Chapter 2 Linear Equations and Inequalities in One...
21
Chapter 2 Linear Equations and Inequalities in One Variable Exercise Set 2.1 2. A solution set is a set containing all of the solutions for a given equation. 4. { } or ∅ 6. Subtract b from both sides. 8. 3 3 15 3 3 3 15 3 3 18 3 18 3 3 6 x x x x x − = − + = + = = = Check: () ? ? 36 3 15 18 3 15 15 15 − = − = = 10. 7 5 17 7 7 5 17 7 5 10 5 10 5 5 2 u u u u u − = − − = − − = − = − − =− Check: ( ) ( ) ? ? ? 7 5 2 17 7 10 17 7 10 17 17 17 − − = −− = + = = 12. 3 4 4 4 3 3 4 4 3 4 4 4 4 4 4 4 8 p p p p p p p p p + = − − + = − − = − + = − + = Check: () () ? ? 38 4 48 4 24 4 32 4 28 28 + = − + = − = 14. 6 5 3 4 6 3 5 3 3 4 3 5 4 3 5 5 4 5 3 9 3 9 3 3 3 q q q q q q q q q q q − = + − − = − + − = − + = + = = = Check: () () ? ? 63 5 33 4 18 5 9 4 13 13 − = + − = + = 16. ( ) 3 2 12 3 6 12 3 6 6 12 6 3 6 3 6 3 3 2 h h h h h h + = + = + − = − = = = Check: ( ) () ? ? 32 2 12 34 12 12 12 + = = =
Transcript of Chapter 2 Linear Equations and Inequalities in One...
Chapter 2
Linear Equations and Inequalities in One Variable
Exercise Set 2.1
2. A solution set is a set containing all of the solutions for a given equation.
4. { } or ∅
6. Subtract b from both sides.
8. 3 3 15
3 3 3 15 3
3 18
3 18
3 36
x
x
x
x
x
− =− + = +
=
=
=
Check:
( )?
?
3 6 3 15
18 3 15
15 15
− =
− =
=
10. 7 5 17
7 7 5 17 7
5 10
5 10
5 52
u
u
u
u
u
− =− − = −
− =− =− −
= −
Check:
( )
( )
?
?
?
7 5 2 17
7 10 17
7 10 17
17 17
− − =
− − =
+ =
=
12. 3 4 4 4
3 3 4 4 3 4
4 4
4 4 4 4
8
p p
p p p p
p
p
p
+ = −− + = − −
= −+ = − +
=
Check:
( ) ( )?
?
3 8 4 4 8 4
24 4 32 4
28 28
+ = −
+ = −
=
14. 6 5 3 4
6 3 5 3 3 4
3 5 4
3 5 5 4 5
3 9
3 9
3 33
q q
q q q q
q
q
q
q
q
− = +− − = − +
− =− + = +
=
=
=
Check:
( ) ( )?
?
6 3 5 3 3 4
18 5 9 4
13 13
− = +
− = +
=
16. ( )3 2 12
3 6 12
3 6 6 12 6
3 6
3 6
3 32
h
h
h
h
h
h
+ =+ =
+ − = −=
=
=
Check:
( )
( )
?
?
3 2 2 12
3 4 12
12 12
+ =
=
=
Instructor’s Solutions Manual 9
18. ( )2 3 2 4
2 3 6 4
5 6 4
5 6 6 4 6
5 10
5 10
5 52
b b
b b
b
b
b
b
b
+ − =+ − =
− =− + = +
=
=
=
Check:
( ) ( )
( )
?
?
?
2 2 3 2 2 4
4 3 0 4
4 0 4
4 4
+ − =
+ =
+ =
=
20. ( )2 9 11 7 1a a+ = + −
2 9 11 7 7
2 9 4 7
2 2 9 4 7 2
9 4 5
9 4 4 4 5
5 5
5 5
5 51
a a
a a
a a a a
a
a
a
a
a
+ = + −+ = +
− + = + −= +
− = − +=
=
=
Check:
( ) ( )
( )
?
?
?
2 1 9 11 7 1 1
2 9 11 7 0
11 11 0
11 11
+ = + −
+ = +
= +
=
22. Multiply both sides by the LCD, 3.
( )
15 7 10 5
31
3 5 3 7 10 53
15 21 30 15
15 6 30
15 6 30
15 5 30
15 30 5 30 30
45 5
45 5
5 59
y y y
y y y
y y y
y y
y y y y
y
y
y
y
y
− = + −
⎛ ⎞− = + −⎜ ⎟⎝ ⎠− = + −− = +
− − = − +− = +
− − = + −− =−
=
− =
Check:
( ) ( ) ( )?
?
19 5 7 9 10 5 9
3
3 5 63 10 45
8 8
− − = − + − −
− − = − + +
− = −
24. Multiply both sides by the LCD, 12. 2 1 1 7
3 6 4 122 1 1 7
12 123 6 4 12
2 1 1 712 12 12 12
3 6 4 128 2 3 7
10 3 7
10 3 3 7 3
10 10
10 10
10 101
v v
v v
v v
v v
v
v
v
v
v
+ − =
⎛ ⎞ ⎛ ⎞+ − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⋅ + ⋅ − ⋅ = ⋅
+ − =− =
− + = +=
=
=
Check:
( ) ( )?
?
?
2 1 1 71 1
3 6 4 122 1 1 7
3 6 4 128 2 3 7
12 12 12 127 7
12 12
+ − =
+ − =
+ − =
=
10 Chapter 2 Linear Equations and Inequalities in One Variable
26. Multiply both sides by 10 to clear decimals.
( ) ( )0.2 0.7 0.7 3.6
10 0.2 0.7 10 0.7 3.6
2 7 7 36
2 7 7 7 7 36
5 7 36
5 7 7 36 7
5 29
5 29
5 55.8
p p
p p
p p
p p p p
p
p
p
p
p
− = −− = −− = −
− − = − −− − = −
− − + = − +− = −− −=− −
=
Check:
( ) ( )?
?
0.2 5.8 0.7 0.7 5.8 3.6
1.16 0.7 4.06 3.6
0.46 0.46
− = −
− = −
=
28. Multiply both sides by 10 to clear decimals.
( ) ( )
0.3 1.4 0.7 2.9 0.5
1.7 0.7 2.9 0.5
10 1.7 0.7 10 2.9 0.5
17 7 29 5
17 5 7 29 5 5
22 7 29
22 7 7 29 7
22 22
22 22
22 221
z z z
z z
z z
z z
z z z z
z
z
z
z
z
+ + = −+ = −+ = −+ = −
+ + = − ++ =
+ − = −=
=
=
Check:
( ) ( ) ( )?
