Intermediate Algebra (12th Edition)

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

Chapter 1 - Mixed Review Exercises - Page 132: 17


$x=\left\{ \dfrac{11}{3},1 \right\}$

Work Step by Step

Since for any $a\gt0$, $|x|=a$ implies $x=a$ or $x=-a$, then the given equation, $ |3x-7|=4 ,$ is equivalent to \begin{array}{l}\require{cancel} 3x-7=4 \text{ OR } 3x-7=-4 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 3x-7=4 \\\\ 3x=4+7 \\\\ 3x=11 \\\\ x=\dfrac{11}{3} \\\\\text{ OR }\\\\ 3x-7=-4 \\\\ 3x=-4+7 \\\\ 3x=3 \\\\ x=\dfrac{3}{3} \\\\ x=1 .\end{array} Hence, the solutions are $ x=\left\{ \dfrac{11}{3},1 \right\} .$
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