Inequalities

CHAPTER 2. SOLVING LINEAR EQUATIONS
108
3. Solve the equation. To solve the equation, first combine like terms on the
left-hand side of the equation.
M + (6M + 3500) = 45500
7M + 3500 = 45500
Subtract 3500 from both sides of the equation and simplify.
7M + 3500 − 3500 = 45500 − 3500
7M = 42000
Divide both sides of the equation by 7.
7M
42000
=
7
7
M = 6000
4. Answer the question. The amount invested in the mutual fund is M =
$6, 000. The amount invested in the certificate of deposit is:
Certificate of deposit = 6M + 3500
= 6(6000) + 3500
= 36000 + 3500
= 39500
Hence, the amount invested in the certificate of deposit is $39,500.
5. Look back. Note that the two amounts total
6000 + 39500 = 45500,
so we have the correct solution.
2.6
Inequalities
1. Draw a number line, indicating the scale below the line. For each value 4, 3,
−4, 7/8, and −8/3, plot the corresponding point on the line and label it with
its value.
−4
−5
−4
− 83
−3
7
8
−2
−1
0
1
2
3
4
3
4
5
The position of the numbers on the number line give us the ordering from
smallest to largest: −4, −8/3, 7/8, 3, and 4.
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2.6. INEQUALITIES
109
3. Draw a number line, indicating the scale below the line. For each value −5,
5, 4, 2/3, and 8/3, plot the corresponding point on the line and label it with
its value.
2
3
−5
−5
−4
−3
−2
−1
0
8
3
1
2
3
4
5
4
5
The position of the numbers on the number line give us the ordering from
smallest to largest: −5, 2/3, 8/3, 4, and 5.
5. {x : x ≥ −7} is read “the set of all x such that x is greater than or equal to
−7”; that is, the set of all x that lie to the “right of” −7 or including −7 on
the number line. Note that this does include the number −7.
−7
7. {x : x < 2} is read “the set of all x such that x is less than 2”; that is, the
set of all x that lie to the “left of” 2 on the number line. Note that this does
not include the number 2.
2
9. (−∞, 2) = {x : x < 2}, and so it is read “the set of all x such that x is less
than 2”; that is, the set of all x that lie to the “left of” 2 on the number line.
Note that this does not include the number 2.
2
11. (6, ∞) = {x : x > 6}, and so it is read “the set of all x such that x is
greater than 6”; that is, the set of all x that lie to the “right of” 6 on the
number line. Note that this does not include the number 6.
6
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110
CHAPTER 2. SOLVING LINEAR EQUATIONS
13. {x : x < 7} is read “the set of all x such that x is greater than 7”; that is,
the set of all x that lie to the “right of” 7 on the number line. Note that this
does not include the number 7.
7
15. [0, ∞) = {x : x ≥ 0}, and so it is read “the set of all x such that x is
greater than or equal to 0”; that is, the set of all x that lie to the “right of” 0
or including 0 on the number line. Note that this does include the number 0.
0
17. {x : x ≤ −2} is read “the set of all x such that x is less than or equal to
−2”; that is, the set of all x that lie to the “left of” −2 or including −2 on the
number line. Note that this does include the number −2.
−2
19. (−∞, 3] = {x : x ≤ 3}, and so it is read “the set of all x such that x is
less than or equal to 3”; that is, the set of all x that lie to the “left of” 3 or
including 3 on the number line. Note that this does include the number 3.
3
21. Consider the shaded region on the given number line.
9
Note that every number to the left of 9, including 9, (i.e., “less than or equal
to 9”) is shaded. Using set-builder notation, the shaded region is described by
{x : x ≤ 9}.
23. Consider the shaded region on the given number line.
Second Edition: 2012-2013
2.6. INEQUALITIES
111
−8
Note that every number to the left of −8 (i.e., “less than −8”) is shaded. Using
set-builder notation, the shaded region is described by {x : x < −8}.
25. Consider the shaded region on the given number line.
−2
Note that every number to the right of −2 (i.e., “greater than −2”) is shaded.
Using set-builder notation, the shaded region is described by {x : x > −2}.
27. Consider the shaded region on the given number line.
−3
Note that every number to the right of −3, including −3, (i.e., “greater than
or equal to −3”) is shaded. Using set-builder notation, the shaded region is
described by {x : x ≥ −3}.
