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 - Page 120: 89


$x=\left\{ -\dfrac{5}{3},\dfrac{1}{3} \right\}$

Work Step by Step

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