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: 53



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \dfrac{1}{3}x+\dfrac{2}{5}=\dfrac{4}{5}+\dfrac{3}{5}x-\dfrac{2}{3} ,$ use the properties of equality to isolate the variable. Then do checking of the solution. $\bf{\text{Solution Details:}}$ The $LCD$ of the denominators, $\{ 3,5 \}$ is $ 15 $ since it is the least number that can be divided evenly (no remainder) by all the denominators. Multiplying both sides by the $LCD= 15 $ and using the properties of equality results to \begin{array}{l}\require{cancel} \dfrac{1}{3}x+\dfrac{2}{5}=\dfrac{4}{5}+\dfrac{3}{5}x-\dfrac{2}{3} \\\\ 15\left( \dfrac{1}{3}x+\dfrac{2}{5} \right) =15\left( \dfrac{4}{5}+\dfrac{3}{5}x-\dfrac{2}{3} \right) \\\\ 5x+6=12+9x-10 \\\\ 5x-9x=12-10-6 \\\\ -4x=-4 \\\\ x=\dfrac{-4}{-4} \\\\ x=1 .\end{array} Checking: If $x=1,$ then \begin{array}{l}\require{cancel} \dfrac{1}{3}x+\dfrac{2}{5}=\dfrac{4}{5}+\dfrac{3}{5}x-\dfrac{2}{3} \\\\ \dfrac{1}{3}(1)+\dfrac{2}{5}=\dfrac{4}{5}+\dfrac{3}{5}(1)-\dfrac{2}{3} \\\\ \dfrac{1}{3}+\dfrac{2}{5}=\dfrac{4}{5}+\dfrac{3}{5}-\dfrac{2}{3} \\\\ \dfrac{5}{15}+\dfrac{6}{15}=\dfrac{12}{15}+\dfrac{9}{15}-\dfrac{10}{15} \\\\ \dfrac{11}{15}=\dfrac{11}{15} \text{ (TRUE) } .\end{array} Hence, the solution is $ x=1 .$
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