29. Consider the shaded region on the given number line.
4
Note that every number to the right of 4 (i.e., “greater than 4”) is shaded.
Using set-builder notation, the shaded region is described by (4, ∞).
31. Consider the shaded region on the given number line.
−2
Note that every number to the left of −2 (i.e., “less than −2”) is shaded. Using
interval notation, the shaded region is described by (−∞, −2).
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CHAPTER 2. SOLVING LINEAR EQUATIONS
112
33. Consider the shaded region on the given number line.
5
Note that every number to the left of 5, including 5, (i.e., “less than or equal
to 5”) is shaded. Using interval notation, the shaded region is described by
(−∞, 5].
35. Consider the shaded region on the given number line.
1
Note that every number to the right of 1, including 1, (i.e., “greater than or
equal to 1”) is shaded. Using interval notation, the shaded region is described
by [1, ∞).
37. To “undo” adding 10, we subtract 10 from both sides of the inequality.
x + 10 < 19
x + 10 − 10 < 19 − 10
Original inquality.
Subtract 10 from both sides.
x<9
Simplify both sides.
Shade all real numbers that are “less than” or “left of” 9 on a number line.
9
Thus, using set-builder and interval notation, the solution is:
{x : x < 9} = (−∞, 9)
39. To “undo” multiplying by 4, we divide both sides of the inequality by 4.
Because we are dividing by a positive number, we do not reverse the inequality
sign.
4x < 8
8
4x
<
4
4
x<2
Original inquality.
Divide both sides by 4.
Simplify both sides.
Shade all real numbers that are “less than” or “left of” 2 on a number line.
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2.6. INEQUALITIES
113
2
Thus, using set-builder and interval notation, the solution is:
{x : x < 2} = (−∞, 2)
41. To “undo” multiplying by −2, we divide both sides of the inequality by
−2. Because we are dividing by a negative number, we reverse the inequality
sign.
−2x ≤ −2
−2x
−2
≥
−2
−2
Original inquality.
Divide both sides by −2
and reverse the inequality sign.
Simplify both sides.
x≥1
Shade all real numbers that are “greater than or equal to” or “right of or
including” 1 on a number line.
1
Thus, using set-builder and interval notation, the solution is:
{x : x ≥ 1} = [1, ∞)
43. To “undo” subtracting 18, we add 18 to both sides of the inequality.
x − 18 > −10
x − 18 + 18 > −10 + 18
Original inquality.
Add 18 to both sides.
x>8
Simplify both sides.
Shade all real numbers that are “greater than” or “right of” 8 on a number
line.
8
Thus, using set-builder and interval notation, the solution is:
{x : x > 8} = (8, ∞)
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CHAPTER 2. SOLVING LINEAR EQUATIONS
114
45. We need to isolate terms containing x on one side of the inequality. We
begin by adding 9x to both sides of the inequality.
−5x − 6 ≥ 4 − 9x
−5x − 6 + 9x ≥ 4 − 9x + 9x
Original inquality.
Add 9x to both sides.
4x − 6 ≥ 4
Simplify both sides.
We continue to isolate terms containing x on one side of the inequality. We
next add 6 to both sides of the inequality.
4x − 6 + 6 ≥ 4 + 6
Add 6 to both sides.
4x ≥ 10
Simplify both sides.
To “undo” multiplying by 4, divide both sides by 4. Since we are dividing both
sides by a positive number, we do not reverse the inequality sign.
4x
10
≥
4
4
5
x≥
2
Divide both sides by 4.
Reduce to lowest terms.
Shade all real numbers that are “greater than or equal to” or “right of or
including” 5/2 on a number line.
5/2
Thus, using set-builder and interval notation, the solution is:
{x : x ≥ 5/2} = [5/2, ∞)
47. To “undo” subtracting 6, add 6 to both sides of the inequality.
16x − 6 ≤ 18
16x − 6 + 6 ≤ 18 + 6
16x ≤ 24
Original inquality.
Add 6 to both sides.
Simplify both sides.
To “undo” multiplying by 16, divide both sides by 16. Since we are dividing
both sides by a positive number, we do not reverse the inequality sign.
16x
24
≤
16
16
3
x≤
2
Divide both sides by 16.
Reduce to lowest terms.
Shade all real numbers that are “less than or equal to” or “left of or including”
3/2 on a number line.
