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: 45

Answer

$x=\left\{ -\dfrac{14}{3}, \dfrac{16}{3} \right\}$

Work Step by Step

Using the properties of equality, the given equation, $ |3x-1|+10=25 ,$ is equivalent to \begin{array}{l}\require{cancel} |3x-1|+10-10=25-10 \\ |3x-1|=15 .\end{array} Since for any $c\gt0$, $|x|=c$ implies $x=c \text{ or } x=-c,$ the equation above implies \begin{array}{l}\require{cancel} 3x-1=15 \\\\\text{OR}\\\\ 3x-1=-15 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 3x-1=15 \\ 3x-1+1=15+1 \\ 3x=16 \\ \dfrac{3x}{3}=\dfrac{16}{3} \\ x=\dfrac{16}{3} \\\\\text{OR}\\\\ 3x-1=-15 \\ 3x-1+1=-15+1 \\ 3x=-14 \\ \dfrac{3x}{3}=-\dfrac{14}{3} \\ x=-\dfrac{14}{3} .\end{array} Hence, the solutions are $ x=\left\{ -\dfrac{14}{3}, \dfrac{16}{3} \right\} .$
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