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: 47

Answer

no solution

Work Step by Step

Using the properties of equality, the given equation, $ 5|6-5x|=15x-35 ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{5|6-5x|}{5}=\dfrac{15x-35}{5} \\\\ |6-5x|=3x-7 .\end{array} Removing the absolute value sign, the expression above is equivalent to \begin{array}{l}\require{cancel} 6-5x=3x-7 \\\\\text{ OR }\\\\ 6-5x=-(3x-7) .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 6-5x=3x-7 \\ -5x-3x=-7-6 \\ -8x=-13 \\ \dfrac{-8x}{-8}=\dfrac{-13}{-8} \\ x=\dfrac{13}{8} \\\\\text{ OR }\\\\ 6-5x=-(3x-7) \\ 6-5x=-3x+7 \\ -5x+3x=7-6 \\ -2x=1 \\ \dfrac{-2x}{-2}=\dfrac{1}{-2} \\ x=-\dfrac{1}{2} .\end{array} Since the right side of the original equation is not a constant, then checking of solutions is required. Substituting $ x=\dfrac{13}{8} $ in the original equation results to \begin{array}{l}\require{cancel} 5|6-5x|=15x-35 \\\\ 5\left| 6-5\left( \dfrac{13}{8} \right) \right|=15\left( \dfrac{13}{8} \right)-35 \\\\ 5\left| 6-\dfrac{65}{8} \right|=\dfrac{195}{8}-35 \\\\ 5\left| \dfrac{48}{8}-\dfrac{65}{8} \right|=\dfrac{195}{8}-\dfrac{280}{8} \\\\ 5\left| -\dfrac{17}{8} \right|=-\dfrac{85}{8} \\\\ 5\left(\dfrac{17}{8} \right)=-\dfrac{85}{8} \\\\ \dfrac{85}{8}=-\dfrac{85}{8} \text{ (FALSE)} .\end{array} Since the substitution above ended with a FALSE statement, then $ x=\dfrac{13}{8} $ is not a solution. Substituting $ x=-\dfrac{1}{2} $ in the original equation results to \begin{array}{l}\require{cancel} 5|6-5x|=15x-35 \\\\ 5\left| 6-5\left( -\dfrac{1}{2} \right) \right|=15\left( -\dfrac{1}{2} \right)-35 \\\\ 5\left| 6+\dfrac{5}{2} \right|=-\dfrac{15}{2}-35 \\\\ 5\left| \dfrac{12}{2}+\dfrac{5}{2} \right|=-\dfrac{15}{2}-\dfrac{70}{2} \\\\ 5\left| \dfrac{17}{2} \right|=-\dfrac{85}{2} \\\\ 5\left(\dfrac{17}{2} \right)=-\dfrac{85}{2} \\\\ \dfrac{85}{2}=-\dfrac{85}{2} \text{ (FALSE)} .\end{array} Since the substitution above ended with a FALSE statement, then $ x=-\dfrac{1}{2} $ is not a solution. Hence, there is no solution.
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