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 - Introduction to Algebraic Expressions - Review Exercises: Chapter 1: 31

Answer

$\dfrac{3}{5}$

Work Step by Step

Using the $LCD= 15 $, the given expression, $ \dfrac{2}{3}-\dfrac{1}{15} $, simplifies to \begin{array}{l}\require{cancel} \dfrac{5(2)-1(1)}{15} \\\\= \dfrac{10-1}{15} \\\\= \dfrac{9}{15} \\\\= \dfrac{\cancel{3}\cdot3}{\cancel{3}\cdot5} \\\\= \dfrac{3}{5} .\end{array}
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