2. Linear Inequalities and Absolute Value
Chapter 3
Linear Inequalities and Absolute Value

3. Absolute Value Equations and Inequalities
3.3
Absolute Value Equations
and Inequalities

4. 3.3 Absolute Value Equations and Inequalities
Objectives
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.
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5. 3.3 Absolute Value Equations and Inequalities
In Section1.1 we saw that the absolute value of a number x, written |x|,
represents the distance from x to 0 on the number line.
For example, the solutions of |x| = 5 are 5 and –5.
5 units from zero.
5 units from zero. 5 –5
x = –5 or x = 5
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6. 3.3 Absolute Value Equations and Inequalities
Because the absolute value represents the distance from 0, it is
reasonable to interpret the solutions of |x| > 5 to be all numbers that are
more than 5 units from 0.
More than
5 units from zero
More than
5 units from zero 5 –5
The set (-∞, –5) U (5, ∞) fits this description. Because the graph consists of
two separate intervals, the solution set is described using or as x < –5 or
x > 5.
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7. 3.3 Absolute Value Equations and Inequalities
The solution set of |x| < 5 consists of all numbers that are less than 5
units from 0 on the number line. Another way of thinking of this is to think
of all numbers between –5 and 5. This set of numbers is given by (–5, 5),
as shown in the figure below. Here, the graph shows that –5 < x < 5, which
means x > –5 and x < 5.
Less than 5 units from zero 5 –5
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8. 3.3 Absolute Value Equations and Inequalities
The equation and inequalities just described are examples of absolute
value equations and inequalities. They involve the absolute value of a
variable expression and generally take the form
|ax + b| = k,
|ax + b| > k, or
|ax + b| < k,
where k is a positive number.
From the previous examples, we see that
|x| = 5 has the same solution set as x = –5 or x = 5,
|x| > 5 has the same solution set as x < –5 or x > 5,
|x| < 5 has the same solution set as x > –5 and x < 5.
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9. 3.3 Absolute Value Equations and Inequalities
Summary:
Solving Absolute Value Equations and Inequalities
Let k be a positive real number, and p and q be real numbers.
1. To solve |ax + b| = k, solve the following compound equation.
ax + b = k or ax + b = –k.
The solution set is usually of the form {p, q}, which includes two
numbers. p q
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10. 3.3 Absolute Value Equations and Inequalities
Summary:
Solving Absolute Value Equations and Inequalities
Let k be a positive real number, and p and q be real numbers.
2. To solve |ax + b| > k, solve the following compound inequality.
ax + b > k or ax + b < –k.
The solution set is of the form (-∞, p) U (q, ∞), which consists of two
separate intervals. p q
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11. 3.3 Absolute Value Equations and Inequalities
Summary:
Solving Absolute Value Equations and Inequalities
Let k be a positive real number, and p and q be real numbers.
3. To solve |ax + b| < k, solve the three-part inequality
–k < ax + b < k
The solution set is of the form (p, q), a single interval. p q
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12. 3.3 Absolute Value Equations and Inequalities
EXAMPLE 1
Solving an Absolute Value Equation
Solve |2x + 3| = 5.
For |2x + 3| to equal 5, 2x + 3 must be 5 units from 0 on the number line.
This can happen only when 2x + 3 = 5 or 2x + 3 = –5. Solve this compound
equation as follows.
2x + 3 = 5 or 2x + 3 = –5
2x = 2 or
2x = –8
x = 1 or
x = –4
Check by substituting 1 and then –4 in the original absolute value equation
to verify that the solution set is {–4, 1}. –5 –4 –3 –2 –1 1 2 3 4 5
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13. 3.3 Absolute Value Equations and Inequalities
Writing Compound Statements
NOTE
Some people prefer to write the compound statements in Cases 1 and 2 of
the summary on the previous slides as the equivalent forms
ax + b = k or –(ax + b) = k
and ax + b > k or –(ax + b) > k.
These forms produce the same results.
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14. 3.3 Absolute Value Equations and Inequalities
EXAMPLE 2
Solving an Absolute Value Inequality with >
Solve |2x + 3| > 5.
By part 2 of the summary, this absolute value inequality is rewritten as
2x + 3 > 5 or 2x + 3 < –5,
because 2x + 3 must represent a number that is more than 5 units from 0 on
either side of the number line. Now, solve the compound inequality.
