## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$x=\left\{ -\dfrac{1}{3},3 \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $5-2|3x-4|=-5 ,$ isolate first the absolute value expression. Then use the definition of an absolute value equality. Finally, use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Using the proeprties of equality to isolate the absolute value expression, the given is equivalent to \begin{array}{l}\require{cancel} 5-2|3x-4|=-5 \\\\ -2|3x-4|=-5-5 \\\\ -2|3x-4|=-10 \\\\ |3x-4|=\dfrac{-10}{-2} \\\\ |3x-4|=5 .\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} 3x-4=5 \\\\\text{OR}\\\\ 3x-4=-5 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 3x-4=5 \\\\ 3x=5+4 \\\\ 3x=9 \\\\ x=\dfrac{9}{3} \\\\ x=3 \\\\\text{OR}\\\\ 3x-4=-5 \\\\ 3x=-5+4 \\\\ 3x=-1 \\\\ x=-\dfrac{1}{3} .\end{array} Hence, $x=\left\{ -\dfrac{1}{3},3 \right\} .$