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2.6. INEQUALITIES
115
3/2
Thus, using set-builder and interval notation, the solution is:
{x : x ≤ 3/2} = (−∞, 3/2]
49. We need to isolate terms containing x on one side of the inequality. We
begin by adding 4x to both sides of the inequality.
−14x − 6 ≥ −10 − 4x
−14x − 6 + 4x ≥ −10 − 4x + 4x
Original inquality.
Add 4x to both sides.
−10x − 6 ≥ −10
Simplify both sides.
We continue to isolate terms containing x on one side of the inequality. We
next add 6 to both sides of the inequality.
−10x − 6 + 6 ≥ −10 + 6
−10x ≥ −4
Add 6 to both sides.
Simplify both sides.
To “undo” multiplying by −10, divide both sides by −10. Since we are dividing
both sides by a negative number, we reverse the inequality sign.
−10x
−4
≤
−10
−10
x≤
Divide both sides by −10
and reverse the inequality sign.
2
5
Reduce to lowest terms.
Shade all real numbers that are “less than or equal to” or “left of or including”
2/5 on a number line.
2/5
Thus, using set-builder and interval notation, the solution is:
{x : x ≤ 2/5} = (−∞, 2/5]
51. To “undo” adding 18, subtract 18 from both sides of the inequality.
5x + 18 < 38
5x + 18 − 18 < 38 − 18
5x < 20
Original inquality.
Subtract 18 from both sides.
Simplify both sides.
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CHAPTER 2. SOLVING LINEAR EQUATIONS
116
To “undo” multiplying by 5, divide both sides by 5. Since we are dividing both
sides by a positive number, we do not reverse the inequality sign.
5x
20
<
5
5
x<4
Divide both sides by 5.
Simplify both sides.
Shade all real numbers that are “less than” or “left of” 4 on a number line.
4
Thus, using set-builder and interval notation, the solution is:
{x : x < 4} = (−∞, 4)
53. We need to isolate terms containing x on one side of the inequality. We
begin by adding 6x to both sides of the inequality.
−16x − 5 ≥ −11 − 6x
−16x − 5 + 6x ≥ −11 − 6x + 6x
Original inquality.
Add 6x to both sides.
−10x − 5 ≥ −11
Simplify both sides.
We continue to isolate terms containing x on one side of the inequality. We
next add 5 to both sides of the inequality.
−10x − 5 + 5 ≥ −11 + 5
Add 5 to both sides.
−10x ≥ −6
Simplify both sides.
To “undo” multiplying by −10, divide both sides by −10. Since we are dividing
both sides by a negative number, we reverse the inequality sign.
−10x
−6
≤
−10
−10
x≤
Divide both sides by −10
and reverse the inequality sign.
3
5
Reduce to lowest terms.
Shade all real numbers that are “less than or equal to” or “left of or including”
3/5 on a number line.
3/5
Thus, using set-builder and interval notation, the solution is:
{x : x ≤ 3/5} = (−∞, 3/5]
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2.6. INEQUALITIES
117
55. We need to isolate terms containing x on one side of the inequality. We
begin by adding 8x to both sides of the inequality.
2x − 9 ≥ 5 − 8x
Original inquality.
2x − 9 + 8x ≥ 5 − 8x + 8x
10x − 9 ≥ 5
Add 8x to both sides.
Simplify both sides.
We continue to isolate terms containing x on one side of the inequality. We
next add 9 to both sides of the inequality.
10x − 9 + 9 ≥ 5 + 9
Add 9 to both sides.
10x ≥ 14
Simplify both sides.
To “undo” multiplying by 10, divide both sides by 10. Since we are dividing
both sides by a positive number, we do not reverse the inequality sign.
10x
14
≥
10
10
7
x≥
5
Divide both sides by 10.
Reduce to lowest terms.
Shade all real numbers that are “greater than or equal to” or “right of or
including” 7/5 on a number line.
7/5
Thus, using set-builder and interval notation, the solution is:
{x : x ≥ 7/5} = [7/5, ∞)
57. To “undo” subtracting 4, add 4 to both sides of the inequality.
−10x − 4 ≤ 18
−10x − 4 + 4 ≤ 18 + 4
−10x ≤ 22
Original inquality.
Add 4 to both sides.
Simplify both sides.
