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: 71



Work Step by Step

Using the properties of equality, the solution to the given equation, $ 2r+8=6r+10 $, is \begin{array}{l} 2r-6r=10-8 \\\\ -4r=2 \\\\ r=\dfrac{2}{-4} \\\\ r=-\dfrac{1}{2} .\end{array} CHECKING: \begin{array}{l} 2\left( -\dfrac{1}{2} \right)+8=6\left( -\dfrac{1}{2} \right)+10 \\\\ -1+8=-3+10 \\\\ 7=7 \text{ (TRUE)} .\end{array} Hence, the solution is $ r=-\dfrac{1}{2} $.
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