# Chapter 9 - Inequalities and Problem Solving - 9.3 Absolute-Value Equations and Inequalities - 9.3 Exercise Set: 45

$x=\left\{ -\dfrac{9}{2},\dfrac{11}{2} \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ Given that $f(x)=\left|\dfrac{1-2x}{5} \right| ,$ to find $x$ for which $f(x)=2 ,$ use substitution. Then use the definition of absolute value equality. Finally, use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Replacing $f(x)$ with $2 ,$ then \begin{array}{l}\require{cancel} f(x)=\left|\dfrac{1-2x}{5} \right| \\\\ 2=\left|\dfrac{1-2x}{5} \right| \\\\ \left|\dfrac{1-2x}{5} \right|=2 .\end{array} Since for any $c\gt0$, $|x|=c$ implies $x=c \text{ or } x=-c,$ the given equation is equivalent to \begin{array}{l}\require{cancel} \dfrac{1-2x}{5}=2 \\\\\text{OR}\\\\ \dfrac{1-2x}{5}=-2 .\end{array} Solving each equation above results to \begin{array}{l}\require{cancel} \dfrac{1-2x}{5}=2 \\\\ 5\cdot\dfrac{1-2x}{5}=5\cdot2 \\\\ 1-2x=10 \\\\ -2x=10-1 \\\\ -2x=9 \\\\ x=\dfrac{9}{-2} \\\\ x=-\dfrac{9}{2} \\\\\text{OR}\\\\ \dfrac{1-2x}{5}=-2 \\\\ 5\cdot\dfrac{1-2x}{5}=5\cdot(-2) \\\\ 1-2x=-10 \\\\ -2x=-10-1 \\\\ -2x=-11 \\\\ x=\dfrac{-11}{-2} \\\\ x=\dfrac{11}{2} .\end{array} Hence, $x=\left\{ -\dfrac{9}{2},\dfrac{11}{2} \right\} .$

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