers that are both greater than

−3 1 —

3 and less than −6, the inequality has no solution. Note

that the original inequality states that 22 < 14, so there is no

solution.

26. 2m − 1 ≥ 5 or 5m > −25

5m

— 5 >

−25 —

5

m > −5

+ 1 + 1 2m ≥ 6

2m — 2 ≥

6 —

2

m ≥ 3 or m > −5

Because all numbers greater than or equal to 3 are also

greater than −5, every number greater than −5 makes at

least one of the inequalities true. So, the solution of the

compound inequality is m > −5.

3 7−9 −5 −1

27. −y + 3 ≤ 8 and y + 2 > 9

− 2 − 2

y > 7

− 3 − 3 −y ≤ 5

−y

— −1

≥ 5 —

−1

y ≥ −5 and y > 7

All numbers greater than 7 are also greater than −5,

and make both inequalities true. So, the solution of the

compound inequality is y > 7.

3 7 11−5 −1

28. x − 8 ≤ 4 or 2x + 3 > 9

− 3 − 3

2x > 6

2x — 2 >

6 —

2

x > 3

+ 8 + 8 x ≤ 12

x ≤ 12 or

The graphs of these two inequalities overlap to cover the

whole number line. So, all real numbers are a solution of one

inequality or the other, and therefore all real numbers are a

solution of the compound inequality.

0 4 8 12 16

29. 2n + 19 ≤ 10 + n or −3n + 3 < −2n + 33

− n − n + 3n + 3n

n + 19 ≤ 10 3 < n + 33

− 19 − 19 −33 −33

n ≤ −9 or −30 < n

The graphs of these two inequalities overlap to cover the

whole number line. So, all real numbers are a solution of one

inequality or the other, and therefore all real numbers are a

solution of the compound inequality.

0−40 −30 −20 −10


Chapter 2

30. 3x − 18 < 4x − 23 and x − 16 < −22

+ 16 + 16

x < −6

− 3x − 3x

−18 < x − 23

+ 23 + 23

5 < x and x < −6

Because there is no overlap between the numbers greater

than 5 and less than −6, there are no numbers that make both

inequalities true. So, the compound inequality has

no solution.

31. 4(x − 6) ? 2(x − 10) and 5(x + 2) ≥ 2(x + 8)

5(x) + 5(2) ≥ 2(x) + 2(8)

5x + 10 ≥ 2x + 16

− 2x − 2x

3x + 10 ≥ 16

− 10 − 10

3x ≥ 6

3x — 3 ≥

6 —

3

x ≥ 2

4(x) − 4(6) ? 2(x) − 2(10)

4x − 24 ? 2x − 20

− 2x − 2x

2x − 24 ? −20

+ 24 + 24

2x ? 4

2x — 2 ?

4 —

2

x ? 2 and

In order for the solution of the compound inequality to be

only one value, the fi rst inequality must be

4(x − 6) ≤ 2(x − 10). Then the solutions of the inequalities

are x ≤ 2 and x ≥ 2, and 2 is the only number that satisfi es

both inequalities. So, for the compound inequality to have

a solution of one value, x = 2, the inequality symbol to use

is ≤.

32. Sample answer: A lifeguard training class will be offered as

long as at least 6 people sign up to take it, and there is only

enough equipment for a maximum of 10 people. So, the

number of people p in the class must be 6 ≤ p ≤ 10, which is

modeled by the graph.

33. x + 7 > 5 x + 5 > 7 5 + 7 > x

− 7 − 7 − 5 − 5 12 > x

x > −2 x > 2

no; A value of 1 does not make the inequality x > 2 true.

34. a. Let P be the annual profi ts (in millions of dollars) from

2006 to 2013.

50 ≤ P ≤ 90

b. P = R − C

50 ≤ P ≤ 90

50 ≤ R − C ≤ 90

50 ≤ R − 125 ≤ 90

+ 125 + 125 + 125

175 ≤ R ≤ 215

no; The solution is 175 ≤ R ≤ 215. So, the company’s

annual revenue from 2006 to 2013 was at least $175

million and no more than $215 million. Because $160

million is too small to be in this range, it is not possible

that the company had this annual revenue from 2006

to 2013.


35. ∣ d — 9 ∣ = 6

d — 9 = 6 or

d — 9 = −6

9 ⋅ d — 9 = 9 ⋅ 6 9 ⋅

d — 9 = 9 ⋅ (−6)

d = 54 d = −54

The solutions are d = −54 and d = 54.

0 40 80−80 −40

54−54

36. 7 ∣ 5p − 7 ∣ = −21

7 ∣ 5p − 7 ∣

— 7 =

−21 —

7

∣ 5p − 7 ∣ = −3

Because absolute value expressions must be greater than or

equal to 0, the expression ∣ 5p − 7 ∣ cannot equal − 3. So, the

equation has no solution.


Chapter 2

37. ∣ r + 2 ∣ = ∣ 3r − 4 ∣ r + 2 = 3r − 4 or r + 2 = −(3r − 4)

r + 2 = −3r + 4

+ 3r + 3r

4r + 2 = 4

− 2 − 2

4r = 2

4r — 4 =

2 —

4

r = 1 —

2

− r − r

2 = 2r − 4

+ 4 + 4

6 = 2r

6 —

2 =

2r —

2

3 = r

Check: ∣ r + 2 ∣ = ∣ 3r − 4 ∣ ∣ r + 2 ∣ = ∣ 3r − 4 ∣

∣ 1 — 2 + 2 ∣ =? ∣ 3 ( 1 —

2 ) − 4 ∣

∣ 1 — 2 +

4 —

2 ∣ =? ∣ 3 —

2 −

8 —

2 ∣

5 —

2 =

? ∣ −

5 — 2 ∣

5 —

2 =

5 —

2 ✓

∣ 3 + 2 ∣ =? ∣ 3(3) − 4 ∣

∣ 5 ∣ =? ∣ 9 − 4 ∣

5 =?

∣ 5 ∣ 5 = 5 ✓

The solutions are r = 1 —

2 and r = 3.