2x + 3 > 5 or 2x + 3 < –5
2x > 2 or
2x < –8
x > 1 or
x < –4
Check these solutions. The solution set is (–∞, –4) U (1, ∞). Notice that the
graph consists of two intervals. –5 –4 –3 –2 –1 1 2 3 4 5
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15. 3.3 Absolute Value Equations and Inequalities
EXAMPLE 3
Solving an Absolute Value Inequality with <
Solve |2x + 3| < 5.
The expression 2x + 3 must represent a number that is less than 5 units
from 0 on either side of the number line. 2x + 3 must be between –5 and 5.
As part 3 of the summary shows, this is written as the three-part inequality
–5 < 2x + 3 < 5.
–8 < 2x < 2
Subtract 3 from each part.
–4 < x < 1
Divide each part by 2.
Check that the solution set is (–4, 1), so the graph consists of the single
interval shown below. –5 –4 –3 –2 –1 1 2 3 4 5
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16. 3.3 Absolute Value Equations and Inequalities
Caution
CAUTION
When solving absolute value equations and inequalities of the types in
Examples 1, 2, and 3, remember the following.
The methods described apply when the constant is alone on one side
of the equation or inequality and is positive.
Absolute value equations and absolute value inequalities of the form
|ax + b| > k translate into “or” compound statements.
Absolute value inequalities of the form |ax + b| < k translate into
“and” compound statements, which may be written as three-part
inequalities.
An “or” statement cannot be written in three parts. It would be incorrect
to use –5 > 2x + 3 > 5 in Example 2, because this would imply that
–5 > 5, which is false.
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17. 3.3 Absolute Value Equations and Inequalities
EXAMPLE 4
Solving an Absolute Value Equation That
Requires Rewriting
Solve the equation |x – 7| + 6 = 9.
First, rewrite so that the absolute value expression is alone on one side
of the equals sign by subtracting 6 from each side.
|x – 7| + 6 – 6 = 9 – 6
Subtract 6.
|x – 7| = 3
Now use the method shown in Example 1.
x – 7 = 3 or x – 7 = –3
x = 10 or
x = 4
Check that the solution set is {4, 10} by substituting 4 and then 10 into the
original equation.
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18. 3.3 Absolute Value Equations and Inequalities
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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19. 3.3 Absolute Value Equations and Inequalities
EXAMPLE 5
Solving an Equation with Two Absolute
Values
Solve the equation |y + 4| = |2y – 7|.
This equation is satisfied either if y + 4 and 2y – 7 are equal to each other,
or if y + 4 and 2y – 7 are negatives of each other. Thus,
y + 4 = 2y – or y + 4 = –(2y – 7).
Solve each equation.
y + 4 = 2y – or y + 4 = –(2y – 7)
11 = y
3y = 3
y = 1
Check that the solution set is {1, 11}.
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20. 3.3 Absolute Value Equations and Inequalities
Special Cases for Absolute Value
Special Cases for Absolute Value
The absolute value of an expression can never be negative: |a| ≥ 0
for all real numbers a.
The absolute value of an expression equals 0 only when the expression is equal to 0.
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21. 3.3 Absolute Value Equations and Inequalities
EXAMPLE 6
Solving Special Cases of Absolute Value
Equations
Solve each equation.
(a) |2n + 3| = –7
See Case 1 in the preceding slide. Since the absolute value of an
expression can never be negative, there are no solutions for this equation.
The solution set is Ø.
(b) |6w – 1| = 0
See Case 2 in the preceding slide. The absolute value of the expres-
sion 6w – 1 will equal 0 only if
6w – 1 = 0. 1 6
The solution of this equation is . Thus, the solution set of the original
equation is { }, with just one element. Check by substitution. 1 6
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22. 3.3 Absolute Value Equations and Inequalities
EXAMPLE 7
Solving Special Cases of Absolute Value
Inequalities
Solve each inequality.
(a) |x| ≥ –2
The absolute value of a number is always greater than or equal to 0.
Thus, |x| ≥ –2 is true for all real numbers. The solution set is (–∞, ∞).
(b) |x + 5| – 1 < –8
Add 1 to each side to get the absolute value expression alone on one
side.
|x + 5| < –7
There is no number whose absolute value is less than –7, so this inequality
has no solution. The solution set is Ø.
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23. 3.3 Absolute Value Equations and Inequalities
EXAMPLE 7
Solving Special Cases of Absolute Value
Inequalities
Solve each inequality.
(c) |x – 9| + 2 ≤ 2
Subtracting 2 from each side gives
|x – 9| ≤ 0
The value of |x – 9| will never be less than 0. However, |x – 9| will equal 0
when x = 9. Therefore, the solution set is {9}.
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