?
0.3 1 1.4 1 0.7 2.9 0.5 1
0.3 1.4 0.7 2.9 0.5
2.4 2.4
+ + = −
+ + = −
=
30. 7 42 27 64 9
20 42 64 9
20 9 42 64 9 9
11 42 64
11 42 42 64 42
11 22
11 22
11 112
d d d
d d
d d d d
d
d
d
d
d
− − = − −− − = − −
− + − = − − +− − = −
− − + = − +− = −− −
=− −
=
Check:
( ) ( ) ( )?
?
7 2 42 27 2 64 9 2
14 42 54 64 18
82 82
− − = − −
− − = − −
− = −
32. ( ) ( )12 2 3 11 5 2 7c c c− − = − −
12 2 6 11 10 35
10 6 46 10
10 10 6 46 10 10
20 6 46
20 6 6 46 6
20 40
20 40
20 202
c c c
c c
c c c c
c
c
c
c
c
− + = − ++ = −
+ + = − ++ =
+ − = −=
=
=
Check:
( ) ( ) ( )( )( ) ( )
( )
?
?
?
?
12 2 2 2 3 11 5 2 2 7
24 2 1 11 5 4 7
24 2 11 5 3
26 11 15
26 26
− − = − −
− − = − −
+ = − −
= +
=
34. ( ) ( )18 8 3 5 4 27s s− − = − + +
18 24 8 5 20 27
6 8 5 7
6 8 5 5 5 7
6 13 7
6 6 13 7 6
13 13
13 13
13 131
s s
s s
s s s s
s
s
s
s
s
− + = − − +− + = − +
− + + = − + +− + =
− + + = +=
=
=
Instructor’s Solutions Manual 11
Check:
( ) ( )
( ) ( )
?
?
?
18 8 3 1 5 1 4 27
18 8 2 5 5 27
18 16 25 27
2 2
− − = − + +
− = − +
− = − +
=
36. ( ) ( )5 3 4 5 5 5 3m m m+ − − = + −
5 3 4 5 5 25 3
8 5 22
8 5 22
8 4 22
8 22 4 22 22
14 4
14 4
4 47
2
m m m
m m
m m m m
m
m
m
m
m
+ − + = + −+ = +
− + = − += +
− = + −− =−
=
− =
Check:
( )
?
?
?
?
7 7 75 3 4 5 5 5 3
2 2 2
35 28 73 5 5 5 3
2 2 2
35 33 19 5 3
2 2
35 153 19 3
2 29 9
2 2
⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − − − = − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠
⎛ ⎞ ⎛ ⎞− + − − − = − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞− + − − = −⎜ ⎟⎝ ⎠
− + + = −
=
38. Multiply both sides by the LCD, 10. 4 1
6 75 2
4 110 6 10 7
5 2
8 60 70 5
8 5 60 70 5 5
13 60 70
13 60 60 70 60
13 130
13 130
13 1310
k k
k k
k k
k k k k
k
k
k
k
k
− = −
⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠− = −
+ − = − +− =
− + = +=
=
=
Check:
( ) ( )?
?
4 110 6 7 10
5 2
8 6 7 5
2 2
− = −
− = −
=
40. Multiply both sides by the LCD, 12.
( )
( )
( )
2 36 3
3 42 3
12 6 12 33 4
8 6 36 9
8 48 36 9
8 8 48 36 9 8
48 36
48 36 36 36
84
d d
d d
d d
d d
d d d d
d
d
d
− = +
⎛ ⎞⋅ − = +⎜ ⎟⎝ ⎠
− = +− = +
− − = + −− = +
− − = − +− =
Check:
( ) ( )
( )
?
?
2 384 6 3 84
3 42
90 3 633
60 60
− − = + −
− = + −
− = −
42. Multiply both sides by 10 to clear decimals.
( )( ) ( )
3.3 0.6 1.1 4 2
10 3.3 0.6 10 4.4 1.1 2
33 6 44 11 20
33 6 24 11
33 6 11 24 11 11
33 5 24
33 33 5 24 33
5 9
5 9
5 51.8
a a
a a
a a
a a
a a a a
a
a
a
a
a
− = − −
− = − −− = − −− = −
− + = − ++ =
− + = −= −−
=
= −
Check:
( ) ( )( )( )
?
?
3.3 0.6 1.8 1.1 4 1.8 2
3.3 1.08 1.1 5.8 2
4.38 4.38
− − = − − −
+ = −
=
12 Chapter 2 Linear Equations and Inequalities in One Variable
44. Multiply both sides by 100 to clear decimals.
( ) ( )
2.28 1.6 4.6 0.05 0.25
2.6 2.28 0.05 4.35
100 2.6 2.28 100 0.05 4.35
260 228 5 435
260 5 228 5 5 435
255 228 435
255 228 228 435 228
255 663
255 663
255 2552.6
x x x
x x
x x
x x
x x x x
x
x
x
x
x
− + = + −− = +− = +− = +
− − = − +− =
− + = +=
=
=
Check:
( ) ( )?
?
