Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 9 - Inequalities and Problem Solving - 9.3 Absolute-Value Equations and Inequalities - 9.3 Exercise Set: 34

Answer

$z=\left\{ -3,3 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 5|z|+2=17 ,$ isolate first the absolute value expression. Then use the definition of an absolute value equality. $\bf{\text{Solution Details:}}$ Using the properties of equality, the given equation is equivalent to \begin{array}{l}\require{cancel} 5|z|+2=17 \\\\ 5|z|=17-2 \\\\ 5|z|=15 \\\\ |z|=\dfrac{15}{5} \\\\ |z|=3 .\end{array} Since for any $c\gt0$, $|x|=c$ implies $x=c \text{ or } x=-c,$ the given equation is equivalent to \begin{array}{l}\require{cancel} z=3 \\\\\text{OR}\\\\ z=-3 .\end{array} Hence, $ z=\left\{ -3,3 \right\} .$
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