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 R - Elementary Algebra Review - R.2 Equations, Inequalities, and Problem Solving - R.2 Exercise Set - Page 946: 12

Answer

$z=-\dfrac{16}{3}$

Work Step by Step

Using the properties of equality, the value of the variable that satisfies the given expression, $ \dfrac{2}{3}=-\dfrac{z}{8} ,$ is \begin{array}{l}\require{cancel} \dfrac{z}{8}=-\dfrac{2}{3} \\\\ 8\left( \dfrac{z}{8} \right)=-\dfrac{2}{3}(8) \\\\ z=-\dfrac{16}{3} .\end{array}
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