1. 9.2 Absolute Value Equations and Inequalities

2. Use the distance definition of absolute value.
Objective 1
Use the distance definition of absolute value.
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3. Use the distance definition of absolute value.
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. We need to understand the concept of absolute value in order to solve equations or inequalities involving absolute values. We solve them by solving the appropriate compound equation or inequality.
Distance is 5, so |5| = 5.
Distance is 5, so |5| = 5.
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4. Use the distance definition of absolute value.
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5. Solve equations of the form |ax + b| = k, for k > 0.
Objective 2
Solve equations of the form
|ax + b| = k, for k > 0.
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6. Use the distance definition of absolute value.
Remember that because absolute value refers to distance from the origin, an absolute value equation will have two parts.
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7. CLASSROOM EXAMPLE 1
Solving an Absolute Value Equation
Solve |3x – 4| = 11.
3x – 4 =  or x – 4 = 11
3x – =  x – =
3x = 7 3x = 15
x = 5
Check by substituting and 5 into the original absolute value equation to verify that the solution set is
Solution:
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8. Solve inequalities of the form |ax + b| < k and of the form
Objective 3
Solve inequalities of the form
|ax + b| < k and of the form
|ax + b| > k, for k > 0.
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9. ] [ Solve |3x – 4|  11. 3x – 4 ≤ 11 or 3x – 4  11
CLASSROOM EXAMPLE 2
Solving an Absolute Value Inequality with >
Solve |3x – 4|  11.
3x – 4 ≤  or x – 4  11
3x – ≤  x – 
3x ≤  x  15
x  5
Check the solution. The solution set is
The graph consists of two intervals.
Solution: 8 -4 -2 2 4 6 -5 -1 3 7 -3 5 1 ] [
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10. [ ] Solve |3x – 4| ≤ 11. 11 ≤ 3x – 4 ≤ 11 11 + 4 ≤ 3x – 4 ≤ 11+ 4
CLASSROOM EXAMPLE 3
Solving an Absolute Value Inequality with <
Solve |3x – 4| ≤ 11.
11 ≤ 3x – 4 ≤ 11
 ≤ 3x – 4 ≤ 11+ 4
7 ≤ 3x ≤ 15
Check the solution. The solution set is
The graph consists of a single interval.
Solution: 8 -4 -2 2 4 6 -5 -1 3 7 -3 5 1 [ ]
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11. Solve inequalities of the form |ax + b| < k and of the form |ax + b| > k, for k > 0.
When solving absolute value equations and inequalities of the types in Examples 1, 2, and 3, remember the following:
1. The methods describe apply when the constant is alone on one side of the equation or inequality and is positive.
2. Absolute value equations and absolute value inequalities of the form |ax + b| > k translate into “or” compound statements.
3. Absolute value inequalities of the form |ax + b| < k translate into “and” compound statements, which may be written as three-part inequalities.
4. An “or” statement cannot be written in three parts.
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12. Solve absolute value equations that involve rewriting.
Objective 4
Solve absolute value equations that involve rewriting.
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13. First get the absolute value alone on one side of the equals sign.
CLASSROOM EXAMPLE 4
Solving an Absolute Value Equation That Requires Rewriting
Solve |3x + 2| + 4 = 15.
First get the absolute value alone on one side of the equals sign.
|3x + 2| + 4 = 15
|3x + 2| + 4 – 4 = 15 – 4
|3x + 2| = 11
3x + 2 =  or x + 2 = 11
3x = 13 3x = 9
x = 3
The solution set is
Solution:
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14. Solution set: (, 7)  (3, )
CLASSROOM EXAMPLE 5
Solving Absolute Value Inequalities That Require Rewriting
Solve the inequality.
|x + 2| – 3 > 2
|x + 2| > 5
x + 2 > or x + 2 <  5
x > x < 7
Solution set: (, 7)  (3, )
Solution:
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15. Solve the inequality. |x + 2| – 3 < 2 |x + 2| < 5
CLASSROOM EXAMPLE 5
Solving Absolute Value Inequalities That Require Rewriting (cont’d)
Solve the inequality.
|x + 2| – 3 < 2
|x + 2| < 5
5 < x + 2 < 5
7 < x < 3
Solution set: (7, 3)
Solution:
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16. Solve equations of the form |ax + b| = |cx + d| .
Objective 5
Solve equations of the form
|ax + b| = |cx + d| .
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17. Solving |ax + b| = |cx + d|
Solve equations of the form |ax + b| = |cx + d|.
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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18. Check that the solution set is
CLASSROOM EXAMPLE 6
Solving an Equation with Two Absolute Values
Solve |4x – 1| = |3x + 5|.
4x – 1 = 3x or x – 1 = (3x + 5)
4x – 6 = 3x or x – 1 = 3x – 5
 6 =  x or x =  4
x = or
Check that the solution set is
Solution:
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19. Solve special cases of absolute value equations and inequalities.
Objective 6
Solve special cases of absolute value equations and inequalities.
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20. Special Cases of Absolute Value
Solve special cases of absolute value equations and inequalities.
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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21. Solve each equation. |6x + 7| = – 5
CLASSROOM EXAMPLE 7
Solving Special Cases of Absolute Value Equations
Solve each equation.
|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 .
Solution:
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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22. Solve each inequality. |x | > – 1
CLASSROOM EXAMPLE 8
Solving Special Cases of Absolute Value Inequalities
Solve each inequality.
|x | > – 1
The absolute value of a number is always greater than or equal to 0. The solution set is (, ).
|x – 10| – 2 ≤ –3
|x – 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 .
|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}.
Solution:
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