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: 79



Work Step by Step

Using the properties of equality, the solution to the given equation, $ 2(3t+1)-5=t-(t+2) $, is \begin{array}{l} 6t+2-5=t-t-2 \\\\ 6t-3=-2 \\\\ 6t=-2+3 \\\\ 6t=1 \\\\ t=\dfrac{1}{6} .\end{array} CHECKING: \begin{array}{l} 2\left(3\cdot\dfrac{1}{6}+1 \right)-5=\dfrac{1}{6}-\left( \dfrac{1}{6}+2 \right) \\\\ 2\left(\dfrac{1}{2}+\dfrac{2}{2} \right)-5=\dfrac{1}{6}-\left( \dfrac{1}{6}+\dfrac{12}{6} \right) \\\\ 2\left(\dfrac{3}{2} \right)-5=\dfrac{1}{6}-\left( \dfrac{13}{6} \right) \\\\ 3-5=\dfrac{1}{6}-\dfrac{13}{6} \\\\ -2=-\dfrac{12}{6} \\\\ -2=-2 \text{ (TRUE)} .\end{array} Hence, the solution is $ t=\dfrac{1}{6} $.
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