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 - Page 599: 54


set of all real numbers

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ |y-2|=|2-y| ,$ use the definition of an absolute value 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} y-2=2-y \\\\\text{OR}\\\\ y-2=-(2-y) .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} y-2=2-y \\\\ y+y=2+2 \\\\ 2y=4 \\\\ y=\dfrac{4}{2} \\\\ y=2 \\\\\text{OR}\\\\ y-2=-(2-y) \\\\ y-2=-2+y \\\\ y-y=-2+2 \\\\ 0=0 \text{ (TRUE)} .\end{array} Since the solution above ended with a TRUE statement, then the solution set is the set of all real numbers.
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