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: 90


$x=\left\{ -\dfrac{13}{4},\dfrac{11}{4} \right\}$

Work Step by Step

Using the properties of equality, the given equation, $ \left|\dfrac{2}{3}x+\dfrac{1}{6}\right|+\dfrac{1}{2}=\dfrac{5}{2} ,$ is equivalent to \begin{array}{l}\require{cancel} \left|\dfrac{2}{3}x+\dfrac{1}{6}\right|=\dfrac{5}{2}-\dfrac{1}{2} \\\\ \left|\dfrac{2}{3}x+\dfrac{1}{6}\right|=\dfrac{4}{2} \\\\ \left|\dfrac{2}{3}x+\dfrac{1}{6}\right|=2 .\end{array} Since for any $a\gt0$, $|x|=a$ implies $x=a$ OR $x=-a$, then the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{2}{3}x+\dfrac{1}{6}=2 \text{ OR } \dfrac{2}{3}x+\dfrac{1}{6}=-2 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} \dfrac{2}{3}x+\dfrac{1}{6}=2 \\\\ 6\left(\dfrac{2}{3}x+\dfrac{1}{6}\right)=(2)6 \\\\ 4x+1=12 \\\\ 4x=12-1 \\\\ 4x=11 \\\\ x=\dfrac{11}{4} \\\\\text{ OR }\\\\ 6\left(\dfrac{2}{3}x+\dfrac{1}{6}\right)=(-2)6 \\\\ 4x+1=-12 \\\\ 4x=-12-1 \\\\ 4x=-13 \\\\ x=-\dfrac{13}{4} .\end{array} Hence, the solutions are $ x=\left\{ -\dfrac{13}{4},\dfrac{11}{4} \right\} .$
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