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: 59

Answer

$x=-\dfrac{60}{37}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \dfrac{2}{5}x-\dfrac{3}{2}x=\dfrac{3}{4}x+3 ,$ remove first the fraction by multiplying both sides by the $LCD.$ Then use the properties of equality to isolate the variable. Do checking of the solution. $\bf{\text{Solution Details:}}$ The $LCD$ of the denominators, $\{ 5,2,4,1 \},$ is $ 20 $ since this is the least number that can be evenly divided (no remainder) by all the denominators. Multiplying both sides by the $LCD,$ the given equation is equivalent to \begin{array}{l}\require{cancel} \dfrac{2}{5}x-\dfrac{3}{2}x=\dfrac{3}{4}x+3 \\\\ 20\left( \dfrac{2}{5}x-\dfrac{3}{2}x \right) =20\left( \dfrac{3}{4}x+3 \right) \\\\ 8x-30x=15x+60 .\end{array} Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} 8x-30x=15x+60 \\\\ 8x-30x-15x=60 \\\\ -37x=60 \\\\ x=\dfrac{60}{-37} \\\\ x=-\dfrac{60}{37} .\end{array} Checking: If $x=-\dfrac{60}{37},$ then \begin{array}{l}\require{cancel} \dfrac{2}{5}x-\dfrac{3}{2}x=\dfrac{3}{4}x+3 \\\\ \dfrac{2}{5}\left( -\dfrac{60}{37} \right) -\dfrac{3}{2} \left( -\dfrac{60}{37} \right)=\dfrac{3}{4}\left( -\dfrac{60}{37} \right)+3 \\\\ \dfrac{2}{\cancel5}\left( -\dfrac{\cancel5(12)}{37} \right) -\dfrac{3}{\cancel2} \left( -\dfrac{\cancel2(30)}{37} \right)=\dfrac{3}{\cancel4}\left( -\dfrac{\cancel4(15)}{37} \right)+3 \\\\ -\dfrac{24}{37}+\dfrac{90}{37}=-\dfrac{45}{37}+\dfrac{111}{37} \\\\ \dfrac{66}{37}=\dfrac{66}{37} \text{ (TRUE) } .\end{array} Hence, the solution is $ x=-\dfrac{60}{37} .$
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