0 1 2 3 4

12

38. ∣ 1 — 2 w − 6 ∣ = ∣ w + 7 ∣

1 —

2 w − 6 = w + 7 or

1 —

2 w − 6 = −(w + 7)

1 —

2 w − 6 = −w − 7

+ w + w

3 — 2 w − 6 = −7

+ 6 + 6

3 — 2 w = −1

2 — 3 ⋅

3 —

2 w = 2 —

3 ⋅ (−1)

w = − 2 — 3

− 1 —

2 w −

1 —

2 w

−6 = 1 —

2 w + 7

− 7 − 7

−13 = 1 —

2 w

2 ⋅ (−13) = 2 ⋅ 1 —

2 w

−26 = w

Check:

∣ 1 — 2 w − 6 ∣ = ∣ w + 7 ∣ ∣ 1 —

2 w − 6 ∣ = ∣ w + 7 ∣

∣ 1 — 2 ( −

2 — 3 ) − 6 ∣ =? ∣ −

2 — 3 + 7 ∣

∣ − 1 — 3 − 6 ∣ =? ∣ −

2 — 3 +

21 —

3 ∣

∣ − 1 — 3 −

18 —

3 ∣ =? ∣ 19

— 3 ∣

∣ − 19

— 3 ∣ =? 19

— 3

19 —

3 =

19 —

3 ✓

∣ 1 — 2 (−26) − 6 ∣ =? ∣ −26 + 7 ∣

∣ −13 − 6 ∣ =? ∣ −19 ∣

∣ −19 ∣ =? 19

19 = 19 ✓

The solutions are w = −26 and w = − 2 — 3 .

0−32 −24 −16 −8

−26 −23

39. Mean:

1 + 1 + 2 + 5 + 6 + 8 + 10 + 12 + 12 + 13

———— 10

= 70

— 10

= 7

Absolute deviations: 7 − 1 = 6, 7 − 1 = 6, 7 − 2 = 5,

7 − 5 = 2, 7 − 6 = 1, 8 − 7 = 1, 10 − 7 = 3, 12 − 7 = 5,

12 − 7 = 5, 13 − 7 = 6

Mean absolute deviation:

6 + 6 + 5 + 2 + 1 + 1 + 3 + 5 + 5 + 6

———— 10

= 40

— 10

= 4

So, the data values deviate from the mean by an average of 4

and are clustered close together.

40. Mean:

24 + 26 + 28 + 28 + 30 + 30 + 32 + 32 + 34 + 36

————— 10

= 300

— 10

= 30

Absolute deviations: 30 − 24 = 6, 30 − 26 = 4,

30 − 28 = 2, 30 − 28 = 2, 30 − 30 = 0, 30 − 30 = 0,

32 − 30 = 2, 32 − 30 = 2, 34 − 30 = 4, 36 − 30 = 6

Mean absolute deviation:

6 + 4 + 2 + 2 + 0 + 0 + 2 + 2 + 4 + 6

———— 10

= 28

— 10

= 2.8

So, the data values deviate from the mean by an average of

2.8 and are clustered close together.


1. a. Sample answer: The inequality ∣ x + 2 ∣ ≤ 3 is true when

the expression x + 2 has a value less than or equal to 3

and greater than or equal to −3.

x + 2 ≥ −3 and x + 2 ≤ 3

b. x + 2 ≥ −3 and x + 2 ≤ 3

− 2 − 2 − 2 − 2 x ≥ −5 and x ≤ 1

The solution is x ≥ −5 and x ≤ 1 (−5 ≤ x ≤ 1).

c. Sample answer: Write a compound inequality representing

the distance between the absolute value expression and 0.

2. a. 0 2 4 6 8 10−2−4−6−8−10

x + 2 = 0

−2 + 2 =?

0

0 = 0 ✓

b.

0 2 4 6 8 10−2−4−6−8−10

−5 1

All of the values between −5 and −1 are solutions of the

absolute value inequality ∣ x + 2 ∣ ≤ 3 and the compound

inequality from Exploration 1.

c. Plot the distances to determine the endpoints of the

solution.


Chapter 2

3. a. Ax-6-5-4-3-2-1012

B|x + 2|432101234

21

345678910

The solutions given by the spreadsheet are −5 and 1 and

the numbers in between, because they are the values of x

that make ∣ x + 2 ∣ less than or equal to 3.

b. The spreadsheet method yields the same solutions as the

other two methods.

c. Sample answer: You can use a spreadsheet to quickly

calculate the values of an absolute value expression for

many values of the variable, and fi nd the ones that give the

expected solution.

4. Sample answer: You can solve an absolute value inequality

algebraically, graphically on a number line, or numerically

with a spreadsheet.

5. Sample answer: The algebraic method allows you to

calculate exact values for the boundaries of the solution

sets for all types of absolute value inequalities, even when

the boundary points are not whole numbers. The graphical

method provides a visual representation of how the endpoint

of the solution set are related to each other and to the other

numbers in the inequality, but it can be tedious to create

and work with the graph especially if you do not know

approximate values of the boundary points and/or they are

not whole numbers. The numerical method helps you see

how the values of an absolute value expression change as the

value of the variable changes, but it can be time-consuming

especially if you do not know approximate values of the

boundary points, and it may not provide an exact answer if

the boundary points are not whole numbers.

2.6 Monitoring Progress (pp. 89–90)

1. ∣ x ∣ ≤ 3.5

x ≥ −3.5 and x ≤ 3.5

The solution is −3.5 ≤ x ≤ 3.5.

0 2 4−2−4

−3.5 3.5

2. ∣ k − 3 ∣ < −1

By defi nition, the absolute value of an expression must be

greater than or equal to 0. The expression ∣ k − 3 ∣ cannot be

less than −1. So, the inequality has no solution.

3. ∣ 2w − 1 ∣ < 11

2w − 1 > −11 and 2w − 1 < 11

+ 1 + 1 + 1 + 1 2w > −10 2w < 12

2w — 2 >

−10 —

2

2w —

2 <

12 —

2

w > −5 and w < 6

The solution is −5 < w < 6.

0 4 8−4−8

−5 6

4. ∣ x + 3 ∣ > 8

x + 3 < −8 or x + 3 > 8

− 3 − 3 − 3 − 3 x < −11 or x > 5

The solution is x < −11 or x > 5.

0 6−6−12

5−11

5. ∣ n + 2 ∣ − 3 ≥ −6

+ 3 + 3 ∣ n + 2 ∣ ≥ −3 By defi nition, the absolute value of an expression must be

greater than or equal to 0. The expression ∣ n + 2 ∣ will always

be greater than −3. So, all real numbers are solutions.

