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: 60



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \dfrac{5}{16}y+\dfrac{3}{8}y=2+\dfrac{1}{4}y ,$ 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, $\{ 16,8,1,4 \},$ is $ 16 $ 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{5}{16}y+\dfrac{3}{8}y=2+\dfrac{1}{4}y \\\\ 16\left( \dfrac{5}{16}y+\dfrac{3}{8}y \right) =16\left( 2+\dfrac{1}{4}y \right) \\\\ 5y+6y=32+4y .\end{array} Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} 5y+6y=32+4y \\\\ 5y+6y-4y=32 \\\\ 7y=32 \\\\ y=\dfrac{32}{7} .\end{array} Checking: If $y=\dfrac{32}{7},$ then \begin{array}{l}\require{cancel} \dfrac{5}{16}y+\dfrac{3}{8}y=2+\dfrac{1}{4}y \\\\ \dfrac{5}{16}\left( \dfrac{32}{7} \right)+\dfrac{3}{8}\left( \dfrac{32}{7} \right)=2+\dfrac{1}{4}\left( \dfrac{32}{7} \right) \\\\ \dfrac{5}{\cancel{16}}\left( \dfrac{\cancel{16}(2)}{7} \right)+\dfrac{3}{\cancel8}\left( \dfrac{\cancel8(4)}{7} \right)=2+\dfrac{1}{\cancel4}\left( \dfrac{\cancel4(8)}{7} \right) \\\\ \dfrac{10}{7}+\dfrac{12}{7}=2+\dfrac{8}{7} \\\\ \dfrac{10}{7}+\dfrac{12}{7}=\dfrac{14}{7}+\dfrac{8}{7} \\\\ \dfrac{22}{7}=\dfrac{22}{7} \text{ (TRUE) } .\end{array} Hence, the solution is $ y=\dfrac{32}{7} .$
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