To “undo” multiplying by −10, divide both sides by −10. Since we are dividing
both sides by a negative number, we reverse the inequality sign.
−10x
22
≥
−10
−10
x≥−
11
5
Divide both sides by −10
and reverse the inequality sign.
Reduce to lowest terms.
Shade all real numbers that are “greater than or equal to” or “right of or
including” −11/5 on a number line.
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CHAPTER 2. SOLVING LINEAR EQUATIONS
118
−11/5
Thus, using set-builder and interval notation, the solution is:
{x : x ≥ −11/5} = [−11/5, ∞)
59. To “undo” adding 4, subtract 4 from both sides of the inequality.
−12x + 4 < −56
Original inquality.
−12x + 4 − 4 < −56 − 4
−12x < −60
Subtract 4 from both sides.
Simplify both sides.
To “undo” multiplying by −12, divide both sides by −12. Since we are dividing
both sides by a negative number, we reverse the inequality sign.
−60
−12x
>
−12
−12
Divide both sides by −12
and reverse the inequality sign.
Simplify both sides.
x>5
Shade all real numbers that are “greater than” or “right of” 5 on a number
line.
5
Thus, using set-builder and interval notation, the solution is:
{x : x > 5} = (5, ∞)
61. We need to isolate terms containing x on one side of the inequality. We
begin by subtracting 6x from both sides of the inequality.
15x + 5 < 6x + 2
15x + 5 − 6x < 6x + 2 − 6x
9x + 5 < 2
Original inquality.
Subtract 6x from both sides.
Simplify both sides.
We continue to isolate terms containing x on one side of the inequality. We
next subtract 5 from both sides of the inequality.
9x + 5 − 5 < 2 − 5
9x < −3
Second Edition: 2012-2013
Subtract 5 from both sides.
Simplify both sides.
2.6. INEQUALITIES
119
To “undo” multiplying by 9, divide both sides by 9. Since we are dividing both
sides by a positive number, we do not reverse the inequality sign.
−3
9x
<
9
9
1
x<−
3
Divide both sides by 9.
Reduce to lowest terms.
Shade all real numbers that are “less than” or “left of” −1/3 on a number line.
−1/3
Thus, using set-builder and interval notation, the solution is:
{x : x < −1/3} = (−∞, −1/3)
63. The common denominator is 8. Clear the fractions from the inequality by
multiplying both sides of the inequality by 8.
9
3
x>
2
8
3
9
8
x >
8
2
8
Original inequality.
Multiply both sides by 8.
12x > 9
Cancel and multiply.
To “undo” multiplying by 12, divide both sides of the inequality by 12. Because
we are dividing by a positive number, we do not reverse the inequality.
9
12x
>
12
12
3
x>
4
Divide both sides by 12.
Simplify.
Shade all real numbers that are “greater than” or “right of” 3/4 on a number
line.
3/4
Thus, using set-builder and interval notation, the solution is:
{x : x > 3/4} = (3/4, ∞)
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120
65. The common denominator is 10. We will now clear the fractions from the
inequality by multiplying both sides of the inequality by 10.
3
9
<
2
5 3
9
10 x +
< 10
2
5
3
9
10x + 10
< 10
2
5
x+
10x + 15 < 18
Original inequality.
Multiply both sides by 10.
On the left, distribute the 10.
Multiply.
To “undo” the effect of adding 15, subtract 15 from both sides of the inequality.
10x + 15 − 15 < 18 − 15
10x < 3
Subtract 15 from both sides.
Simplify both sides.
To “undo” the effect of multiplying by 10, divide both sides of the inequality
by 10.
3
10x
<
Divide both sides by 10.
10
10
3
x<
Simplify both sides.
10
Shade all real numbers that are “less than” or “left of” 3/10 on a number line.
3/10
Thus, using set-builder and interval notation, the solution is:
{x : x < 3/10} = (−∞, 3/10)
67. Thus, using set-builder and interval notation, the solution is:
{x : x ≥ 5/7} = [5/7, ∞)
69. The common denominator is 56. We will now clear the fractions from the
inequality by multiplying both sides of the inequality by 56.
x−
3
9
≥−
8
7
3
9
56 x −
≥ 56 −
8
7
3
9
≥ 56 −
56x − 56
8
7
56x − 21 ≥ −72
Second Edition: 2012-2013
Original inequality.
Multiply both sides by 56.