0 2 4−2−4

6. 3 ∣ d + 1 ∣ − 7 ≥ −1

+ 7 + 7 3 ∣ d + 1 ∣ ≥ 6

3 ∣ d + 1 ∣

— 3 ≥ 6 —

3

∣ d + 1 ∣ ≥ 2 d + 1 ≤ −2 or d + 1 ≥ 2

− 1 − 1 − 1 − 1 d ≤ −3 or d ≥ 1

The solution is d ≤ −3 or d ≥ 1.

0 2−2−4

−3 1


Chapter 2

7. The mean price is 6640

— 10

= $664.

Let x represent a price you are willing to pay.

∣ x − 664 ∣ ≤ 75

−75 ≤ x − 664 ≤ 75

+ 664 + 664 + 664

589 ≤ x ≤ 739

The solution is 589 ≤ x ≤ 739. So, the prices you will

consider must be at least $589 and at most $739. Five prices

meet your condition: $650, $660, $670, $650, and $725.

2.6 Exercises (pp. 91–92)


1. yes; By defi nition, the absolute value of an expression must

be greater than or equal to 0. The expression ∣ 4x − 2 ∣ will

always be greater than −6. So, all real numbers are solutions.

2. Solving the inequality ∣ w − 9 ∣ ≤ 2 requires a compound

inequality joined by “and”. Solving the inequality

∣ w − 9 ∣ ≥ 2 requires a compound inequality joined by “or”.


3. ∣ x ∣ < 3

x > −3 and x < 3


0 2 4−2−4

3−3

4. ∣ y ∣ ≥ 4.5

y ≤ −4.5 or y ≥ 4.5

The solution is y ≤ −4.5 or y ≥ 4.5.

0 4 8−4−8

−4.5 4.5

5. ∣ d + 9 ∣ > 3

d + 9 < −3 or d + 9 > 3

− 9 − 9 − 9 − 9 d < −12 or d > −6

The solution is d < −12 or d > −6.

0−16 −12 −4−8

−6

6. ∣ h − 5 ∣ ≤ 10

−10 ≤ h − 5 ≤ 10

+ 5 + 5 + 5 −5 ≤ h ≤ 15

The solution is −5 ≤ h ≤ 15.

0 10 20−10−20

15−5

7. ∣ 2s − 7 ∣ ≥ −1

By defi nition, the absolute value of an expression must

be greater than or equal to 0. The expression ∣ 2s − 7 ∣ will always be greater than −1. So, all real numbers

are solutions.

0 2 4−2−4

8. ∣ 4c + 5 ∣ > 7

4c + 5 < −7 or 4c + 5 > 7

− 5 − 5 − 5 − 5 4c < −12 4c > 2

4c — 4 <

−12 —

4

4c —

4 >

2 —

4

c < −3 or c > 1 —

2

The solution is c < −3 or c > 1 —

2 .

0 2−2−4

−3 12

9. ∣ 5p + 2 ∣ < − 4


greater than or equal to 0. The expression ∣ 5p + 2 ∣ cannot be


10. ∣ 9 − 4n ∣ < 5

−5 < 9 − 4n < 5

− 9 − 9 − 9

−14 < −4n < −4

−14

— −4

> −4n

— −4

> −4

— −4

3.5 > n > 1

The solution is 1 < n < 3.5.

0 1 2 3 4

3.5

11. ∣ 6t − 7 ∣ − 8 ≥ 3

+ 8 + 8

∣ 6t − 7 ∣ ≥ 11

6t − 7 ≤ −11 or 6t − 7 ≥ 11

+ 7 + 7 + 7 + 7

6t ≤ −4 6t ≥ 18

6t — 6 ≤

−4 —

6

6t —

6 ≥ 18

— 6

t ≤ − 2 —

3 or t ≥ 3

The solution is t ≤ − 2 —

3 or t ≥ 3.

0 2 4−2

− 323


Chapter 2

12. ∣ 3j − 1 ∣ + 6 > 0

− 6 − 6 ∣ 3j − 1 ∣ > −6


greater than or equal to 0. The expression ∣ 3j − 1 ∣ will always be greater than −6. So, all real numbers

are solutions.

0 2 4−2−4

13. 3 ∣ 14 − m ∣ > 18

3 ∣ 14 − m ∣

— 3 >

18 —

3

∣ 14 − m ∣ > 6

14 − m < −6 or 14 − m > 6

− 14 − 14 − 14 − 14

−m < −20 −m > −8

−m — −1

> −20

— −1

−m

— −1

< −8

— −1

m > 20 or m < 8

The solution is m < 8 or m > 20.

0 8 16 24 32

20

14. −4 ∣ 6b − 8 ∣ ≤ 12

−4 ∣ 6b − 8 ∣

—— −4

≥ 12

— −4

∣ 6b − 8 ∣ ≥ −3


greater than or equal to 0. The expression ∣ 6b − 8 ∣ will

always be greater than −3. So, all real numbers are solutions

of the original inequality.

0 2 4−2−4

15. 2 ∣ 3w + 8 ∣ −13 ≤ −5

+ 13 + 13

2 ∣ 3w + 8 ∣ ≤ 8

2 ∣ 3w + 8 ∣

— 2 ≤ 8 —

2

∣ 3w + 8 ∣ ≤ 4

−4 ≤ 3w + 8 ≤ 4

− 8 − 8 − 8

−12 ≤ 3w ≤ −4

−12

— 3 ≤

3w —

3 ≤

−4 —

3

−4 ≤ w ≤ − 4 —

3

The solution is −4 ≤ w ≤ − 4 —

3 .

0 2−2−4−6

−43

16. −3 ∣ 2 − 4u ∣ + 5 < −13

− 5 − 5

−3 ∣ 2 − 4u ∣ < −18

−3 ∣ 2 − 4u ∣

—— −3

> −18

— −3

∣ 2 − 4u ∣ > 6

2 − 4u < −6 or 2 − 4u > 6

− 2 − 2 − 2 − 2

−4u < −8 −4u > 4

−4u — −4

> −8

— −4

−4u — −4

< 4 —

−4

u > 2 or u < −1

The solution is u < −1 or u > 2.