2.6 2.28 1.6 2.6 4.6 0.05 2.6 0.25
2.6 2.28 4.16 4.6 0.13 0.25
4.48 4.48
− + = + −
− + = + −
=
46. ( ) ( )16 3 2 1 5 1 13 2
16 48 2 1 5 5 13 2
18 49 18 7
49 7
q q q q
q q q q
q q
− + − = + + +− + − = + + +
− = +− ≠
No solution
48. ( )5 2 3 5 5 8
5 10 3 5 5 8
8 5 5 8
5 5
p p p
p p p
p p
− + − + = − −− − − + = − −
− − = − −− = −
All real numbers
50. Mistake: The distributive property was not used correctly. Correct: ( )4 3 1 14
4 3 3 14
1 3 14
3 13
13
3
p
p
p
p
p
+ − =+ − =
+ ==
=
52. Mistake: Subtracted before distributing into the parentheses. Correct: ( )7 2 2 3 5
7 2 4 3 5
2 5
5
2
d
d
d
d
− + − =− − − =
− =
= −
54. Mistake: Did not distribute the − with the 4. Correct:
( )3 4 2 3 2 6
3 8 12 2 6
5 12 2 6
5 5 12 2 5 6
12 3 6
12 6 3 6 6
6 3
6 3
3 32
x x x
x x x
x x
x x x x
x
x
x
x
x
− − = − +− + = − +− + = − +
− + + = − + += +
− = + −=
=
=
56. 2 2 2
2 2 2 2 2
2 2 2
c a b
c a a a b
c a b
= +
− = − +
− =
58. I Prt
I Prt
Pr PrI
tPr
=
=
=
60. 2
2
2 2
2
A rh
A rh
r rA
hr
ππ
π π
π
=
=
=
62. 2
2
2
2
2 2
2
1
31
3 33
3
3
3
V r h
V r h
V r h
V r h
r rV
hr
π
π
π
ππ π
π
=
⋅ = ⋅
=
=
=
64. 2
2
2
t
t
t
ωα
α ωα ω
= −
= −
− =
66. ( )
( )( )( )
1
1
1 1
1
B P rt
P rtB
rt rt
BP
rt
= +
+=
+ +
=+
68. 2 2
2 2 2 2
2 2
2 2
2 22
2
P r d
P r r r d
P r d
P r d
P rd
ππ π πππ
π
= +− = − +− =−
=
− =
Instructor’s Solutions Manual 13
70. 4 3 9
4 4 3 9 4
3 9 4
3 9 4
3 39 4
3
x y
x x y x
y x
y x
xy
+ =− + = −
= −−
=
−=
72. 20
20 0 0
20
20
20
16
16
16
16
16 16
16
h t h
h h t h h
h h t
h h t
h ht
= − +
− = − + −
− = −
− −=
− −−
=−
74. 21
2E mv mgy= +
( )
( )( )
( )
2
2
2
2
2
2
1
21
21
2 22
2
2
2
E mgy mv mgy mgy
E mgy mv
E mgy mv
E mgy mv
E mgy mv
m m
E mgyv
m
− = + −
− =
− = ⋅
− =
−=
−=
76. Mistake: Subtracted b instead of dividing by b. Correct: A bh
A bh
b bA
hb
=
=
=
78. Mistake: Did not multiply both sides of the
equation by 2. Correct: 1
21
2 22
2
2
2
A bh
A bh
A bh
A bh
h hA
bh
=
⋅ = ⋅
=
=
=
Exercise Set 2.2
2. Answers will vary. Some possible answers: Draw a picture. Make a table. Underline key words. Search for a related example.
4. n + 2, n + 4
6. ( )0.30 40 n+ or ( )12 0.30n+ ml
8. ( )
( )
( ) ( )
532
95
89 329
9 9 589 32
5 5 9160.2 32
128.2
C F
F
F
F
F°
= −
− = −
− = ⋅ −
− = −− =
10. Find the area of the house: 35 40
1400
A
A
= ⋅=
Find the area of the lot: ( )( )
185 65 110
20.5 150 110
8250
A
A
A
= +
==
Subtract the area of the house from the area of the lot to find the area to be landscaped.
8250 – 1400 = 6850 sq. ft.
12.
( )( )1725 0.03 5
1725 0.15
1725 1.15
$1500
B P Prt
P P
P P
P
P
= += += +==
14. Let w be the number. Translate to an equation and solve for w.
11 6 7
11 7 6 7 7
18 6
18 6
6 63
w
w
w
w
w
= −+ = − +
=
=
=
14 Chapter 2 Linear Equations and Inequalities in One Variable
16. Let n be the number. Translate to an equation and solve for n.
( )2 3 8
2 6 8
2 6 6 8 6
2 2
2 2
2 21
n
n
n
n
n
n
− = −− = −
− + = − += −−=
= −
18. Let a be the number. Translate to an equation and solve for a.
( )6 3 5 4
6 3 15 4
21 3 4
21 3 3 4 3
21 7
21 7
7 73
a a
a a
a a
a a a a
a
a
a
− − =− + =
− =− + = +
=
=
=
20. Let p represent the wholesale price. Because $22,678 is the MSRP, it is the result of adding 15% of the wholesale price to the wholesale price. Translate to an equation and solve for p. 0.15 22,678
1.15 22,678
1.15 22,678
1.15 1.1519,720
p p
p
p
p
+ ==
=
=
The wholesale price is $19,720.
22. Let p represent the original price. Because $24.50 is the discounted price, it is the result of subtracting 40% of the original price from the original price. Translate to an equation and solve for p.
0.40 24.50
0.60 24.50
0.60 24.50
0.60 0.6040.83
p p
p
p
p
− ==
=
≈
The original price is approximately $40.83.
24. Let x represent the previous month’s revenue. Because the current month’s revenue of $24,360 is a 12.5% decrease over the previous month, it is the result of subtracting 12.5% of x from x. Translate to an equation and solve.
0.125 24,360
0.875 24,360
0.875 24,360
0.875 0.87527,840
x x
x
x
x
− ==
=
=
The previous month’s revenue was $27,840.
26. The two consecutive integers are x and x + 1.
( )1 167
2 1 167
2 1 1 167 1
2 166
2 166
2 283
x x
x
x
x
x
x
+ + =+ =
+ − = −=
=
=
The integers are 83 and 83 + 1 = 84.
28. The consecutive odd integers are x, x + 2, and x + 4.
( ) ( )2 4 165
3 6 165
3 159
53
x x x
x
x
x
+ + + + =+ =
==
The integers are 53, 53 + 2 = 55, and 53 + 4 = 57.
30. If two angles are complementary their sum is 90 degrees. Let one of the angles be x and the other be 3x – 5.
( )3 5 90
4 5 90
4 95
23.75
x x
x
x
x
+ − =− =
==
The angles are 23.75º and 3(23.75) – 5 = 66.25º.
32. If two angles are supplementary their sum is 180 degrees. Let one of the angles be x and the other angle be 2x – 15.
( )2 15 180
3 15 180
3 195
65
x x
x
x
x
+ − =− =
==
The angles are 65º and 2(65) – 15 = 115º.
34.
( )
Categories Value Number Amount
1 gallon 13 13
1/2 gallon 14.50 7 7 14.50
x x
x x− −
Instructor’s Solutions Manual 15
36. Categories Rate Time Distance
Janet 4545
10Paul 50 10
50
dd
dd
++
38.
( )( )
Concen- Vol. of Vol. of
Solutions trate Solution Acid
45% 0.45 0.45
35% 0.35 80 0.35 80
40% 0.40 80 0.40 80
x x
x x− −
40. Translate the information from the table in Exercise 34 to an equation, and solve.
( )13 7 14.50 155.50
13 101.5 7 155.50
6 101.5 155.50
6 101.5 101.5 155.50 101.5
6 54
6 54
6 69
x x
x x
x
x
x
x
x
+ − =+ − =
+ =+ − = −
=
=
=
$9 for the 1 gal. and $5.50 for the ½ gal.
