Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 1 - Summary Exercises - Reviewing a Chapter - Solving Linear and Absolute Value Equations and Inequalities: 39

Answer

$\left( -\infty, -4 \right) \cup \left( 7, \infty\right)$

Work Step by Step

Since for any $a\gt0$, $|x|\gt a$ implies $x\gt a$ or $x\lt-a,$ then the given inequality, $ |2x-3|\gt11 ,$ is equivalent to \begin{array}{l}\require{cancel} 2x-3\gt11 \text{ OR } 2x-3\lt-11 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} 2x-3\gt11 \\\\ 2x\gt11+3 \\\\ 2x\gt14 \\\\ x\gt\dfrac{14}{2} \\\\ x\gt7 \\\\\text{ OR }\\\\ 2x-3\lt-11 \\\\ 2x\lt-11+3 \\\\ 2x\lt-8 \\\\ x\lt-\dfrac{8}{2} \\\\ x\lt-4 .\end{array} Hence, the solution is the interval $ \left( -\infty, -4 \right) \cup \left( 7, \infty\right) .$
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