## Intermediate Algebra (12th Edition)

$x=\dfrac{23}{3}$
$\bf{\text{Solution Outline:}}$ Using $f(x)=-\dfrac{3}{2}x+\dfrac{7}{2}$ from Exercise $80,$ replace $f(x)$ with $-8$ and then solve for $x.$ $\bf{\text{Solution Details:}}$ The expression $f(x)=-8$ means the value of $x$ when $f(x)$ is $-8.$ Replacing $f(x)$ with $-8$ and then solving for $x$ results to\begin{array}{l}\require{cancel} f(x)=-\dfrac{3}{2}x+\dfrac{7}{2} \\\\ -8=-\dfrac{3}{2}x+\dfrac{7}{2} \\\\ 2(-8)=2\left( -\dfrac{3}{2}x+\dfrac{7}{2} \right) \\\\ -16=-3x+7 \\\\ 3x=7+16 \\\\ 3x=23 \\\\ x=\dfrac{23}{3} .\end{array}
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