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 94: 62



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \dfrac{1}{5}(4x-1)=7 ,$ use the properties of equality to isolate the variable. Do checking of the solution. $\bf{\text{Solution Details:}}$ Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{5}(4x-1)=7 \\\\ 5\cdot\dfrac{1}{5}(4x-1)=5\cdot7 \\\\ 4x-1=35 \\\\ 4x=35+1 \\\\ 4x=36 \\\\ x=\dfrac{36}{4} \\\\ x=9 .\end{array} Checking: If $x=9,$ then \begin{array}{l}\require{cancel} \dfrac{1}{5}(4x-1)=7 \\\\ \dfrac{1}{5}(4\cdot9-1)=7 \\\\ \dfrac{1}{5}(36-1)=7 \\\\ \dfrac{1}{5}(35)=7 \\\\ 7=7 \text{ (TRUE) } .\end{array} Hence, the solution is $ x=9 .$
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