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M098
Carson Elementary and Intermediate Algebra 3e
Section 8.2
Objectives
1.
Solve equations involving absolute value
Vocabulary
Prior Knowledge
Absolute Value:
 Absolute value is the distance a number is from 0.

Absolute value is never negative.
- (2x + 3) means the opposite of (2x + 3). Be sure to distribute the minus sign to both terms in the
parentheses. - (2x + 3) = – 2x – 3
New Concepts
As we begin these new problem types, keep in mind the absolute value concept. Absolute value will never be
a negative number. It is always either 0 or a positive number. (Be careful not to say “Absolute value is always
positive” because 0 by definition is neither positive nor negative so | 0 | is neither positive nor negative.)
For Example:
| something | > -5
Always true. The absolute value will always be greater than a negative number.
No solution necessary. The answer is all real numbers: ( -∞, ∞).
| something | = -1
Can't happen! Absolute value is never negative. No solution necessary. The
answer is the empty set: Ø.
| something | < -2
Can't happen! To be less than a negative number, the absolute value would have
to be negative. No solution necessary. The answer is the empty set: Ø.
| something | ≥ 0
Always true. Absolute value is either 0 or a positive number. No solution
necessary. The answer is all real numbers: (-∞, ∞)
Solve equations involving absolute value.
By looking at a basic absolute value equation like | x | = 3, we see that there will be two possible
solutions: 3 and -3. To solve absolute value equations then, we need to set up two cases. The quantity
in the absolute value can equal either the number on the right or the opposite of that number.
In order to use the Absolute Value Property Rule, the equation must be in the proper form. The absolute
value must be isolated on one side of the equation.
Example 1:
|x+3|–1=5
|x+3|=6
V. Zabrocki 2011
Because this equation contains an absolute value, we
must first isolate the absolute value by adding 1 to both
sides.
page 1
M098
Carson Elementary and Intermediate Algebra 3e
x+3=6
x+3=–6
x=3
x = -9
Section 8.2
Separate the equation into the positive case and the
negative case.
Solve each equation by subtracting 3. The solutions are 3
and -9.
Example 2:
–3 = 1 – | 4u – 2 |
– 4 = – | 4u – 2 |
Because this equation contains an absolute value, we
must first isolate the absolute value by subtracting 1 from
both sides.
4 = | 4u – 2 |
Multiply through by -1 to make the absolute value positive.
4 = 4u – 2
- 4 = 4u – 2
6 = 4u
6
u 
4
3
u 
2
-2 = 4u
2
u  
4
1
u  
2
Separate the equation into the positive case and the
negative case.
Solve each equation. Always reduce fractions to lowest
terms.
The solutions are 
1
3
and .
2
2
Example 3:
| 6p + 5 | + 4 = 1
| 6p + 5 | = -3
Because this equation contains an absolute value, we
must first isolate the absolute value by subtracting 4 from
both sides.
No solution - Ø
Absolute value is never negative.
Two absolute values: Again, by looking at a simple case like | x | = | 3 |, we can see that there will be two
solutions: 3 or -3. Once again, if the equation contains two absolute value terms, there will be two
cases.
Example 4:
| p – 1 | = | 2p + 8 |
p – 1 = 2p + 8
p – 1 = -(2p + 8)
p – 1 = -2p – 8
-p = 9
p = -9
3p = -7
7
p  
3
V. Zabrocki 2011
Separate the equation into the positive case and the
negative case. Distribute the negative to both terms in
the negative case.
Solve each equation. The solutions are -9 and 
7
.
3
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