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 1-2 - Cumulative Review - Page 152: 34


$n=\{ -5,5 \}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 3|n|+10=25 ,$ isolate first the absolute value expression. Then use the definition of an absolute value equality to solve for the variable. $\bf{\text{Solution Details:}}$ Using the properties of equality, the given expression is equivalent to \begin{array}{l}\require{cancel} 3|n|+10=25 \\\\ 3|n|=25-10 \\\\ 3|n|=15 \\\\ |n|=\dfrac{15}{3} \\\\ |n|=5 .\end{array} Since for any $c\gt0$, $|x|=c$ implies $x=c \text{ or } x=-c,$ the equation above is equivalent to \begin{array}{l}\require{cancel} n=5 \\\\\text{OR}\\\\ n=-5 .\end{array} Hence, $ n=\{ -5,5 \} .$
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