## College Algebra (11th Edition)

Published by Pearson

# Chapter 1 - Section 1.8 - Absolute Value Equations and Inequalities - 1.8 Exercises - Page 154: 33

#### Answer

$\left( -\infty,0 \right)\cup \left( 6,\infty \right)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $4\left|x-3 \right|\gt12 ,$ use the properties of inequality to isolate the absolute value expression. Then remove the absolute value sign using the properties of absolute value inequality. Then use the properties of inequality to isolate the variable. $\bf{\text{Solution Details:}}$ Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} \left|x-3 \right|\gt\dfrac{12}{4} \\\\ \left|x-3 \right|\gt3 .\end{array} For any $a\gt0,$ $|x|\ge a$ implies $x\ge a$ OR $x\le -a$ (Note that the symbols $\ge$ may be replaced with $\gt$ and the symbol $\le$ with $\lt.$) Hence, the inequality above is equivalent to \begin{array}{l}\require{cancel} x-3\gt3 \\\\\text{OR}\\\\ x-3\lt-3 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} x-3\gt3 \\\\ x\gt3+3 \\\\ x\gt6 \\\\\text{OR}\\\\ x-3\lt-3 \\\\ x\lt-3+3 \\\\ x\lt0 .\end{array} Hence, the solution is the interval $\left( -\infty,0 \right)\cup \left( 6,\infty \right) .$

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