42. Translate the information from the table in Exercise 36 to an equation, and solve.
( )
10
45 5050 45 10
50 45 450
50 45 45 45 450
5 450
5 450
5 590
d d
d d
d d
d d d d
d
d
d
+=
= += +
− = − +=
=
=
Paul has traveled 90 miles. That has taken him 90
2 hr.45
= So, it is 6:00 + 2 hr = 8:00 PM.
44. Translate the information from the table in Exercise 38 to an equation, and solve.
( ) ( )0.45 0.35 80 0.40 80
0.45 28 0.35 32
0.1 28 32
0.1 28 28 32 28
0.1 4
0.1 4
0.1 0.140
x x
x x
x
x
x
x
x
+ − =+ − =
+ =+ − = −
=
=
=
40 L of each must be mixed to form 80 L of a 40% acid solution.
46. Create a table.
( )
how many value total amount
desktop 10 10
laptop 4 260 4 260
x x
x x+ +
Translate the information in the table to an equation and solve.
( )10 4 260 20,080
10 4 1040 20,080
14 1040 20,080
14 1040 1040 20,080 1040
14 19,040
14 19,040
14 141360
x x
x x
x
x
x
x
x
+ + =+ + =
+ =+ − = −
=
=
=
Desktops are $1360 and laptops are $1620.
48. Create a table.
( )how many value total amount
Monopoly 2 8.95 8.95 2
Trivial 15.95 15.95
Pursuit
x x
x x
Translate the information in the table to an equation and solve.
( )8.95 2 15.95 575.45
17.9 15.95 575.45
33.85 575.45
33.85 575.45
33.85 33.8517
x x
x x
x
x
x
+ =+ =
=
=
=
He ordered 17 Trivial Pursuit games and 34 Monopoly games.
16 Chapter 2 Linear Equations and Inequalities in One Variable
50. Create a table. rate time distance
east 60 60
west 48 48
x x
x x
Translate the information in the table to an equation and solve. 60 48 270
108 270
108 270
108 1082.5
x x
x
x
x
+ ==
=
=
The cars will be 270 miles apart in 2.5 hours.
52. Create a table.
( )rate time distance
Bonnie 65 0.25 65 0.25
Bob 70 70
x x
x x
+ +
Translate the information in the table to an equation and solve.
( )65 0.25 70
65 16.25 70
65 65 16.25 70 65
16.25 5
16.25 5
5 53.25
x x
x x
x x x x
x
x
x
+ =+ =
− + = −=
=
=
Bob will catch up in 3.25 hours at 5:00 PM.
54. Create a table.
( )
( )
vol. of vol. ofconcentrate
solution fertilizer
40% 0.40 10 0.40 10
50% 0.50 0.50
46% 0.46 10 0.46 10
x x
x x+ +
Translate the information in the table to an equation and solve.
( ) ( )0.40 10 0.50 0.46 10
4 0.50 0.46 4.6
4 0.50 0.46 0.46 0.46 4.6
4 0.04 4.6
4 4 0.04 4.6 4
0.04 0.6
0.04 0.6
0.04 0.0415
x x
x x
x x x x
x
x
x
x
x
+ = ++ = +
+ − = − ++ =
− + = −=
=
=
He should use 15 oz. of 50% solution.
56. Create a table.
( )
( )
vol. of vol. ofconcentrate
solution HCl
10% 0.10 50 0.10 50
25% 0.25 0.25
20% 0.20 50 0.20 50
x x
x x+ +
Translate the information in the table to an equation and solve.
( ) ( )0.10 50 0.25 0.20 50
5 0.25 0.20 10
5 0.25 0.20 0.20 0.20 10
5 0.05 10
5 5 0.05 10 5
0.05 5
0.05 5
0.05 0.05100
x x
x x
x x x x
x
x
x
x
x
+ = ++ = +
+ − = − ++ =
− + = −=
=
=
100 ml. of 10%
Exercise Set 2.3
2. For x < a, the value of a is not included in the solution set. For x a≤ , the value of a is included in the solution set.
4. All real numbers less than or equal to 4 are part of the solution set.
6. The direction of an inequality symbol changes when both sides of the inequality are multiplied or divided by a negative number.
8. a) { }7n n ≤ b) ( ],7−∞
c)
10. a) { }3x x > − b) ( )3,− ∞
c)
12. a) 2
3a a⎧ ⎫≥ −⎨ ⎬⎩ ⎭
b) 2
,3
⎡ ⎞− ∞⎟⎢ ⎠⎣
c)
Instructor’s Solutions Manual 17
14. a) { }3.2s s ≤ − b) ( ], 3.2−∞ −
c)
16. 6 1
6 6 1 6
7
x
x
x
+ ≥ −+ − ≥ − −
≥ −
a) { }7x x ≥ − b) [ )7,− ∞
c)
18. 4 24
4 24
4 46
x
x
x
− < −− −>− −
>
a) { }6x > b) ( )6,∞
c)
20. 9 7 11
9 7 7 11 7
9 18
9 18
9 92
y
y
y
y
y
− >− + > +
>
>
>
a) { }2y y > b) ( )2,∞
c)
22. 3 4 17
3 4 4 17 4
3 13
3 13
3 313
3
c
c
c
c
c
− + <− + − < −
− <−
>− −
> −
a) 13
3c c⎧ ⎫> −⎨ ⎬⎩ ⎭
b) 13
,3
⎛ ⎞− ∞⎜ ⎟⎝ ⎠
c)
24. 1
45 5
15 4 5
5 5
20 1
20 1 1 1
19
q
q
q
q
q
> +
⎛ ⎞⋅ > ⋅ +⎜ ⎟⎝ ⎠> +
− > − +>
a) { }19q q < b) ( ),19−∞
c)
26. 5 3 13 4
5 4 3 13 4 4
3 13
3 3 13 3
16
b b
b b b b
b
b
b
− ≥ +− − ≥ + −
− ≥− + ≥ +
≥
a) { }16b b ≥ b) [ )16,∞
c)
28. 4 7 2 19
