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: 25


$x=\left\{ -\dfrac{32}{3},8 \right\}$

Work Step by Step

Since for any $a\gt0$, $|x|=a$ implies $x=a$ OR $x=-a$, then the given equation, $ \left|1+\dfrac{3}{4}x\right|=7 ,$ is equivalent to \begin{array}{l}\require{cancel} 1+\dfrac{3}{4}x=7 \text{ OR } 1+\dfrac{3}{4}x=-7 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 1+\dfrac{3}{4}x=7 \\\\ 4\left(1+\dfrac{3}{4}x\right)=(7)4 \\\\ 4+3x=28 \\\\ 3x=28-4 \\\\ 3x=24 \\\\ x=\dfrac{24}{3} \\\\ x=8 \\\\\text{ OR }\\\\ 1+\dfrac{3}{4}x=-7 \\\\ 4\left(1+\dfrac{3}{4}x\right)=(-7)4 \\\\ 4+3x=-28 \\\\ 3x=-28-4 \\\\ 3x=-32 \\\\ x=-\dfrac{32}{3} .\end{array} Hence, the solutions are $ x=\left\{ -\dfrac{32}{3},8 \right\} .$
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