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: 19



Work Step by Step

The given equation, $ |x-1|=5x+10 ,$ is equivalent to \begin{array}{l}\require{cancel} x-1=5x+10 \\\\\text{OR}\\\\ x-1=-(5x+10) .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-1=5x+10 \\ x-5x=10+1 \\ -4x=11 \\ \dfrac{-4x}{-4}=\dfrac{11}{-4} \\ x=-\dfrac{11}{4} \\\\\text{OR}\\\\ x-1=-(5x+10) \\ x-1=-5x-10 \\ x+5x=-10+1 \\ 6x=-9 \\ \dfrac{6x}{6}=-\dfrac{9}{6} \\ x=-\dfrac{3}{2} .\end{array} Since the right side of the given equation is not a constant, then checking of solution/s is required. Substituting $ x=-\dfrac{11}{4} $ in the original equation results to \begin{array}{l}\require{cancel} |x-1|=5x+10 \\ \left|-\dfrac{11}{4}-1\right|=5\left(-\dfrac{11}{4}\right)+10 \\ \left|-\dfrac{11}{4}-\dfrac{4}{4}\right|=-\dfrac{55}{4}+10 \\ \left|-\dfrac{15}{4}\right|=-\dfrac{55}{4}+\dfrac{40}{4} \\ \left|-\dfrac{15}{4}\right|=-\dfrac{15}{4} \\ \dfrac{15}{4}=-\dfrac{15}{4} \text{ (FALSE)} .\end{array} Since the substitution above resulted in a FALSE statement, then $ x=-\dfrac{11}{4} ,$ is not a solution (an extraneous solution). Substituting $ x=-\dfrac{3}{2} $ in the original equation results to \begin{array}{l}\require{cancel} |x-1|=5x+10 \\ \left|-\dfrac{3}{2}-1\right|=5\left(-\dfrac{3}{2}\right)+10 \\ \left|-\dfrac{3}{2}-\dfrac{2}{2}\right|=-\dfrac{15}{2}+10 \\ \left|-\dfrac{5}{2}\right|=-\dfrac{15}{2}+\dfrac{20}{2} \\ \left|-\dfrac{5}{2}\right|=\dfrac{5}{2} \\ \dfrac{5}{2}=\dfrac{5}{2} \text{ (TRUE)} .\end{array} Hence, $ x=-\dfrac{3}{2} .$
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