4 7 2 2 2 19
4 5 19
4 4 5 19 4
5 15
5 15
5 53
k k
k k k k
k
k
k
k
k
− < − +− + < − + +
− <− − < −
− <−
>− −
> −
a) { }3k k > − b) ( )3,− ∞
c)
18 Chapter 2 Linear Equations and Inequalities in One Variable
30. ( )6 3 2 10 2
18 12 10 2
18 2 2
18 2 2 2 2
18 0
18 0
18 180
a
a
a
a
a
a
a
+ − ≥+ − ≥
+ ≥+ − ≥ −
≥
≥
≥
a) { }0a a ≥ b) [ )0,∞
c)
32. ( ) ( )4 10 17 3 13v v− ≤ + +
4 40 17 51 13
4 40 17 64
4 17 40 17 17 64
13 40 64
13 40 40 64 40
13 104
13 104
13 138
v v
v v
v v v v
v
v
v
v
v
− ≤ + +− ≤ +
− − ≤ − +− − ≤
− − + ≤ +− ≤− ≥− −
≥ −
a) { }8v v ≥ − b) [ )8,− ∞
c)
34. ( )1 4 13 28
7 7 4x x− > − +
3 4 14
7 7 43 4 4 4 1
47 7 7 7 4
14
41
4 4 4417
4
x x
x x x x
x
x
x
− > − +
+ − > − + +
− >
− + > +
>
a) 17
4x x
⎧ ⎫>⎨ ⎬⎩ ⎭
b) 17
,4
⎛ ⎞∞⎜ ⎟⎝ ⎠
c)
36. ( ) ( )1 110 5 0
5 2p p+ − + >
( ) ( )
( ) ( )
1 110 10 5 10 0
5 2
2 10 5 5 0
2 20 5 25 0
3 5 0
3 5 5 0 5
3 5
3 5
3 35
3
p p
p p
p p
p
p
p
p
p
⎛ ⎞+ − + > ⋅⎜ ⎟⎝ ⎠
+ − + >+ − − >
− − >− − + > +
− >− <− −
< −
a) 5
3p p
⎧ ⎫< −⎨ ⎬⎩ ⎭
b) 5
,3
⎛ ⎞−∞ −⎜ ⎟⎝ ⎠
c)
38. 1.2 1.4 1.5 0.5
1.2 1.5 1.4 1.5 1.5 0.5
0.3 1.4 0.5
0.3 1.4 1.4 0.5 1.4
0.3 0.9
0.3 0.9
0.3 0.33
b b
b b b b
b
b
b
b
b
− ≥ −− − ≥ − −− − ≥ −
− − + ≥ − +− ≥−
≤− −
≤ −
a) { }3b b ≤ − b) ( ], 3−∞ −
c)
40. ( )3.2 3.6 1.8 0.3 6 0.4f f f+ − ≤ + −
3.2 3.6 1.8 0.3 1.8 0.4
1.4 3.6 0.3 1.4
1.4 0.3 3.6 0.3 0.3 1.4
1.1 3.6 1.4
1.1 3.6 3.6 1.4 3.6
1.1 2.2
1.1 2.2
1.1 1.12
f f f
f f
f f f f
f
f
f
f
f
+ − ≤ + −+ ≤ +
− + ≤ − ++ ≤
+ − ≤ −≤ −−
≤
≤ −
a) { }2f f ≤ − b) ( ], 2−∞ −
c)
Instructor’s Solutions Manual 19
42. ( ) ( )2 2 4 11 4 2 6
4 8 11 4 8 6
4 3 4 2
4 4 3 4 4 2
3 2
a a
a a
a a
a a a a
+ − ≤ − ++ − ≤ − +
− ≤ −− − ≤ − −
− ≤ −
Because there are no variable terms left in the inequality and the inequality is true, every real number is a solution for the original inequality. a) { }| is a real numbera a
b) ( ),−∞ ∞
c)
44. ( ) ( ) ( )3 2 6 2 4 5 3
6 18 2 8 5 15
4 10 4 15
4 4 10 4 4 15
10 15
x x x x
x x x x
x x
x x x x
− − − > + −− − + > + −
− > +− − > − +
− >
Because there are no variable terms left in the inequality and the inequality is false, there are no solutions for the original inequality. a) { } or ∅
b) No interval notation c)
46. 2
83
3 2 38
2 3 212
x
x
x
> −
⋅ > − ⋅
> −
48. 3 2 10
3 2 2 10 2
3 12
3 12
3 34
x
x
x
x
x
− <− + < +
<
<
<
50. 3 2 17
3 3 2 17 3
2 14
2 14
2 27
x
x
x
x
x
− ≥− − ≥ −
− ≥− ≤− −
≤ −
52. ( )3 4 8 16
3 12 8 16
3 4 16
3 4 4 16 4
3 12
3 12
3 34
x
x
x
x
x
x
x
− + ≥ −− + ≥ −
− ≥ −− + ≥ − +
≥ −−≥
≥ −
54. Let x represent Eric’s score on the fourth paper. 70 62 75
704207
704
207 280
207 207 280 207
73
x
x
x
x
x
+ + +≤
+≤
+ ≤− + ≤ −
≤
Eric must repeat the course if his score on the fourth paper is lower than 73.
56. Let x represent Paulette’s score on her third game. 186 178
1753364
1753
364 525
364 364 525 364
161
x
x
x
x
x
+ +≥
+ ≥
+ ≥− + ≥ −
≥
Paulette must bowl 161 or higher.
58. Let w represent the width of the spa. 4 14
4 14
4 43.5
w
w
w
≥
≥
≥
The width must be 3.5 ft. or more.
60. Let d represent the diameter of the pipe. 12
12
12
3.8 cm
C
d
d
d
π
π
≤≤
≤
≤
The diameter of the pipe can be approximately 3.8 cm or less.
62. Let t represent Therese’s time. 480 75
480 75
75 756.4
t
t
t
≤
≤
≤
She will complete her trip in 6.4 hours or more.
20 Chapter 2 Linear Equations and Inequalities in One Variable
64. To find profit, we subtract costs from revenue.
( ) ( )65 10,000 25 15,000 0
65 10,000 25 15,000 0
40 5000 0
40 5000
125
n n
n n
n
n
n
+ − + ≥+ − − ≥
− ≥≥≥
125 or more homes must be cleaned to break even or make a profit.
66. a) ( )
( )
52795 32
95
27639
1535 C
t
t
t
< −
<
<
b) 1535 Ct °≥
68. force mass acceleration
24,500 9.8
24,500 9.8
9.8 9.82500 kg
m
m
m
= ×− ≤ −− −
≥− −
≥
The mass that can be safely lifted is about 2500 kg or less.
Puzzle Problem
There is no conflict in what they said; therefore, because “at least” one lied, they both must have lied; therefore, the brother is older.
