Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 1 - Expressions, Equations, and Inequalities - 1-6 Absolute Value Equations and Inequalities - Practice and Problem-Solving Exercises - Page 46: 21

Answer

$x=\dfrac{3}{2}$

Work Step by Step

The given equation, $ |3x+5|=5x+2 ,$ is equivalent to \begin{array}{l}\require{cancel} 3x+5=5x+2 \\\\\text{OR}\\\\ 3x+5=-(5x+2) .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 3x+5=5x+2 \\ 3x-5x=2-5 \\ -2x=-3 \\ \dfrac{-2x}{-2}=\dfrac{-3}{-2} \\ x=\dfrac{3}{2} \\\\\text{OR}\\\\ 3x+5=-(5x+2) \\ 3x+5=-5x-2 \\ 3x+5x=-2-5 \\ 8x=-7 \\ \dfrac{8x}{8}=-\dfrac{7}{8} \\ x=-\dfrac{7}{8} .\end{array} Since the right side of the given equation is not a constant, then checking of solution/s is required. Substituting $ x=\dfrac{3}{2} $ in the original equation results to \begin{array}{l}\require{cancel} |3x+5|=5x+2 \\ \left| 3\left(\dfrac{3}{2}\right)+5 \right|=5\left(\dfrac{3}{2}\right)+2 \\ \left| \dfrac{9}{2}+5 \right|=\dfrac{15}{2}+2 \\ \left| \dfrac{9}{2}+\dfrac{10}{2} \right|=\dfrac{15}{2}+\dfrac{4}{2} \\ \left| \dfrac{19}{2} \right|=\dfrac{19}{2} \\ \dfrac{19}{2}=\dfrac{19}{2} \text{ (TRUE)} .\end{array} Substituting $ x=-\dfrac{7}{8} $ in the original equation results to \begin{array}{l}\require{cancel} |3x+5|=5x+2 \\ \left| 3\left(-\dfrac{7}{8}\right)+5 \right|=5\left(-\dfrac{7}{8}\right)+2 \\ \left| -\dfrac{21}{8}+5 \right|=-\dfrac{35}{8}+2 \\ \left| -\dfrac{21}{8}+\dfrac{40}{8} \right|=-\dfrac{35}{8}+\dfrac{16}{8} \\ \left| \dfrac{19}{8} \right|=-\dfrac{19}{8} \\ \dfrac{19}{8}=-\dfrac{19}{8} \text{ (FALSE)} .\end{array} Since the substitution above resulted in a FALSE statement, then $ x=-\dfrac{7}{8} ,$ is not a solution (an extraneous solution). Hence, the solution is $ x=\dfrac{3}{2} .$
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