Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 3.3 - 1.

Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 3.3 - 1


Linear Inequalities and Absolute Value

Chapter 3


3.3

Absolute Value Equations

and Inequalities

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.3 - 4

3.3 Absolute Value Equations and Inequalities

Objectives

1. Use the distance definition of absolute value.

2. Solve equations of the form |ax + b| = k, for k > 0.

3. Solve inequalities of the form |ax + b| < k and of the form |ax + b| > k, for k > 0.

4. Solve absolute value equations that involve rewriting.

5. Solve equations of the form |ax + b| = |cx + d|.

6. Solve special cases of absolute value equationsand inequalities.



Absolute Value

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.

0 5–5

For example, the solutions of |x| = 5 are 5 and –5.

5 units from zero. 5 units from zero.

x = –5 or x = 5



Absolute Value

0 5–5

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.

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.

More than

5 units from zero

More than

5 units from zero



Absolute Value

0 5–5

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



Absolute Value

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, |ax + b| < k,or

where k is a positive number.

|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.

From the previous examples, we see that


3.3 Absolute Value Equations and InequalitiesSummary:

Solving Absolute Value Equations and Inequalities

1. To solve |ax + b| = k, solve the following compound equation.

Let k be a positive real number, and p and q be real numbers.

ax + b = k or ax + b = –k.

The solution set is usually of the form {p, q}, which includes two

numbers.

p q




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




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



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

x = 1

2x = –8

x = –4

or

or

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 0 1 2 3 4 5



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.



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

2x > 2

x > 1

2x < –8

x < –4

or

or

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.

–5 –4 –3 –2 –1 0 1 2 3 4 5

2x + 3 > 5 or 2x + 3 < –5,

Check these solutions. The solution set is (–∞, –4) U (1, ∞). Notice that the

graph consists of two intervals.



EXAMPLE 3 Solving an Absolute Value Inequality with <

Solve |2x + 3| < 5.

–5 < 2x + 3 < 5.

–8 < 2x < 2

–4 < x < 1

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 –4 –3 –2 –1 0 1 2 3 4 5

Check that the solution set is (–4, 1), so the graph consists of the single

interval shown below.

Subtract 3 from each part.

Divide each part by 2.


CAUTION

When solving absolute value equations and inequalities of the types inExamples 1, 2, and 3, remember the following.1. The methods described 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-partinequalities.

4. An “or” statement cannot be written in three parts. It would be incorrectto use –5 > 2x + 3 > 5 in Example 2, because this would imply that–5 > 5, which is false.


Caution



EXAMPLE 4 Solving an Absolute Value Equation That

Requires Rewriting

Solve the equation |x – 7| + 6 = 9.

|x – 7| + 6 – 6 = 9 – 6

|x – 7| = 3

x – 7 = 3 or x – 7 = –3

First, rewrite so that the absolute value expression is alone on one side

of the equals sign by subtracting 6 from each side.

Now use the method shown in Example 1.

Subtract 6.

x = 10

Check that the solution set is {4, 10} by substituting 4 and then 10 into the

original equation.

or x = 4



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).



EXAMPLE 5 Solving an Equation with Two Absolute

Values

Solve the equation |y + 4| = |2y – 7|.

y + 4 = 2y – 7 or y + 4 = –(2y – 7).

y + 4 = 2y – 7 or 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,

Solve each equation.

11 = y

Check that the solution set is {1, 11}.

3y = 3

y = 1



Special Cases for Absolute Value

Special Cases for Absolute Value

1. The absolute value of an expression can never be negative: |a| ≥ 0

for all real numbers a.

2. The absolute value of an expression equals 0 only when the expression is equal to 0.



EXAMPLE 6 Solving Special Cases of Absolute Value

Equations

Solve each equation.

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.

(a) |2n + 3| = –7

See Case 2 in the preceding slide. The absolute value of the expres-sion 6w – 1 will equal 0 only if

6w – 1 = 0.

(b) |6w – 1| = 0

The solution of this equation is . Thus, the solution set of the original

equation is { }, with just one element. Check by substitution.

16

16




Inequalities

Solve each inequality.

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 (–∞, ∞).

(a) |x| ≥ –2

Add 1 to each side to get the absolute value expression alone on oneside.

|x + 5| < –7

(b) |x + 5| – 1 < –8

There is no number whose absolute value is less than –7, so this inequalityhas no solution. The solution set is Ø.




Inequalities

Solve each inequality.

Subtracting 2 from each side gives

|x – 9| ≤ 0

(c) |x – 9| + 2 ≤ 2

The value of |x – 9| will never be less than 0. However, |x – 9| will equal 0when x = 9. Therefore, the solution set is {9}.

Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 3.3 - 1.

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