Collaborative Exercises
1. F + L
2. Yes, because 40% off the price of lenses means a customer pays 60% of the regular price, which would be 0.6L.
3. 40% of $90 = 36 or without discount: 120 + 90 =$210; with: 120 + 0.6(90) = $174; Anna saves 210 – 174 = $36
4. ( )0.6 90 125
54 125
54 54 125 54
71
F
F
F
F
+ ≤+ ≤
+ − ≤ −≤
5. Senior discount: ( )0.75 140 260 $300+ = ; 40%
off lenses sale: ( )140 0.6 260 $296+ = ; the 40%
sale is slightly better.
6. ( )60 0.60 0.75 60
60 0.60 45 0.75
60 0.60 0.75 45 0.75 0.75
60 0.15 45
60 60 0.15 45 60
0.15 15
0.15 15
0.15 0.15$100
L L
L L
L L L L
L
L
L
L
L
+ < ++ < +
+ − < + −− <
− − < −− < −− −>− −
>
7. 0.85(F + L)
8. ( )80 0.60 0.85 80
80 0.60 68 0.85
80 0.60 0.85 68 0.85 0.85
80 0.25 68
80 80 0.25 68 80
0.25 12
0.25 12
0.25 0.2548
L L
L L
L L L L
L
L
L
L
L
+ < ++ < +
+ − < + −− <
− − < −− < −− −>− −
>
Exercise Set 2.4
2. and and or
4. For two sets A and B, the union of A and B, symbolized by A B∪ , is a set containing each element in either A or B.
6. We graph the region included in either of the two inequalities.
8. { }{ }
a. 2, 4,6
b. 2,4,6,8,10
10. { }{ }
a. 14,16
b. 8,10,12,14,16,18
12.
{ }a.
b. , , , , , , ,a e h l m o t v
∅ 14. { }{ }
a. , ,
b. , , , , ,
c d e
a b c d e f
16. 3 7n− ≤ < 18. 0 3m≤ <
20. 1 1r− < ≤ 22. 7 15t< <
24. 3 9x< <
26. 4 0x− ≤ <
Instructor’s Solutions Manual 21
28. 2 1x− < < −
30. 0 8x< ≤
32. 4 7x< < a)
b) { }4 7x x< < c) ( )4,7
34. 3 1
4
x
x
− ≥≥
and 2 10
8
x
x
+ <<
a)
b) { }4 8x x≤ < c) [ )4,8
36. 4 8
2
x
x
− ≥≤ −
and 3 15
5
x
x
− ≤ −≥
a)
b) { } or ∅ c) no interval notation
38. 3 5 1
3 6
2
x
x
x
+ ≥ −≥ −≥ −
and 5 2 13
5 15
3
x
x
x
− <<<
a)
b) { }2 3x x− ≤ < c) [ )2,3−
40. 6 4 14
6 18
3
x
x
x
− + < −− < −
>
and 3 2
1
1
x
x
x
− − ≥ −− ≥
≤ −
a)
b) { } or ∅ c) no interval notation
42. 2 2 2
0 4
x
x
− ≤ − ≤≤ ≤
a)
b) { }0 4x x≤ ≤ c) [ ]0,4
44. 8 3 4 14
12 3 18
4 6
x
x
x
< − << << <
a)
b) { }4 6x x< < c) ( )4,6
46. 0 2 5 13
2 5 15
23
5
x
x
x
≤ − + ≤≤ ≤
≤ ≤
a)
b) 2
35
x x⎧ ⎫≤ ≤⎨ ⎬⎩ ⎭
c) 2
,35⎡ ⎤⎢ ⎥⎣ ⎦
48. 3 5 2 1
8 2 4
4 2
x
x
x
− < − <− < − < −
> >
a)
b) { }2 4x x< < c) ( )2,4
50. 4 4 2
0 6
0 6
6 0
x
x
x
x
− ≤ − − ≤≤ − ≤≥ ≥ −
− ≤ ≤
a)
b) { }6 0x x− ≤ ≤ c) [ ]6,0−
52. 3n ≤ or 4n >
54. 2m ≤ or 8m >
56. 3r > − or 1r ≥ −
22 Chapter 2 Linear Equations and Inequalities in One Variable
58. 6t > or 13t <
60. 4 2
2
a
a
− ≤ −≤
or 4 2
6
a
a
− ≥≥
a)
b) { }2 or 6a a a≤ ≥
c) ( ] [ ), 2 6,−∞ ∪ ∞
62. 3 5 14
3 9
3
t
t
t
− ≤ −≤ −≤ −
or 5 5 5
5 10
2
t
t
t
+ ≥ −≥ −≥ −
a)
b) { }3 or 2t t t≤ − ≥ −
c) ( , 3] [ 2, )−∞ − ∪ − ∞
64. 2 3 1
2 2
1
x
x
x
− − ≤ −− ≤
≥ −
or 2 3 5
2 8
4
x
x
x
− − ≥− ≥
≤ −
a)
b) { }4 or 1x x x≤ − ≥ −
c) ( , 4] [ 1, )−∞ − ∪ − ∞
66. 8 2 2
2 6
3
q
q
q
− ≤− ≤ −
≥
or 2 4
2
q
q
− ≥ −≤
a)
b) { }2 or 3q q q≤ ≥
c) ( , 2] [3, )−∞ ∪ ∞
68. 3
3 64
33
43 12
4
x
x
x
x
+ ≥
≥
≥≥
or 3
3 65
33
53 15
5
x
x
x
x
− ≤ −
≤ −
≤ −≤ −
a)
b) { }5 or 4x x x≤ − ≥
c) ( , 5] [4, )−∞ − ∪ ∞
70. ( )3 1 6
3 3 6
3 3
1
y
y
y
y
− − ≤− + ≤
− ≤≥ −
or ( )4 2 4
4 8 4
4 4
1
y
y
y
y
− − ≤− + ≤
− ≤ −≥
a)
b) { }1y y ≥ − c) [ )1,− ∞
72. ( )5 1 8 8
5 5 8 8
5 3 8
5 5
1
d
d
d
d
d
− + <− + <
+ <<<
or ( )5 1 2 18
5 5 2 18
5 3 18
5 15
3
d
d
d
d
d
+ − <+ − <
+ <<<
a)
b) { }3d d < c) ( ),3−∞
74. ( )4 2 1 1
4 8 1 1
4 7 1
4 8
2
k
k
k
k
k
− − − ≥ −− + − ≥ −
− + ≥ −− ≥ −
≤
or ( )4 2 1 5
4 8 1 5
4 7 5
4 2
1
2
k
k
k
k
k
− − − ≤− + − ≤
− + ≤− ≤ −
≥
a)