0 2 4−2−4

−1

17. 6 ∣ −f + 3 ∣ + 7 > 7

− 7 − 7

6 ∣ −f + 3 ∣ > 0

6 ∣ −f + 3 ∣

— 6 >

0 —

6

∣ −f + 3 ∣ > 0

−f + 3 > 0 or −f + 3 < 0

− 3 − 3 − 3 − 3 −f > −3 −f < −3

−f — −1

< −3

— −1

−f

— −1

> −3

— −1

f < 3 or f > 3

So, the solution is f < 3 or f > 3. Or, in other words, all real

numbers except 3 are solutions of the inequality.

0 2 4 6

3


Chapter 2

18. 2 — 3 ∣ 4v + 6 ∣ − 2 ≤ 10

+ 2 + 2

2 —

3 ∣ 4v + 6 ∣ ≤ 12

3 —

2 ⋅

2 —

3 ∣ 4v + 6 ∣ ≤

3 —

2 ⋅ 12

∣ 4v + 6 ∣ ≤ 18

−18 ≤ 4v + 6 ≤ 18

− 6 − 6 − 6

−24 ≤ 4v ≤ 12

−24

— 4 ≤

4v —

4 ≤

12 —

4

−6 ≤ v ≤ 3

The solution is −6 ≤ v ≤ 3.

0 4−4−8

−6 3

19. Let w be the acceptable numbers of words.

∣ w − 500 ∣ ≤ 30

−30 ≤ w − 500 ≤ 30

+ 500 + 500 + 500

470 ≤ w ≤ 530

The solution is 470 ≤ w ≤ 530. So, essay contest entries must

have at least 470 words and no more than 530 words.

20. Let T be the normal body temperatures (in degrees Celsius)

of a camel throughout the day.

∣ T − 37 ∣ ≤ 3

−3 ≤ T − 37 ≤ 3

+ 37 + 37 + 37

34 ≤ T ≤ 40

The solution is 34 ≤ T ≤ 40. So, a normal body temperature

for a camel is at least 34° C and no more than 40° C.

21. After writing the absolute value inequality, the fi rst step

should be to write a compound inequality with a second

inequality that has the expression greater than −20.

∣ x − 5 ∣ < 20

−20 < x − 5 < 20

+ 5 + 5 + 5

−15 < x < 25


22. Because the absolute value expression is greater than 13, the

compound inequality should use the word “or”, and it should

have the expression less than −13 or greater than 13.

∣ x + 4 ∣ > 13

x + 4 < − 13 or x + 4 > 13

− 4 − 4 − 4 − 4

x < − 17 or x > 9

The solution is x < − 17 or x > 9.

23. A number is less than 6 units from 0.

∣ n − 0 ∣ < 6

∣ n ∣ < 6

n > −6 and n < 6

The solution is −6 < n < 6.

24. A number is more than 9 units from 3.

∣ n − 3 ∣ > 9

n − 3 < −9 or n − 3 > 9

+ 3 + 3 + 3 + 3

n < −6 or n > 12

The solution is n < −6 or n > 12.

25. Half of a number is at most 5 units from 14.

∣ 1 — 2 n − 14 ∣ ≤ 5

−5 ≤ 1 —

2 n − 14 ≤ 5

+ 14 + 14 + 14

9 ≤ 1 —

2 n ≤ 19

2 ⋅ 9 ≤ 2 ⋅ 1 —

2 n ≤ 2 ⋅ 19

18 ≤ n ≤ 38

The solution is 18 ≤ n ≤ 38.

26. Twice a number is no less than 10 units from −1.

∣ 2n − (−1) ∣ ≥ 10

∣ 2n + 1 ∣ ≥ 10

2n + 1 ≤ −10 or 2n + 1 ≥ 10

− 1 − 1 − 1 − 1

2n ≤ −11 2n ≥ 9

2n — 2 ≤

−11 —

2

2n —

2 ≥

9 —

2

n ≤ − 11

— 2 or n ≥

9 —

2

The solution is n ≤ − 11

— 2 or n ≥

9 —

2 .


Chapter 2

27. The mean weight is

0.58 + 0.63 + 0.65 + 0.53 + 0.61

——— 5 =

3 —

5 = 0.6.

Let w be the weights of the gaskets that get thrown out.

∣ w − 0.6 ∣ > 0.06

w − 0.6 < −0.06 or w − 0.6 > 0.06

+ 0.6 + 0.6 + 0.6 + 0.6

w < 0.54 or w > 0.66

The solution is w < 0.54 or w > 0.66. So, gaskets that weigh

less than 0.54 pound or more than 0.66 pound should be

thrown out. The only gasket from this batch that should be

thrown out is the one that weighs 0.53 pound.

28. The mean acceleration is

10.56 + 9.52 + 9.73 + 9.80 + 9.78 + 10.91

———— 6 =

60.3 —

6

= 10.05.

∣ x − 10.05 ∣ ≤ d

Order the data values: 9.52, 9.73, 9.78, 9.80, 10.56, 10.91.

To fi nd the greatest possible value for d, let x equal the least

and greatest values, and solve for d.

Let x = 9.52. Let x = 10.91.

∣ 9.52 − 10.05 ∣ ≤ d ∣ 10.91 − 10.05 ∣ ≤ d

∣ −0.53 ∣ ≤ d ∣ 0.86 ∣ ≤ d

0.53 ≤ d 0.86 ≤ d

Because 0.86 > 0.53, the students can state that the absolute

deviation of each measured value x from the mean is at

most 0.86.

29. The difference between the areas of the fi gures is less than 2.

∣ 1 — 2 (4)(x + 6) − 2(6) ∣ < 2

∣ 1 — 2 (4)(x + 6) − 2(6) ∣ < 2

∣ 2(x + 6) − 12 ∣ < 2

∣ 2(x) + 2(6) − 12 ∣ < 2

∣ 2x + 12 − 12 ∣ < 2

∣ 2x ∣ < 2

−2 < 2x < 2

−2

— 2 <

2x —

2 <

2 —

2

−1 < x < 1

In order for the difference between the areas of the fi gures to

be less than 2, the values of x must be greater than −1 and

less than 1.

30. The difference between the perimeters of the fi gures

is less than

or equal to 3.

