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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.
Copyright © Big Ideas Learning, LLC Algebra 1 63All rights reserved. Worked-Out Solutions
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.
Copyright © Big Ideas Learning, LLC Algebra 1 63All rights reserved. Worked-Out Solutions
64 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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.
Copyright © Big Ideas Learning, LLC Algebra 1 65All rights reserved. Worked-Out Solutions
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 ✓
The solution is y = 2.
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 .
66 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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 .
2.2 Explorations (p. 61)
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:
Attempts Completions Yards Touchdowns Interceptions
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:
Attempts Completions Yards Touchdowns Interceptions
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.
Copyright © Big Ideas Learning, LLC Algebra 1 67All rights reserved. Worked-Out Solutions
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
The solution is x ≥ −3.
+ 2 + 2
6 > x
The solution is x < 6.
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)
Vocabulary and Core Concept Check
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.
Monitoring Progress and Modeling with Mathematics
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
68 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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
The solution is m ≤ 7.
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
Copyright © Big Ideas Learning, LLC Algebra 1 69All rights reserved. Worked-Out Solutions
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.
70 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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
Copyright © Big Ideas Learning, LLC Algebra 1 71All rights reserved. Worked-Out Solutions
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.
Maintaining Mathematical Profi ciency
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 ✓
The solution is x = 4.
44. −3y = −18 Check: −3y = −18
−3y — −3
= −18
— −3
−3(6) =?
−18
y = 6 −18 = −18 ✓
The solution is y = 6.
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.
2.3 Explorations (p. 67)
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
The solution is x < 2.
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
The solution is x < 4.
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
The solution is x ≤ 2.
Dividing each side of an inequality by the same positive
number produces an equivalent inequality.
72 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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
The solution is x ≥ −1.
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
The solution is x > −4.
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
The solution is x ≥ −2.
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
The solution is x ≥ 3.
7x — 7 <
−21 —
7
x < −3
The solution is x < −3.
c. 10 < −5x d. −3x ≤ 0
−3x — −3
≥ 0 —
−3
x ≥ 0
The solution is x ≥ 0.
10
— −5
> −5x
— −5
−2 > x
The solution is x < −2.
2.3 Monitoring Progress (pp. 68 – 70)
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
The solution is 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
Copyright © Big Ideas Learning, LLC Algebra 1 73All rights reserved. Worked-Out Solutions
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.
2.3 Exercises (pp. 71 – 72)
Vocabulary and Core Concept Check
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
Monitoring Progress and Modeling with Mathematics
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
The solution is 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
The solution is 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
74 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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
The solution is x < −9.
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.
Copyright © Big Ideas Learning, LLC Algebra 1 75All rights reserved. Worked-Out Solutions
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.
76 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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
Maintaining Mathematical Profi ciency
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
The solution is 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%.
Copyright © Big Ideas Learning, LLC Algebra 1 77All rights reserved. Worked-Out Solutions
Chapter 2
2.4 Explorations (p. 73)
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 —
9 Divide each side by 9.
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.
+ 6x + 6x Add 6x to each side.
−16 < 8x Simplify.
−16
— 8 <
8x —
8 Divide each side by 8.
−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.
+ 3x + 3x Add 3x to each side.
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.
+ 9x + 9x Add 9x to each side.
−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.
2.4 Monitoring Progress (pp. 74 – 76)
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
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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
The solution is 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)
Vocabulary and Core Concept Check
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.
Monitoring Progress and Modeling with Mathematics
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.
Copyright © Big Ideas Learning, LLC Algebra 1 79All rights reserved. Worked-Out Solutions
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
The solution is 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
80 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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.
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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
The solution is 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.
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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.
Maintaining Mathematical Profi ciency
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.
Copyright © Big Ideas Learning, LLC Algebra 1 83All rights reserved. Worked-Out Solutions
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.
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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.
2.5 Explorations (p. 81)
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.
2.5 Monitoring Progress (pp. 82 – 84)
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
Copyright © Big Ideas Learning, LLC Algebra 1 85All rights reserved. Worked-Out Solutions
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.
2.5 Exercises (pp. 85 – 86)
Vocabulary and Core Concept Check
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.
Monitoring Progress and Modeling with Mathematics
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
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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
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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
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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.
Maintaining Mathematical Profi ciency
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.
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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.
2.6 Explorations (p. 87)
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.
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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
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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)
Vocabulary and Core Concept Check
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”.
Monitoring Progress and Modeling with Mathematics
3. ∣ x ∣ < 3
x > −3 and x < 3
The solution is −3 < 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
By defi nition, the absolute value of an expression must be
greater than or equal to 0. The expression ∣ 5p + 2 ∣ cannot be
less than −4. So, the inequality has no solution.
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
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Chapter 2
12. ∣ 3j − 1 ∣ + 6 > 0
− 6 − 6 ∣ 3j − 1 ∣ > −6
By defi nition, the absolute value of an expression must be
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
By defi nition, the absolute value of an expression must be
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
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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
The solution is −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 .
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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
Copyright © Big Ideas Learning, LLC Algebra 1 95All rights reserved. Worked-Out Solutions
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.
Maintaining Mathematical Profi ciency
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.
96 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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
The solution is 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
Copyright © Big Ideas Learning, LLC Algebra 1 97All rights reserved. Worked-Out Solutions
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
By defi nition, the absolute value of an expression must be
greater than or equal to 0. The expression ∣ k−9 ∣ cannot be
less than −4. So, the inequality has no solution.
98 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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
By defi nition, the absolute value of an expression must be
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
The solution is 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
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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
By defi nition, the absolute value of an expression must be
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.
100 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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.
Copyright © Big Ideas Learning, LLC Algebra 1 101All rights reserved. Worked-Out Solutions
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.
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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.
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