On the left, distribute the 56.
Multiply.
2.6. INEQUALITIES
121
To “undo” the effect of subtracting 21, add 21 to both sides of the inequality.
56x − 21 + 21 ≥ −72 + 21
56x ≥ −51
Add 21 to both sides.
Simplify both sides.
To “undo” the effect of multiplying by 56, divide both sides of the inequality
by 56.
56x
−51
≥
Divide both sides by 56.
56
56
51
x≥−
Simplify both sides.
56
Shade all real numbers that are “greater than or equal to” or “right of or
including” −51/56 on a number line.
−51/56
Thus, using set-builder and interval notation, the solution is:
{x : x ≥ −51/56} = [−51/56, ∞)
71. The common denominator is 35. Clear the fractions from the inequality
by multiplying both sides of the inequality by 35.
6
4
− x≤−
5
7
6
4
35 − x ≤ −
35
5
7
Original inequality.
Multiply both sides by 35.
−42x ≤ −20
Cancel and multiply.
To “undo” multiplying by −42, divide both sides of the inequality by −42.
Because we are dividing by a negative number, we reverse the inequality.
−20
−42x
≥
Divide both sides by −42.
−42
−42
10
x≥
Simplify.
21
Shade all real numbers that are “greater than or equal to” or “right of or
including” 10/21 on a number line.
10/21
Thus, using set-builder and interval notation, the solution is:
{x : x ≥ 10/21} = [10/21, ∞)
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CHAPTER 2. SOLVING LINEAR EQUATIONS
122
73. The common denominator is 45. Clear the fractions from the inequality
by multiplying both sides of the inequality by 45.
6
7
5 2
− x− ≤ −
5
3 9 9 6
7
5 2
45 − x −
−
≤
45
5
3
9 9
−54x − 105 ≤ 25 − 10x
Original inequality.
Multiply both sides by 45.
Cancel and multiply.
We need to isolate all terms containing x on one side of the equation. To
“undo” subtracting 10x, add 10x from both sides of the inequality.
−54x − 105 + 10x ≤ 25 − 10x + 10x
−44x − 105 ≤ 25
Add 10x to both sides.
Simplify both sides.
We continue to isolate terms containing x on one side of the inequality. To
“undo” subtracting 105, add 105 from both sides of the inequality.
−44x − 105 + 105 ≤ 25 + 105
Add 105 to both sides.
−44x ≤ 130
Simplify both sides.
To “undo” the effect of multiplying by −44, divide both sides of the inequality
by −44. Because we are dividing both sides by a negative number, we reverse
the inequality sign.
130
−44x
≥
−44
−44
Divide both sides by −44
and reverse the inequality sign.
65
x≥−
22
Simplify both sides.
Shade all real numbers that are “greater than or equal to” or “right of or
including” −65/22 on a number line.
−65/22
Thus, using set-builder and interval notation, the solution is:
{x : x ≥ −65/22} = [−65/22, ∞)
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2.6. INEQUALITIES
123
75. The common denominator is 14. Clear the fractions from the inequality
by multiplying both sides of the inequality by 14.
9
9
1
7
x+ > x+
2 7
2 7
9
9
1
7
14
x+
>
x+
14
7
2
7
2
18x + 63 > 2x + 49
Original inequality.
Multiply both sides by 14.
Cancel and multiply.
We need to isolate all terms containing x on one side of the equation. To
“undo” adding 2x, subtract 2x from both sides of the inequality.
18x + 63 − 2x > 2x + 49 − 2x
Subtract 2x from both sides.
16x + 63 > 49
Simplify both sides.
We continue to isolate terms containing x on one side of the inequality. To
“undo” adding 63, subtract 63 from both sides of the inequality.
16x + 63 − 63 > 49 − 63
Subtract 63 from both sides.
16x > −14
Simplify both sides.
To “undo” the effect of multiplying by 16, divide both sides of the inequality
by 16. Because we are dividing both sides by a positive number, we do not
reverse the inequality sign.
16x
−14
>
16
16
7
x>−
8
Divide both sides by 16.
Simplify both sides.
Shade all real numbers that are “greater than” or “right of” −7/8 on a number
line.
−7/8
Thus, using set-builder and interval notation, the solution is:
{x : x > −7/8} = (−7/8, ∞)
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CHAPTER 2. SOLVING LINEAR EQUATIONS
124
77. We’ll first clear the decimals by multiplying both sides by 100, which will
move each decimal point two places to the right.
