Intermediate Algebra (12th Edition)

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

Chapter 1 - Section 1.7 - Absolute Value Equations and Inequalities - 1.7 Exercises: 83

Answer

$(-\infty,-3)\cup(2,\infty)$

Work Step by Step

Using the properties of inequality, the given statement, $ |2x+1|+3\gt8 ,$ is equivalent to \begin{array}{l}\require{cancel} |2x+1|\gt8-3 \\\\ |2x+1|\gt5 .\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} 2x+1\gt5 \text{ OR } 2x+1\lt-5 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} 2x+1\gt5 \\\\ 2x\gt5-1 \\\\ 2x\gt4 \\\\ x\gt\dfrac{4}{2} \\\\ x\gt2 \\\\\text{ OR }\\\\ 2x+1\lt-5 \\\\ 2x\lt-5-1 \\\\ 2x\lt-6 \\\\ x\lt-\dfrac{6}{2} \\\\ x\lt-3 .\end{array} Hence, the solution set is the interval $ (-\infty,-3)\cup(2,\infty) .$
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