∣ 2(x + 1) + 2(3) − 4x ∣ ≤ 3

∣ 2(x + 1) + 2(3) − 4x ∣ ≤ 3

∣ 2(x) + 2(1) + 6 − 4x ∣ ≤ 3

∣ 2x + 2 + 6 − 4x ∣ ≤ 3

∣ −2x + 8 ∣ ≤ 3

−3 ≤ −2x + 8 ≤ 3

− 8 − 8 − 8

−11 ≤ −2x ≤ −5

−11

— −2

≥ −2x

— −2

≥ −5

— −2

5.5 ≥ x ≥ 2.5

The solution is 2.5 ≤ x ≤ 5.5. So, in order for the difference

between the perimeters of the fi gures to be less than or equal

to 3, the value of x must be at least 2.5 and at most 5.5.

31. true

32. false; If a is a solution of ∣ x + 3 ∣ > 8, then a is also a

solution of a compound inequality with “or”. So, a may be

a solution of x + 3 > 8 or x + 3 < −8, but not both. So, a

may or may not be a solution of x + 3 > 8.

33. false; If a is a solution of ∣ x + 3 ∣ ≥ 8, then a is also a

solution of the compound inequality x ≤ −11 or x ≥ 5. The

solution of x + 3 ≥ −8 is x ≥ −11. So, if a < −11, then it is

not a solution of the second inequality.

34. true

35. no; The only real number that is not a solution of ∣ n ∣ > 0 is

n = 0, because ∣ 0 ∣ is not greater than 0.

36. Sample answer:

12 cm

15 cm

P = 2ℓ + 2w Check: ∣ P − 60 ∣ ≤ 12

∣ 54 − 60 ∣ <?

12

∣ −6 ∣ <?

12

6 ≤?

12 ✓

= 2(15) + 2(12)

= 30 + 24

= 54


Chapter 2

37. When c < 0, the inequality ∣ ax + b ∣ < c has no solution.

Because the absolute value expression ∣ ax + b ∣ , by

defi nition, must be greater than or equal to 0, the expression

cannot be less than a negative number c.

Also, when c < 0, all real numbers are solutions of

∣ ax + b ∣ > c. Because the absolute value expression

∣ ax + b ∣ , by defi nition, must be greater than or equal to 0, it

will always be greater than a negative number c.

38. An absolute value inequality for the fi rst graph is ∣ x − 2 ∣ ≥ 3.

Sample answer: The shaded values are 3 or more units away

from 2. So, the solutions’ absolute deviation from 2 must be

greater than or equal to 3.

An absolute value inequality for the second graph is

∣ x − 2 ∣ < 3. Sample answer: The shaded values are less than

3 units away from 2. So, the solutions’ absolute deviation

from 2 must be less than 3.

An absolute value inequality for the third graph is

∣ x − 2 ∣ ≤ 3. Sample answer: The shaded values are 3 or less

units away from 2. So, the solutions’ absolute deviation from

2 must be less than or equal to 3.

An absolute value inequality for the fourth graph is

∣ x − 2 ∣ > 3. Sample answer: The shaded values are greater

than 3 units away from 2. So, the solutions’ absolute

deviation from 2 must be greater than 3.

39. Sample answer: If the absolute value of a number is less

than 5, then it must be less than 5 units away from 0 in

either direction on the number line. So, the graph must be an

intersection because the number must be between −5 and 5. If the absolute value of a number is greater than 5, then

it must be more than 5 units from 0 in either direction on

the number line. So, the graph must be a union because the

number must be either less than −5 or greater than 5.

40. The fi rst step is to rewrite each absolute value inequality as a

compound inequality and solve each part.

∣ x − 3 ∣ < 4 and ∣ x + 2 ∣ > 8

−4 < x − 3 < 4 and x + 2 < −8 or x + 2 > 8

+ 3 + 3 + 3 − 2 − 2 − 2 − 2

−1 < x < 7 and x < −10 or x > 6

The solutions are −1 < x < 7 and x < −10 or x > 6.

Because both compound inequalities must be true, fi nd the

intersection of the graphs.

−1 < x < 7

−1 1 3 5 7 9−3

x < −10 or x > 6 6−10

−4 0−8−12 84

−1 < x < 7 and x < −10 or x > 6

3 4 5 6 7 8 9 10

The solution of ∣ x − 3 ∣ < 4 and ∣ x + 2 ∣ > 8 is 6 < x < 7.


41– 44. y

42−2−4

−4

−2

2

4

x

A

B

D

C

41. Point A is in Quadrant I. 42. Point B is on the y-axis.

43. Point C is in Quadrant III. 44. Point D is in Quadrant II.

45. x 0 1 2

5x + 1 5 ⋅ 0 + 1 = 1 5 ⋅ 1 + 1 = 6 5 ⋅ 2 + 1 = 11

x 3 4

5x + 1 5 ⋅ 3 + 1 = 16 5 ⋅ 4 + 1 = 21

46. x −2 −1

−2x − 3 −2(−2) − 3 = 1 −2(−1) − 3 = −1

x 0 1 2

−2x − 3 −2(0) − 3 = − 3 −2(1) − 3 = −5 −2(2) − 3 = −7

2.5 – 2.6 What Did You Learn? (p. 93)

1. Sample answer: A diagram could show the ranges of

elevation with the zone labeled for each range. Then you

could quickly see the boundaries of each range needed for

writing the inequality that represents the elevations in the

given zone.

2. Take the left expression, fi rst inequality symbol, and middle

expression to form the fi rst inequality. Take the middle

expression, second inequality symbol, and last expression to

form the second inequality.

3. Sample answer: You are given the weights of fi ve gaskets

and the desired absolute deviation of the weights, 0.06 pound.

You are asked to fi nd which gasket weights are not within

0.06 pound of the mean weight of the batch, that is which are

outside of the desired weight range.

4. Sample answer: For Exercises 32 and 33, x = −20 is

a solution of ∣ x + 3 ∣ > 8 and ∣ x + 3 ∣ ≥ 8, because

∣ x + 3 ∣ = ∣ −20 + 3 ∣ = ∣ −17 ∣ = 17, and 17 is greater than

8. However, x = −20 is not a solution of either x + 3 > 8 or

x + 3 ≥ 8, because x + 3 = −20 + 3 = −17, which is less

than both 8 and −8.


Chapter 2

Chapter 2 Review (pp. 94 –96)

1. A number d minus 2 is less than −1.

d − 2 < −1

An inequality is d − 2 < −1.

2. Ten is at least the product of a number h and 5.

10 ≥ h ⋅ 5

An inequality is 10 ≥ 5h.

