## Intermediate Algebra: Connecting Concepts through Application

$\text{a) } f(g(2))=-209 \\\\\text{b) } f(6)-g(6)=104 \\\\\text{c) } \dfrac{f(6)}{g(6)}=-\dfrac{67}{37}$
$\bf{\text{Solution Outline:}}$ Substitute the given functions, \begin{array}{l}\require{cancel} f(x)= 12x-5 \\g(x)= -5x-7 ,\end{array} into the required operations. $\bf{\text{Solution Details:}}$ a) \begin{array}{l}\require{cancel} f(g(x))=f(-5x-7) \\\\ f(g(2))=f(-5(2)-7) \\\\ f(g(2))=f(-10-7) \\\\ f(g(2))=f(-17) \\\\ f(g(2))=12(-17)-5 \\\\ f(g(2))=-204-5 \\\\ f(g(2))=-209 \end{array} b) \begin{array}{l}\require{cancel} f(x)-g(x)=(12x-5)-(-5x-7) \\\\ f(6)-g(6)=(12(6)-5)-(-5(6)-7) \\\\ f(6)-g(6)=(72-5)-(-30-7) \\\\ f(6)-g(6)=(67)-(-37) \\\\ f(6)-g(6)=67+37 \\\\ f(6)-g(6)=104 \end{array} c) \begin{array}{l}\require{cancel} \dfrac{f(x)}{g(x)}=\dfrac{12x-5}{-5x-7} \\\\ \dfrac{f(6)}{g(6)}=\dfrac{12(6)-5}{-5(6)-7} \\\\ \dfrac{f(6)}{g(6)}=\dfrac{72-5}{-30-7} \\\\ \dfrac{f(6)}{g(6)}=\dfrac{67}{-37} \\\\ \dfrac{f(6)}{g(6)}=-\dfrac{67}{37} \end{array} Therefore, \begin{array}{l}\require{cancel} \text{a) } f(g(2))=-209 \\\\\text{b) } f(6)-g(6)=104 \\\\\text{c) } \dfrac{f(6)}{g(6)}=-\dfrac{67}{37} .\end{array}