−3.7x − 1.98 ≤ 3.2
−370x − 198 ≤ 320
Original inequality.
Multiply both sides by 100.
To “undo” subtracting 198, add 198 to both sides of the inequality.
−370x − 198 + 198 ≤ 320 + 198
−370x ≤ 518
Add 198 to both sides.
Simplify both sides.
To “undo” multiplying by −370, divide both sides of the inequality by −370.
Because we are dividing by a negative number, so we reverse the inequality
sign.
−370x
518
≥
−370
−370
7
x≥−
5
Simplify both sides.
Reduce to lowest terms.
Shade all real numbers that are “greater than or equal to” or “right of or
including” −7/5 on a number line.
−7/5
Thus, using set-builder and interval notation, the solution is:
{x : x ≥ −7/5} = [−7/5, ∞)
79. We’ll first clear the decimals by multiplying both sides by 10, which will
move each decimal point one place to the right.
−3.4x + 3.5 ≥ 0.9 − 2.2x
−34x + 35 ≥ 9 − 22x
Original inequality.
Multiply both sides by 10.
We need to isolate terms containing x on one side of the inequality. Start by
adding 22x to both sides of the inequality.
−34x + 35 + 22x ≥ 9 − 22x + 22x
−12x + 35 ≥ 9
Add 22x to both sides.
Simplify both sides.
We continue to isolate terms containing x on one side of the inequality. Subtract
35 from both sides of the inequality.
−12x + 35 − 35 ≥ 9 − 35
−12x ≥ −26
Second Edition: 2012-2013
Subtract 35 from both sides.
Simplify both sides.
2.6. INEQUALITIES
125
To “undo” multiplying by −12, divide both sides of the inequality by −12.
Because we are dividing by a negative number, we reverse the inequality sign.
−12x
−26
≤
−12
−12
13
x≤
6
Simplify both sides.
Reduce to lowest terms.
Shade all real numbers that are “less than or equal” or “left of or including ”
13/6 on a number line.
13/6
Thus, using set-builder and interval notation, the solution is:
{x : x ≤ 13/6} = (−∞, 13/6]
81. We’ll first clear the decimals by multiplying both sides by 10, which will
move each decimal point one place to the right.
−1.3x + 2.9 > −2.6 − 3.3x
−13x + 29 > −26 − 33x
Original inequality.
Multiply both sides by 10.
We need to isolate terms containing x on one side of the inequality. Start by
adding 33x to both sides of the inequality.
−13x + 29 + 33x > −26 − 33x + 33x
20x + 29 > −26
Add 33x to both sides.
Simplify both sides.
We continue to isolate terms containing x on one side of the inequality. Subtract
29 from both sides of the inequality.
20x + 29 − 29 > −26 − 29
20x > −55
Subtract 29 from both sides.
Simplify both sides.
To “undo” multiplying by 20, divide both sides of the inequality by 20. Because
we are dividing by a positive number, we do not reverse the inequality sign.
−55
20x
>
20
20
11
x>−
4
Simplify both sides.
Reduce to lowest terms.
Shade all real numbers that are “greater than” or “right of ” −11/4 on a
number line.
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CHAPTER 2. SOLVING LINEAR EQUATIONS
126
−11/4
Thus, using set-builder and interval notation, the solution is:
{x : x > −11/4} = (−11/4, ∞)
83. We’ll first clear the decimals by multiplying both sides by 10, which will
move each decimal point one place to the right.
2.2x + 1.9 < −2.3
22x + 19 < −23
Original inequality.
Multiply both sides by 10.
To “undo” adding 19, subtract 19 from both sides of the inequality.
22x + 19 − 19 < −23 − 19
22x < −42
Subtract 19 from both sides.
Simplify both sides.
To “undo” multiplying by 22, divide both sides of the inequality by 22. Because
we are dividing by a positive number, we do not reverse the inequality sign.
22x
−42
<
22
22
21
x<−
11
Simplify both sides.
Reduce to lowest terms.
Shade all real numbers that are “less than” or “left of ” −21/11 on a number
line.
−21/11
Thus, using set-builder and interval notation, the solution is:
{x : x < −21/11} = (−∞, −21/11)
Second Edition: 2012-2013