b) { }is a real numberk k , or
c) ( ),−∞ ∞
76. 0 1
1
x
x
< −<
and 1 3
4
x
x
− ≤≤
a)
b) { }1 4x x< ≤ c) ( ]1,4
Instructor’s Solutions Manual 23
78. 6 2 8
6 6
1
x
x
x
− < −< −< −
or 4 3 9
4 12
3
x
x
x
− >>>
a)
b) { }1 or 3x x x< − >
c) ( , 1) (3, )−∞ − ∪ ∞
80. 9 3 15
3 5
x
x
≤ − ≤− ≥ ≥ −
a)
b) { }5 3x x− ≤ ≤ − c) [ ]5, 3− −
82. 9 2 7 3
2 2 4
1 2
x
x
x
− < − < −− < <− < <
a)
b) { }1 2x x− < < c) ( )1,2−
84. 4 3 1 7
3 3 6
1 2
x
x
x
≤ − + <≤ − <
− ≥ > −
a)
b) { }2 1x x− < ≤ − c) ( ]2, 1− −
86. 2 2 1
2 1
3
x x
x
x
− ≤ +− ≤
≤
and 1 2 5
1 5
4
4
x x
x
x
x
+ ≤ +− + ≤
− ≤≥ −
a)
b) { }4 3x x− ≤ ≤ c) [ ]4,3−
88. 75 12.5 100
6 8
x
x
≤ ≤≤ ≤
a)
b) { }6 8x x≤ ≤
c) [ ]6,8
90. 76 72 84
70 804
280 232 320
48 88
x
x
x
+ + +≤ <
≤ + <≤ <
a)
b) { }48 88x x≤ < c) [ )48,88
92. a)
b) { }68 72x x° °≤ ≤ c) [ ]68,72
94. a)
b) { }40 mph 70 mphx x≤ ≤
c) [ ]40,70
96. 9
32 32 2125
9 0 180
5 0 100
C
C
C
< + <
< <
< <
a)
b) { }0 C 100 Cx x° °< < c) ( )0,100
Puzzle Problem
Since , then .d
d rt rt
= =
40 40
2 1.7520 km/hr. 22.9 km/hr.
r
r
≤ ≤
≤ ≤
24 Chapter 2 Linear Equations and Inequalities in One Variable
Exercise Set 2.5
2. We are to find numbers that are 5 units from 0.
4. There is no solution.
6. After the absolute value equation is separated into two equations, one with the two expressions equal and the other with the two expressions opposite, one of the equations leads to a contradiction.
8. 5y = − or 5y =
10. 1r = −
Because the absolute value of every real number is a positive number or zero, this equation has no solution.
12. 1 4
5
w
w
− ==
or 1 4
3
w
w
− = −= −
14. 3 6 12
3 6
2
s
s
s
+ ===
or 3 6 12
3 18
6
s
s
s
+ = −= −= −
16. 2 3 5
3 3
1
x
x
x
− =− =
= −
or 2 3 5
3 7
7
3
x
x
x
− = −− = −
=
18. 1 2 2
2 1
1
2
x
x
x
− =− =
= −
or 1 2 2
2 3
3
2
x
x
x
− = −− = −
=
20. 6 5 1p + = −
Because the absolute value of every real number is a positive number or zero, this equation has no solution.
22. 3 5 0
3 5
5
3
m
m
m
− ==
=
24. 3 7 10
3 3
r
r
+ =
=
3 3
1
r
r
==
or 3 3
1
r
r
= −= −
26. 3 1 5
3 6
x
x
+ − =
+ =
3 6
9
x
x
+ = −= −
or 3 6
3
x
x
+ ==
28. 2 2 4
2 2
v
v
− + =
− =
2 2
0
v
v
− = −=
or 2 2
4
v
v
− ==
30. 2 5 2 5
5 2 3
t
t
+ − =
− =
5 2 3
5 1
1
5
t
t
t
− = −= −
= −
or 5 2 3
5 5
1
t
t
t
− ===
32. 3 1 4 8
4 4 8
4 4 8
u
u
u
− = − −
− = − −
= −
4 8 4
4 4
1
u
u
u
− = −==
or 4 8 4
4 12
3
u
u
u
− ===
34. 3 2 5 7
2 5 10
5 5
x
x
x
− − = −
− − = −
− =
5 5
0
x
x
− = −=
or 5 5
10
x
x
− ==
36. 15 2 5 2 5
2 5 2 20
5 2 10
x
x
x
− − = −
− − = −
− =
5 2 10
2 15
15
2
x
x
x
− = −− = −
=
or 5 2 10
2 5
5
2
x
x
x
− =− =
= −
38. ( )4 2 6 18
4 12 2 18
6 12 18
y y
y y
y
− − =
− + =
− =
6 12 18
6 30
5
y
y
y
− ===
or 6 12 18
6 6
1
y
y
y
− = −= −= −
Instructor’s Solutions Manual 25
40. 2 5 3 10
5
p p
p
+ = +− =
or ( )2 5 3 10
2 5 3 10
5 15
3
p p
p p
p
p
+ = − ++ = − −
= −= −
42. 1 2 8
9
p p
p
− = +− =
or ( )1 2 8
1 2 8
7
3
p p
p p
p
− = − +− = − −
= −
44. 3 6 2 3
0 8
0
c c
c
c
− = +==
or ( )3 6 2 3
3 6 2 3
6 4
3
2
c c
c c
c
c
− = − +− = − −
=
=
46. 2 1 1 2
1
2
r r
r
− = −
=
or ( )2 1 1 2
2 1 1 2
1 1
r r
r r
− = − −− = − +− = −
One equation leads to a solution with no variables yet is true. The solution to this absolute value equation is all real numbers.
48. 4 2 2 8
4 8
no solution
q q+ = +=
or ( )4 2 2 8
4 2 2 8
4 12
3
q q
q q
q
q
+ = − ++ = − −
= −= −
This absolute value equation has only one solution, 3− .
