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: 51



Work Step by Step

Multiplying both sides by the $LCD= 15 $, then the solution to the given equation, $ \dfrac{2}{3}+4t=6t-\dfrac{2}{15} $, is \begin{array}{l} 5(2)+15(4t)=15(6t)-1(2) \\\\ 10+60t=90t-2 \\\\ 60t-90t=-2-10 \\\\ -30t=-12 \\\\ t=\dfrac{-12}{-30} \\\\ t=\dfrac{2}{5} .\end{array} CHECKING: \begin{array}{l} \dfrac{2}{3}+4\cdot\dfrac{2}{5}=6\cdot\dfrac{2}{5}-\dfrac{2}{15} \\\\ \dfrac{2}{3}+\dfrac{8}{5}=\dfrac{12}{5}-\dfrac{2}{15} \\\\ \dfrac{10}{15}+\dfrac{24}{15}=\dfrac{36}{15}-\dfrac{2}{15} \\\\ \dfrac{34}{15}=\dfrac{34}{15} \text{ (TRUE)} .\end{array} Hence, the solution is $ t=\dfrac{2}{5} $.
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