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2. Absolute Value Equations and Inequalities
2.6
Absolute Value Equations and Inequalities
Use the distance definition of absolute value.
Solve equations of the form |ax + b| = k, for k > 0.
Solve inequalities of the form |ax + b| < k and of the form |ax + b| > k, for k > 0.
Solve absolute value equations that involve rewriting.
Solve equations of the form |ax + b| = |cx + d| .
Solve special cases of absolute value equations and inequalities. 1 2 3 4 5 6
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Objective 1
Use the distance definition of absolute value.
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The absolute value of a number x, written |x|, is the distance from x to 0 on the number line.
For example, the solutions of |x| = 5 are 5 and 5, as shown below.
Distance is 5, so |5| = 5.
Distance is 5, so |5| = 5.
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Objective 2
Solve equations of the form |ax + b| = k, for k > 0.
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EXAMPLE 1
Solve |3x – 4| = 11.
3x – 4 =  or x – 4 = 11
3x – =  x – =
3x =  x = 15
x = 7/ x = 5
Check by substituting 7/3 and 5 into the original absolute value equation to verify that the solution set is {7/3, 5}.
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Objective 3
Solve inequalities of the form |ax + b| < k and of the form |ax + b| > k, for k > 0.
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EXAMPLE 2
Solve |3x – 4|  11.
3x – 4 ≤  or x – 4  11
3x – ≤  x – 
3x ≤  x  15
x ≤ 7/ x  5
Check the solution. The solution set is (, 7/3]  [5, ). The graph consists of two intervals. 8 -4 -2 2 4 6 -5 -1 3 7 -3 5 1 ] [
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EXAMPLE 3
Solve |3x – 4| < 11.
11< 3x – 4 < 11
 < 3x – 4 < 11+ 4
7 < 3x < 15
7/3 < x < 5
Check the solution. The solution set is (7/3, 5). The graph consists of a single interval. 8 -4 -2 2 4 6 -5 -1 3 7 -3 5 1 ( )
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Objective 4
Solve absolute value equations that involve rewriting.
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EXAMPLE 4
Solve |3a + 2| + 4 = 15.
First get the absolute value alone on one side of the equals sign.
|3a + 2| + 4 = 15
|3a + 2| + 4 – 4 = 15 – 4
|3a + 2| = 11
3a + 2 =  or a + 2 = 11
3a =  a = 9
a = 13/ a = 3
Check that the solution set is {13/3, 3}.
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EXAMPLE 5
Solve each inequality.
a. |x + 2| – 3 > 2.
|x + 2| – 3 = 2
|x + 2| = 5
x + 2 > or x + 2 < 5
x > x < 7
Solution set: (, 7)  (3, ).
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continued
b. |3x + 2| + 4 ≤ 15.
|3x + 2| + 4 ≤ 15
|3x + 2| ≤ 11
11 ≤ 3x + 2 ≤ 11
13 ≤ 3x ≤ 9
13/3 ≤ x ≤ 3
Solution set:
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Objective 5
Solve equations of the form |ax + b| = |cx + d| .
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Solving |ax + b| = |cx + d|
To solve an absolute value equation of the form
|ax + b| = |cx + d|,
solve the compound equation
ax + b = cx + d or ax + b = (cx + d).
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EXAMPLE 5
Solve |4r – 1| = |3r + 5|.
4r – 1 = 3r or r – 1 = (3r + 5)
4r – 6 = 3r or r – 1 = 3r – 5
6 = r or 7r = 4
r = or r = 4/7
Check that the solution set is
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Objective 6
Solve special cases of absolute value equations and inequalities.
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Special Cases of Absolute Value
1. The absolute value of an expression can never be negative; that is, |a|  0 for all real numbers a.
2. The absolute value of an expression equals 0 only when the expression is equal to 0.
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EXAMPLE 6
Solve each equation.
a. |6x + 7| = –5
|6x + 7| = –5
The absolute value of an expression can never be negative, so there are no solutions for this equation. The solution set is .
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continued b.
The expression will equal 0 only if
The solution of the equation is 12.
The solution set is {12}, with just one element.
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EXAMPLE 7
Solve each inequality.
a. |x | > –5
The absolute value of a number is always greater than or equal to 0.
The solution set is (, ).
b. |t – 10| – 2 ≤ –3
|t – 10| ≤ –1 Add 2 to each side.
There is no number whose absolute value is less than –1, so the inequality has no solution. The solution set is .
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continued
c. |x + 2| ≤ 0
The value of |x + 2| will never be less than 0.
|x + 2| will equal 0 when x = –2.
The solution set is {–2}.
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