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 - 1.7 Multiplication and Division of Real Numbers - 1.7 Exercise Set: 124

Answer

$-\dfrac{1}{2}$

Work Step by Step

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