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: 50

Answer

$x=\dfrac{3}{2}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ |x-9|=|x+6| ,$ use the definition of an absolute equality. Then use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Since $|x|=|y|$ implies $x=y \text{ or } x=-y,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x-9=x+6 \\\\\text{OR}\\\\ x-9=-(x+6) .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-9=x+6 \\\\ x-x=6+9 \\\\ 0=15 \text{ (FALSE)} \\\\\text{OR}\\\\ x-9=-(x+6) \\\\ x-9=-x-6 \\\\ x+x=-6+9 \\\\ 2x=3 \\\\ x=\dfrac{3}{2} .\end{array} Hence, $ x=\dfrac{3}{2} .$
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