50. ( )7 2 17
7 2 17
5 17
n
n
n
− − =
− + =
+ =
5 17
22
n
n
+ = −= −
or 5 17
12
n
n
+ ==
52. 2 33
13
3
u
u
u
− = −
= −
= −
or 2 33
53
15
u
u
u
− =
=
=
54. 6 5 2
6 318 15 12
2
5
w
w
w
−=
− =
=
or 6 5 2
6 318 15 12
2
w
w
w
−= −
− = −=
56. 2
1 73
28
3
p
p
+ − =
+ =
28
326
3
p
p
+ = −
= −
or 2
83
22
3
p
p
+ =
=
Exercise Set 2.6
2. x a≤ − or x a≥
4. Shade to the left of – a and to the right of a.
6. when a < 0
8. a)
b) { }3 3y y− ≤ ≤ c) [ ]3,3−
10. 2 5 2
3 7
m
m
− < − << <
a)
b) { }3 7m m< < c) ( )3,7
12. 3 4 8
3 4
p
p
− + ≤
− ≤
4 3 4
1 7
p
p
− ≤ − ≤− ≤ ≤
a)
b) { }1 7p p− ≤ ≤ c) [ ]1,7−
14. 4 12 2 10
4 12 8
x
x
+ + <
+ <
8 4 12 8
20 4 4
5 1
x
x
x
− < + <− < < −− < < −
a)
b) { }5 1x x− < < − c) ( )5, 1− −
26 Chapter 2 Linear Equations and Inequalities in One Variable
16. 5 1 6 8
5 1 14
h
h
− − − <
− − <
14 5 1 14
13 5 15
133
5
h
h
h
− < − − <− < − <
> > −
a)
b) 13
35
h h⎧ ⎫− < <⎨ ⎬⎩ ⎭
c) 13
3,5
⎛ ⎞−⎜ ⎟⎝ ⎠
18. 3 2 7
3 9
u
u
+ < −
< −
a)
b) { } or ∅ c) no interval notation
20. 4 2 3 9
4 2 12
2 3
n
n
n
+ − ≤
+ ≤
+ ≤
3 2 3
5 1
n
n
− ≤ + ≤− ≤ ≤
a)
b) { }5 1n n− ≤ ≤ c) [ ]5,1−
22. a)
b) { }6 or 6h h h≤ − ≥
c) ( ] [ ), 6 6,−∞ − ∞∪
24. 4 3
1
a
a
− < −<
or 4 3
7
a
a
− >>
a)
b) { }1 or 7a a a< >
c) ( ,1) (7, )−∞ ∪ ∞
26. 5 3 6
5 9
x
x
+ − ≥
+ ≥
5 9
14
x
x
+ ≤ −≤ −
or 5 9
4
x
x
+ ≥≥
a)
b) { }14 or 4x x x≤ − ≥
c) ( , 14] [4, )−∞ − ∪ ∞
28. 2 5 1 10
2 5 11
x
x
− − ≥
− ≥
2 5 11
3
x
x
− ≤ −≤ −
or 2 5 11
8
x
x
− ≥≥
a)
b) { }3 or 8x x x≤ − ≥
c) ( , 3] [8, )−∞ − ∪ ∞
30. 6 1 7 4
6 1 11
h
h
− + − >
− + >
6 1 11
2
h
h
− + < −>
or 6 1 11
5
3
h
h
− + >
< −
a)
b) 5
or 23
h h h⎧ ⎫< − >⎨ ⎬⎩ ⎭
c) ( )2,5−∞,−3
⎛ ⎞ ∞⎜ ⎟⎝ ⎠∪
32. 5 2 12
5 10
m
m
+ >
>
2
2 or 2
m
m m
>< − >
a)
b) { }2 or 2m m m< − >
c) ( , 2) (2, )−∞ − ∪ ∞
Instructor’s Solutions Manual 27
34. 3 2 5 1
3 2 6
2 2
x
x
x
− − ≥
− ≥
− ≥
2 2
0
x
x
− ≤ −≤
or 2 2
4
x
x
− ≥≥
a)
b) { }0 or 4x x x≤ ≥
c) ( ,0] [4, )−∞ ∪ ∞
36. 2 4 2 10
2 4 12
x
x
+ − ≥
+ ≥
2 4 12
2 16
8
x
x
x
+ ≤ −≤ −≤ −
or 2 4 12
2 8
4
x
x
x
+ ≥≥≥
a)
b) { }8 or 4x x x≤ − ≥
c) ( ] [ ), 8 4,−∞ − ∪ ∞
38. 3 2 3 3
6 2 0
3 0
y
y
y
− ≤ − + ≤− ≤ − ≤
≥ ≥
a)
b) { }0 3y y≤ ≤ c) [ ]0,3
40. 3 7 2b + > −
This inequality indicates that the absolute value is greater than a negative number. Because the absolute value of every real number is either positive or 0, the solution set is . a)
b) { } is a real numberb b c) ( ),−∞ ∞
42. 5 2 4 1
2 4 4
4 2
u
u
u
− + ≥
− + ≥ −
+ ≤
2 4 2
6 2
u
u
− ≤ + ≤− ≤ ≤ −
a)
b) { }6 2u u− ≤ ≤ − c) [ ]6, 2− −
44. 4 3 6 2 22
4 3 6 24
3 6 6
p
p
p
+ − <
+ <
+ <
6 3 6 6
12 3 0
4 0
p
p
p
− < + <− < <− < <
a)
b) { }4 0p p− < < c) ( )4,0−
46. 4 5 7
5 3
5 3
k
k
k
− + >
− + >
+ < −
Because absolute values cannot be negative, this inequality has no solution.
a)
b) ∅ c) no interval notation
48. 1
7 3 72
110 4
220 8
x
x
x
− ≤ − ≤
− ≤ − ≤
≥ ≥ −
a)
b) { }8 20x x− ≤ ≤ c) [ ]8, 20−
28 Chapter 2 Linear Equations and Inequalities in One Variable
50. 0.5 3 4 5
0.5 3 1
y
y
− + >
− >
0.5 3 1
0.5 2
4
y
y
y
− < −<<
or 0.5 3 1
0.5 4
8
y
y
y
− >>>
a)
b) { }4 or 8y y y< > c) ( , 4) (8, )−∞ ∪ ∞
52. 2
5.3 5.332
03
0
w
w
w
− ≥
− ≥
≤
or 2
5.3 5.332
10.63
15.9
w
w
w
− ≤ −
− ≤ −
≥
a)
b) { }0 or w 15.9w w ≤ ≥
c) ( ] [ ),0 15.9,−∞ ∞∪
54. 8 7 4 3
8 7 1
p
p
+ + >
+ > −
a)
b) { } is a real numberp p c) ( ),−∞ ∞
56. 5x ≤
58. 2x ≥
60. 3 2x − >
62. 1 4x + ≤
64. any negative numberx <
