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 1-14 - Cumulative Review - Final Exam - Page 930: 1

Answer

$\dfrac{7}{15}$

Work Step by Step

The given expression, $ \left|-\dfrac{2}{3}+\dfrac{1}{5}\right| ,$ simplifies to \begin{align*}\require{cancel} & \left|-\dfrac{2}{3}\cdot\dfrac{5}{5}+\dfrac{1}{5}\cdot\dfrac{3}{3}\right| &(\text{change to similar fractions}) \\\\&= \left|-\dfrac{10}{15}+\dfrac{3}{15}\right| \\\\&= \left|-\dfrac{7}{15}\right| \\\\&= \dfrac{7}{15} .\end{align*} Hence, the expression $\left|-\dfrac{2}{3}+\dfrac{1}{5}\right|$ simplifies to $\dfrac{7}{15}$.
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