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 2 - Equations, Inequalities, and Problem Solving - 2.2 Using the Principles Together - 2.2 Exercise Set - Page 95: 87



Work Step by Step

Using the properties of equality, the solution to the given equation, $ \dfrac{1}{6}\left( \dfrac{3}{4}x-2 \right)=-\dfrac{1}{5} $, is \begin{array}{l} \dfrac{3}{24}x-\dfrac{2}{6}=-\dfrac{1}{5} \\\\ \dfrac{1}{8}x-\dfrac{1}{3}=-\dfrac{1}{5} \\\\ \dfrac{1}{8}x=-\dfrac{1}{5}+\dfrac{1}{3} \\\\ \dfrac{1}{8}x=-\dfrac{3}{15}+\dfrac{5}{15} \\\\ \dfrac{1}{8}x=\dfrac{2}{15} \\\\ 8\cdot\dfrac{1}{8}x=\dfrac{2}{15}\cdot8 \\\\ x=\dfrac{16}{15} .\end{array} CHECKING: \begin{array}{l} \dfrac{1}{6}\left( \dfrac{3}{4}\cdot\dfrac{16}{15}-2 \right)=-\dfrac{1}{5} \\\\ \dfrac{1}{6}\left( \dfrac{48}{60}-2 \right)=-\dfrac{1}{5} \\\\ \dfrac{1}{6}\left( \dfrac{4}{5}-2 \right)=-\dfrac{1}{5} \\\\ \dfrac{1}{6}\left( \dfrac{4}{5}-\dfrac{10}{5} \right)=-\dfrac{1}{5} \\\\ \dfrac{1}{6}\left( -\dfrac{6}{5} \right)=-\dfrac{1}{5} \\\\ -\dfrac{1}{5}=-\dfrac{1}{5} \text{ (TRUE)} .\end{array} Hence, the solution is $ x=\dfrac{16}{15} $.
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