3. x > 4 4. y ≤ 2

0 2 4 6 8

0 2 4−2−4

5. −1 ≥ z

0 2 4−2−4

−1

6. p + 4 < 10

− 4 − 4

p < 6

The solution is p < 6.

0 2 4 6 8

7. r − 4 < −6

+ 4 + 4

r < −2

The solution is r < −2.

0 2−2−4−6

8. 2.1 ≥ m − 6.7

+ 6.7 + 6.7

8.8 ≥ m

The solution is m ≤ 8.8.

0 4 8 12 16

8.8

9. 3x > −21

3x — 3 >

−21 —

3

x > −7


0−2−4−6−8

−7

10. −4 ≤ g —

5

5 ⋅ (−4) ≤ 5 ⋅ g —

5

−20 ≤ g The solution is g ≥ −20.

0 10−10−20

11. − 3 —

4 n ≤ 3

− 4 —

3 ⋅ ( −

3 —

4 n ) ≥ −

4 —

3 ⋅ 3

n ≥ −4

The solution is n ≥ −4.

0−2−4−6−8

12. s —

−8 ≥ 11

−8 ⋅ s —

−8 ≤ −8 ⋅ 11

s ≤ −88

The solution is s ≤ −88.

0−40−80−120−160

−88

13. 36 < 2q

36

— 2 <

2q —

2

18 < q

The solution is q > 18.

0 8 16 24 32

18

14. −1.2k > 6

−1.2k — −1.2

< 6 —

−1.2

k < −5

The solution is k < −5.

0−2−4−6−8

−5

15. 3x − 4 > 11

+ 4 + 4

3x > 15

3x — 3 >

15 —

3

x > 5

The solution is x > 5.

0 2 4 6 8

5


Chapter 2

16. −4 < b —

2 + 9

− 9 − 9

−13 < b —

2

2 ⋅ (−13) < 2 ⋅ b —

2

−26 < b

The solution is b > −26.

0−32 −24 −16 −8

−26

17. 7 − 3n ≤ n + 3

+ 3n + 3n

7 ≤ 4n + 3

− 3 − 3

4 ≤ 4n

4 —

4 ≤

4n —

4

1 ≤ n

The solution is n ≥ 1.

0 2 4−2−4

1

18. 2(−4s + 2) ≥ −5s − 10

2(−4s) + 2(2) ≥ −5s − 10

−8s + 4 ≥ −5s − 10

+ 8s + 8s

4 ≥ 3s − 10

+ 10 + 10

14 ≥ 3s

14

— 3 ≥

3s —

3

14

— 3 ≥ s

The solution is s ≤ 14

— 3 .

0 2 4 6 8

143

19. 6(2t + 9) ≤ 12t − 1

6(2t) + 6(9) ≤ 12t − 1

12t + 54 ≤ 12t − 1

− 12t − 12t

54 ≤ −1

The inequality 54 ≤ −1 is false. So, there is no solution.

20. 3r − 8 > 3(r − 6)

3r − 8 > 3(r) − 3(6)

3r − 8 > 3r − 18

− 3r − 3r

−8 > −18

The inequality −8 > −18 is true. So, all real numbers

are solutions.

0 2 4−2−4

21. A number x is more than −6 and at most 8.

x > −6 and x ≤ 8

An inequality is −6 < x ≤ 8.

0 4 8−4−8

−6

22. 19 ≥ 3z + 1 ≥ −5

− 1 − 1 − 1

18 ≥ 3z ≥ −6

18

— 3 ≥

3z —

3 ≥

−6 —

3

6 ≥ z ≥ −2

The solution is −2 ≤ z ≤ 6.

0 4 8−4−8

−2 6

23. r — 4 < −5 or −2r − 7 ≤ 3

+ 7 + 7

−2r ≤ 10

−2r

— −2

≥ 10

— −2

r ≥ −5

4 ⋅ r —

4 < 4 ⋅ (−5)

r < −20

r < −20 or

The solution is r < −20 or r ≥ − 5.

0 10−10−20−30

−5

24. ∣ m ∣ ≥ 10

m ≤ −10 or m ≥ 10

The solution is m ≤ −10 or m ≥ 10.

0 10 20−10−20

25. ∣ k−9 ∣ < −4


greater than or equal to 0. The expression ∣ k−9 ∣ cannot be



Chapter 2

26. 4 ∣ f − 6 ∣ ≤ 12

4 ∣ f − 6 ∣

— 4 ≤

12 —

4

∣ f − 6 ∣ ≤ 3

−3 ≤ f − 6 ≤ 3

+ 6 + 6 + 6

3 ≤ f ≤ 9

The solution is 3 ≤ f ≤ 9.

0 4 8 12 16

3 9

27. 5 ∣ b + 8 ∣ − 7 > 13

+ 7 + 7 5 ∣ b + 8 ∣ > 20

5 ∣ b + 8 ∣

— 5 >

20 —

5

∣ b + 8 ∣ > 4

b + 8 < −4 or b + 8 > 4

− 8 − 8 − 8 − 8 b < −12 or b > −4

The solution is b < −12 or b > −4.

0−16 −12 −4−8

28. ∣ −3g − 2 ∣ + 1 < 6

− 1 − 1 ∣ −3g − 2 ∣ < 5

−5 < −3g − 2 < 5

+ 2 + 2 + 2 −3 < −3g < 7

−3

— −3

> −3g

— −3

> 7 —

−3

1 > g > − 7 —

3

The solution is − 7 —

3 < g < 1.

0 2 4−2−4

1−73

29. ∣ 9 − 2j ∣ + 10 ≥ 2

− 10 − 10

∣ 9 − 2j ∣ ≥ −8


greater than or equal to 0. The expression ∣ 9 − 2j ∣ will always be greater than −8. So, all real numbers

are solutions.

0 4 8 12−4

30. Let h be the acceptable heights of a guard rail.

∣ h − 106 ∣ ≤ 7

−7 ≤ h − 106 ≤ 7

+ 106 + 106 + 106

99 ≤ h ≤ 113

The solution is 99 ≤ h ≤ 113. So, the height of a guardrail

should be at least 99 centimeters and no more than

113 centimeters.

Chapter 2 Test (p. 97)

1. The sum of a number y and 9 is at least −1.

y + 9 ≥ −1

An inequality is y + 9 ≥ −1.

2. A number r is more than 0 or less than or equal to −8.

r > 0 or r ≤ −8

An inequality is r > 0 or r ≤ −8.

