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Solving Absolute Value Equations and Inequalities

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1.7<br />

<strong>Solving</strong> <strong>Absolute</strong> <strong>Value</strong><br />

<strong>Equations</strong> <strong>and</strong> <strong>Inequalities</strong><br />

Goals p Solve absolute value equations <strong>and</strong> inequalities.<br />

p Use absolute values in real-life problems.<br />

Your Notes<br />

VOCABULARY<br />

<strong>Absolute</strong> value of a real number The distance the<br />

number is from 0 on a number line<br />

SOLVING AN ABSOLUTE VALUE EQUATION<br />

The absolute value equation ⏐ax b⏐ c, where c > 0, is<br />

equivalent to the compound statement ax b c<br />

or ax b c .<br />

Example 1<br />

<strong>Solving</strong> an <strong>Absolute</strong> <strong>Value</strong> Equation<br />

⏐4x 2⏐ 6<br />

Original equation<br />

4x 2 6 or 4x 2 6 Expression can be 6<br />

or 6 .<br />

4x 4 or 4x 8 Subtract 2 from each<br />

side.<br />

x 1 or x 2 Divide each side by 4 .<br />

TRANSFORMATIONS OF ABSOLUTE VALUE INEQUALITIES<br />

p The inequality ⏐ax b⏐< c, where c > 0, means that<br />

ax b is between c <strong>and</strong> c. This is equivalent to<br />

c < ax b < c .<br />

p The inequality ⏐ax b⏐> c, where c > 0, means that<br />

ax b is beyond c <strong>and</strong> c. This is equivalent to<br />

ax b < c or ax b > c .<br />

In the first transformation, < can be replaced by ≤ . In the<br />

second transformation, > can be replaced by ≥ .<br />

22 Algebra 2 Notetaking Guide • Chapter 1


Your Notes<br />

Example 2<br />

<strong>Solving</strong> an Inequality of the form⏐ax b⏐ < c<br />

Solve ⏐2x 3⏐ < 3.<br />

Solution<br />

⏐2x 3⏐ < 3 Write original inequality.<br />

3 < 2x 3 < 3 Write equivalent compound inequality.<br />

0 < 2x < 6 Add 3 to each expression.<br />

0 < x < 3 Divide each expression by 2 .<br />

The solutions are all real numbers greater than 0 <strong>and</strong> less<br />

than 3 . Check several solutions in the original inequality.<br />

Graph the solution below.<br />

2 1<br />

0 1 2 3 4 5<br />

Example 3<br />

<strong>Solving</strong> an Inequality of the form ⏐ax b⏐ ≥ c<br />

Solve ⏐2x 7⏐ ≥ 3.<br />

Solution<br />

This absolute value inequality is equivalent to 2x 7 ≤ 3<br />

or 2x 7 ≥ 3 .<br />

Solve first inequality<br />

Solve second inequality<br />

2x 7 ≤ 3 Write inequality. 2x 7 ≥ 3<br />

2x ≤ 10 Subtract 7 2x ≥ 4<br />

from each side.<br />

x ≤ 5 Divide each side x ≥ 2<br />

by 2 .<br />

The solutions are all real numbers less than or equal to 5<br />

or greater than or equal to 2 . Check several solutions in<br />

the original inequality. Graph the solution below.<br />

7<br />

6 5<br />

4 3 2 1<br />

0<br />

Lesson 1.7 • Algebra 2 Notetaking Guide 23


Your Notes<br />

Checkpoint Solve the equation or inequality.<br />

1.⏐5x 2⏐ 3 2.⏐3x 4⏐ < 2 3.⏐2x 3⏐ ≥ 5<br />

1 5 ,1 2 < x < 2 x ≤ 1 or<br />

3<br />

x ≥ 4<br />

Example 4<br />

Write a Model for Tolerance<br />

A dog food manufacturer has a tolerance of 0.25 pound per<br />

bag of dog food advertised as weighing 5 pounds. Write <strong>and</strong><br />

solve an absolute value inequality that describes the<br />

acceptable weights for “5 pound” bags.<br />

Verbal<br />

Model<br />

Actual weight<br />

<br />

Ideal weight<br />

≤<br />

Tolerance<br />

Labels<br />

Actual weight x<br />

(pounds)<br />

Ideal weight 5<br />

(pounds)<br />

Tolerance 0.25<br />

(pounds)<br />

Algebraic<br />

Model<br />

⏐x 5⏐ ≤ 0.25<br />

Write algebraic model.<br />

0.25 ≤ x 5 ≤ 0.25 Write equivalent<br />

compound inequality.<br />

4.75 ≤ x ≤ 5.25 Add 5 to each<br />

expression.<br />

The weights can range between 4.75 pounds <strong>and</strong><br />

5.25 pounds, inclusive.<br />

Checkpoint Complete the following exercise.<br />

Homework<br />

4. A toy manufacturer has a tolerance of 0.1 inch on a ball<br />

that is supposed to have a diameter of 1 inch. Write <strong>and</strong><br />

solve an inequality describing the acceptable diameters<br />

for a ball.<br />

⏐d 1⏐ ≤ 0.1, where d represents the actual<br />

diameter; 0.9 ≤ d ≤ 1.1<br />

24 Algebra 2 Notetaking Guide • Chapter 1

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