## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 1 - Section 1.7 - Absolute Value Equations and Inequalities - 1.7 Exercises: 84

#### Answer

$\left( -\infty,-\dfrac{7}{6} \right)\cup\left( \dfrac{3}{2},\infty \right)$

#### Work Step by Step

Using the properties of inequality, the given statement, $|6x-1|-2\gt6 ,$ is equivalent to \begin{array}{l}\require{cancel} |6x-1|\gt6+2 \\\\ |6x-1|\gt8 .\end{array} Since for any $a\gt0$, $|x|\gt a$ implies $x\gt a$ OR $x\lt-a$, then the equation above is equivalent to \begin{array}{l}\require{cancel} 6x-1\gt8 \text{ OR } 6x-1\lt-8 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} 6x-1\gt8 \\\\ 6x\gt8+1 \\\\ 6x\gt9 \\\\ x\gt\dfrac{9}{6} \\\\ x\gt\dfrac{3}{2} \\\\\text{ OR }\\\\ 6x-1\lt-8 \\\\ 6x\lt-8+1 \\\\ 6x\lt-7 \\\\ x\lt-\dfrac{7}{6} .\end{array} Hence, the solution set is the interval $\left( -\infty,-\dfrac{7}{6} \right)\cup\left( \dfrac{3}{2},\infty \right) .$

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