3. A number k is less than 3 units from 10.

∣ k − 10 ∣ < 3

An inequality is ∣ k − 10 ∣ < 3.

4. x — 2 − 5 ≥ −9

+ 5 + 5

x — 2 ≥ −4

2 ⋅ x —

2 ≥ 2 ⋅ (−4)

x ≥ −8


0−2−4−6−8

5. −4s < 6s + 1

− 6s − 6s

−10s < 1

−10s — −10

> 1 —

−10

s > − 1 —

10

The solution is s > − 1 —

10 .

0−15

15

− 110


Chapter 2

6. 4p + 3 ≥ 2(2p + 1)

4p + 3 ≥ 2(2p) + 2(1)

4p + 3 ≥ 4p + 2

− 4p − 4p

3 ≥ 2

The inequality 3 ≥ 2 is true. So, all real numbers are solutions.

0 2−2

7. −7 < 2c − 1 < 10

+ 1 + 1 + 1

−6 < 2c < 11

−6

— 2

< 2c

— 2 <

11 —

2

−3 < c < 5.5

The solution is −3 < c < 5.5.

−1−2−3−4 0 11 2 3 4 5 6

5.5

8. −2 ≤ 4 − 3a ≤ 13

− 4 − 4 − 4

−6 ≤ −3a ≤ 9

−6

— −3

≥ −3a

— −3

≥ 9 —

−3

2 ≥ a ≥ −3

The solution is −3 ≤ a ≤ 2.

0 2−2−4

−3

9. −5 < 2 − h or 6h + 5 > 71

− 2 − 2 − 5 − 5

−7 < −h 6h > 66

−7

— −1

> −h

— −1

6h

— 6 >

66 —

6

7 > h or h > 11

The solution is h < 7 or h > 11.

0 2 4 6 8 10 12

7 11

10. ∣ 2q + 8 ∣ > 4

2q + 8 < −4 or 2q + 8 > 4

− 8 − 8 − 8 − 8

2q < −12 2q > −4

2q — 2 <

−12 —

2

2q —

2 >

−4 —

2

q < −6 or q > −2

The solution is q < −6 or q > −2.

0−2−4−6−8

11. −2 ∣ y − 3 ∣ − 5 ≥ −4

+ 5 + 5

−2 ∣ y − 3 ∣ ≥ 1

−2 ∣ y − 3 ∣

— −2

≤ 1 —

−2

∣ y − 3 ∣ ≤ − 1 —

2


greater than or equal to 0. The expression ∣ y − 3 ∣ cannot be

less than or equal to − 1 —

2 . So, the inequality has no solution.

12. 4 ∣ −3b + 5 ∣ − 9 < 7

+ 9 + 9

4 ∣ −3b + 5 ∣ < 16

4 ∣ −3b + 5 ∣

—— 4 <

16 —

4

∣ −3b + 5 ∣ < 4

−4 < −3b + 5 < 4

− 5 − 5 − 5

−9 < −3b < −1

−9

— −3

> −3b

— −3

> −1

— −3

3 > b > 1 —

3

The solution is 1 —

3 < b < 3.

0

3

23 11

3 2 223 31

3 4

13

13. Let P be profi t, R be revenue, and E be expenses.

R − E ≥ P

R − 155 ≥ 250

+ 155 + 155

R ≥ 405

Your revenue needs to be at least $405.


Chapter 2

14. Let w be the acceptable widths of a bicycle chain.

∣ w − 0.3 ∣ ≤ 0.0003

−0.0003 ≤ w − 0.3 ≤ 0.0003

+ 0.3 + 0.3 + 0.3

0.2997 ≤ w ≤ 0.3003

The solution is 0.2997 ≤ w ≤ 0.3003. So, a bicycle chain

must be at least 0.2997 inch wide and at most

0.3003 inch wide.

15. If a = c and b < d, then the solution is all real numbers.

If a = c and b > d, then the inequality has no solution.

Now solve for x.

ax + b < cx + d

ax − cx + b < cx − cx + d

ax − cx + b < d

ax − cx + b − b < d − b

ax − cx < d − b

x (a − c) < d − b

If a > c, then a − c > 0. If a < c, then a − c < 0.

x (a − c)

— a − c

< d − b

— a − c

x (a − c)

— a − c

> d − b

— a − c

x < d − b

— a − c

x > d − b

— a − c

16. The numbers that are not solutions are the numbers between

−3 and 2, including 2, but not −3, because −3 is a solution.

So, a compound inequality is −3 < x ≤ 2.

0 2 4−2−4

−3

17. The numbers that are not solutions are the numbers to the

left of −4 and to the right of −1, not including −4 or −1.

They are both solutions because they have closed dots. So, a

compound inequality is x < −4 or x > −1.

0 2 4−2−4−6

−1

18. a. $295 − $175 = $120

Let r be the tax rate.

120 ⋅ r = 7.50

120r — 120

= 7.5

— 120

r = 0.0625, or 6.25%

The tax rate is 0.0625, or 6.25%.

b.

Words: $175 + ( Price of

coat − $175 ) ⋅ ( 1 + Tax rate )

≤

Amount shopper

has to spend

Variable: Let p be the prices of coats that the shopper

can afford.

Inequality: 175 + ( p − 175) ⋅ (1 + 0.0625) ≤ 430

175 + ( p − 175)1.0625 ≤ 430

175 + 1.0625( p) − 1.0625(175) ≤ 430

175 + 1.0625p − 185.9375 ≤ 430

1.0625p − 10.9375 ≤ 430

+ 10.9375 + 10.9375

1.0625p ≤ 440.9375

1.0625p — 1.0625

≤ 440.9375

— 1.0625

p ≤ 415

c.

Words: Price

of

Coat

⋅

( 1

+

Tax

rate in

state B

)

<

$175

+

( Price

of

Coat

−

$175

)

⋅

( 1

+

Tax

rate in

state B

)

Variable: Let p be the prices of coats.

Inequality: p ⋅ (1 + 0.05) < 175 + ( p − 175)(1 + 0.0625)

p(1.05) < 175 + ( p − 175)(1.0625)

1.05p < 175 + 1.0625( p) − 1.0625(175)

1.05p < 175 + 1.0625p − 185.9375

1.05p < 1.0625p − 10.9375

− 1.0625p − 1.0625p

−0.0125p < −10.9375

−0.0125p

— −0.0125

> −10.9375

— −0.0125

p > 875

The 5% tax will be cheaper for coats that cost more than

$875. For example, coats that cost $899, $975, or $1099 will

be cheaper in the state with a 5% sales tax on clothing.

