Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 1 - Section 1.2 - Fractions in Algebra - Exercise Set - Page 30: 121

Answer

not a solution

Work Step by Step

Substituting $x= \dfrac{5}{8} $ in the given equation, $ \dfrac{1}{5}(x+2)=\dfrac{1}{2}\left(x-\dfrac{1}{5}\right) ,$ results to \begin{array}{l}\require{cancel} \dfrac{1}{5}\left(\dfrac{5}{8}+2\right)=\dfrac{1}{2}\left(\dfrac{5}{8}-\dfrac{1}{5}\right) \\\\ \dfrac{1}{5}\left(\dfrac{5}{8}+2\cdot\dfrac{8}{8}\right)=\dfrac{1}{2}\left(\dfrac{5}{8}\cdot\dfrac{5}{5}-\dfrac{1}{5}\cdot\dfrac{8}{8}\right) \Rightarrow\text{ (change to similar fractions)} \\\\ \dfrac{1}{5}\left(\dfrac{5}{8}+\dfrac{16}{8}\right)=\dfrac{1}{2}\left(\dfrac{25}{40}-\dfrac{8}{40}\right) \\\\ \dfrac{1}{5}\left(\dfrac{21}{8}\right)=\dfrac{1}{2}\left(\dfrac{17}{40}\right) \\\\ \dfrac{21}{40}=\dfrac{17}{80} \text{ (FALSE)} .\end{array} Since the substitution above ended with a FALSE statement, then $ x=\dfrac{5}{8} ,$ is not a solution.
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