Intermediate Algebra: Connecting Concepts through Application

$\text{a) } f(g(4))=\dfrac{35}{3} \\\\\text{b) } g(f(4))=12$
$\bf{\text{Solution Outline:}}$ With \begin{array}{l}\require{cancel} f(x)= \dfrac{2}{3}x+\dfrac{1}{3} \\g(x)= 5x-3 ,\end{array} to find $f(g(4)) ,$ find first $g(4) .$ Then substitute the result in $f.$ To find $g(f(4)) ,$ find first $f(4) .$ Then substitute the result in $g.$ $\bf{\text{Solution Details:}}$ a) Replacing $x$ with $4$ in $g$ results to \begin{array}{l}\require{cancel} g(x)=5x-3 \\\\ g(4)=5(4)-3 \\\\ g(4)=20-3 \\\\ g(4)=17 .\end{array} Replacing $x$ with the result above in $f$ results to \begin{array}{l}\require{cancel} f(x)=\dfrac{2}{3}x+\dfrac{1}{3} \\\\ f(17)=\dfrac{2}{3}(17)+\dfrac{1}{3} \\\\ f(17)=\dfrac{34}{3}+\dfrac{1}{3} \\\\ f(17)=\dfrac{35}{3} .\end{array} Hence, $f(g(4))=\dfrac{35}{3} .$ b) Replacing $x$ with $4$ in $f$ results to \begin{array}{l}\require{cancel} f(x)=\dfrac{2}{3}x+\dfrac{1}{3} \\\\ f(4)=\dfrac{2}{3}(4)+\dfrac{1}{3} \\\\ f(4)=\dfrac{8}{3}+\dfrac{1}{3} \\\\ f(4)=\dfrac{9}{3} \\\\ f(4)=3 .\end{array} Replacing $x$ with the result above in $g$ results to \begin{array}{l}\require{cancel} g(x)=5x-3 \\\\ g(3)=5(3)-3 \\\\ g(3)=15-3 \\\\ g(3)=12 .\end{array} Hence, $g(f(4))=12 .$ Therefore, \begin{array}{l}\require{cancel} \text{a) } f(g(4))=\dfrac{35}{3} \\\\\text{b) } g(f(4))=12 .\end{array}