Intermediate Algebra (12th Edition)

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

Chapter 1 - Chapter 1 Test - Page 134: 21

Answer

$\left[ -\dfrac{5}{2},1 \right]$

Work Step by Step

Since for any $a\gt0$, $|x|\lt a$ implies $-a\lt x\lt a,$ then the given inequality, $ |4x+3|\le7 ,$ is equivalent to \begin{array}{l}\require{cancel} -7\le 4x+3\le7 .\end{array} Using the properties of inequality, then \begin{array}{l}\require{cancel} -7-3\le 4x+3-3\le7-3 \\\\ -10\le 4x\le4 \\\\ -\dfrac{10}{4}\le \dfrac{4x}{4}\le\dfrac{4}{4} \\\\ -\dfrac{5}{2}\le x\le1 .\end{array} Hence, the solution is the interval $ \left[ -\dfrac{5}{2},1 \right] .$
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