Chapter 2 Standards Assessment (pp. 98 –99)

1. A; The difference between the actual attendance x and the

expected attendance, 65, is at most 30.


Chapter 2

2. a. ax + 4 ≤ 3x + b

5x + 4 ≤ 3x + b

− 3x − 3x

2x + 4 ≤ b

− 4 − 4 2x ≤ b − 4

2x — 2 ≤

b − 4 —

2

x ≤ b − 4

— 2

Let b − 4

— 2 = −3.

2 ⋅ b − 4

— 2 = 2 ⋅ (−3)

b − 4 = −6

+ 4 + 4 b = −2

So, when a = 5 and b = −2, x ≤ −3.

b. ax + 4 ≤ 3x + b

When a = 3, and b = 5 (or any number greater than

or equal to 4), the solution of the inequality is all real

numbers.

Check: ax + 4 ≤ 3x + b

3x + 4 ≤ 3x + 5

− 3x − 3x

4 ≤ 5 ✓

All real numbers are solutions.

c. ax + 4 ≤ 3x + b

When a = 3 and b = 3 (or any number less than 4), the

inequality has no solution.

Check: ax + 4 ≤ 3x + b

3x + 4 ≤ 3x + 3

− 3x − 3x

4 ≤ 3 ✗

The inequality has no solution.

3. 5x − 6 + x ≥ 2x − 8 2(3x + 8) > 3(2x + 6)

6x + 16 > 6x + 18

16 > 18

6x − 6 ≥ 2x − 8

4x − 6 ≥ −8

4x ≥ −2

x ≥ − 1 — 2

17 < 4x + 5 < 21

12 < 4x < 16

3 < x < 4

x − 8 + 4x ≤ 3(x − 3) + 2x

5x − 8 ≤ 3x − 9 + 2x

5x − 8 ≤ 5x − 9 −8 ≤ −9

9x − 3 < 12 or 6x + 2 > −10

9x < 15 6x > −12

x < 5 —

3 x > −2

or

5(x − 1) ≤ 5x − 3

5x − 5 ≤ 5x − 3

−5 ≤ −3

At least one integer solution No integer solutions

5x − 6 + x ≥ 2x − 8 2(3x + 8) > 3(2x + 6)

9x − 3 < 12 or 6x + 2 > −10 17 < 4x + 5 < 21

5(x − 1) ≤ 5x − 3 x − 8 + 4x ≤ 3(x − 3) + 2x

4. a. The season pass is a better deal if it costs less than paying

$25 per play.

180 < 25x

b. 180 < 25x

180

— 25

< 25x

— 25

7.2 < x

The solution is x > 7.2. So, the season pass is a better deal

if you go to more than 7 plays. The number of plays for

which the season pass is not a better deal are 0, 1, 2, 3, 4,

5, 6, and 7.


Chapter 2

5. 3(2x − 4) = 4(ax − 2)

3(2x) − 3(4) = 4(ax) − 4(2)

6x − 12 = 4ax − 8

6x − 4ax − 12 = 4ax − 4ax − 8

6x − 4ax − 12 = −8

6x − 4ax − 12 + 12 = −8 + 12

6x − 4ax = 4

2x(3 − 2a) = 4

2x(3 − 2a)

— 2 =

4 —

2

x(3 − 2a) = 2

The value of x must be positive if (3 − 2a) is positive.

3 − 2a > 0

− 3 − 3

−2a > −3

−2a — −2

< −3

— −2

a < 1.5

The values a that make the solution positive are the ones less

than 1.5: −2, −1, 0, and 1.

6. 4x − 18 ? −x − 3 and −3x − 9 ? −3

+ 3x + 3x

−9 ? 3x − 3

+ 3 + 3

−6 ? 3x

−6

— 3 ?

3x —

3

−2 ? x

+ x + x

5x − 18 ? −3

+ 18 + 18

5x ? 15

5x — 5 ?

15 —

5

x ? 3 and

In order to match the graph, the solution must be x < 3 and

−2 ≤ x. So, the original compound inequality should be

4x − 18 < −x − 3 and −3x − 9 ≤ −3. Note that these were

solved such that the direction of the inequality symbol did

not have to be reversed, because at no time did you multiply

or divide each side by a negative number.

7. a. Words: Price per

pair of

sneakers⋅

Number

of pairs of

sneakers+

Price per

pair of

socks

⋅ Number

of pairs

of socks≤

Amount of

Gift Card

Variable: Let x be the possible numbers of pairs of

socks you can buy.

Inequality: 80 ⋅ 2 + 12 ⋅ x ≤ 250

160 + 12x ≤ 250

− 160 − 160

12x ≤ 90

12x — 12

≤ 90

— 12

x ≤ 7.5

The solution is x ≤ 7.5. So, you cannot buy 8 pairs of

socks, because 8 is not less than or equal to 7.5. You can

buy at most 7 pairs of socks.

b. 60 + 80x ≤ 250

Because $80 is the price per pair of sneakers, x must be

the number of pairs of sneakers. Also because each pair of

socks costs $12, you must have purchased 60

— 12

= 5 pairs of

socks. So, you can solve this inequality that represents the

possible numbers of pairs of sneakers you can buy with a

$250 gift card, when you are also buying 5 pairs of socks.

8. a. Sample answer: ax + b = cx + d

4x + 3 = 2x + 5

− 2x − 2x

2x + 3 = 5

− 3 − 3 2x = 2

2x

— 2 =

2 —

2

x = 1

When a = 4, b = 3, c = 2, and d = 5, the equation has

one solution: x = 1.

b. Sample answer: ax + b = cx + d

2x − 1 = 2x + 6

− 2x − 2x

−1 ≠ 6

When a = c = 2, b = −1, and d = 6, you get an equation

that is false. So, the equation has no solution.

c. Sample answer: ax + b = cx + d

1x + 4 = 1x + 4

− x − x 4 = 4

When a = c = 1, and b = d = 4, you get an equivalent

equation that is true. So, all real numbers are solutions of

the equation.

chapter 2 · 2016-09-30 · x ≥ 1; sample answer: the inequality is true for all values of x that...

Documents