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

$x=\left\{ -\dfrac{7}{3},-\dfrac{1}{3} \right\}$
$\bf{\text{Solution Outline:}}$ Given that $f(x)=\left|\dfrac{3x+4}{3} \right| ,$ to find $x$ for which $f(x)=1 ,$ 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 $1 ,$ then \begin{array}{l}\require{cancel} f(x)=\left|\dfrac{3x+4}{3} \right| \\\\ 1=\left|\dfrac{3x+4}{3} \right| \\\\ \left|\dfrac{3x+4}{3} \right|=1 .\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{3x+4}{3}=1 \\\\\text{OR}\\\\ \dfrac{3x+4}{3}=-1 .\end{array} Solving each equation above results to \begin{array}{l}\require{cancel} 3\cdot\dfrac{3x+4}{3}=3\cdot1 \\\\ 3x+4=3 \\\\ 3x=3-4 \\\\ 3x=-1 \\\\ x=-\dfrac{1}{3} \\\\\text{OR}\\\\ \dfrac{3x+4}{3}=-1 \\\\ 3\cdot\dfrac{3x+4}{3}=3\cdot(-1) \\\\ 3x+4=-3 \\\\ 3x=-3-4 \\\\ 3x=-7 \\\\ x=-\dfrac{7}{3} .\end{array} Hence, $x=\left\{ -\dfrac{7}{3},-\dfrac{1}{3} \right\} .$