Answer
$(a) f(x)+g(x)=\frac{17}{3}x-\frac{8}{3}$
$(b) f(x)-g(x)=-\frac{13}{3}x+\frac{10}{3}$
$(c)f(x)g(x)=\frac{1}{3}( 10x^2-x-3)$
$(d)\frac{f(x)}{g(x)}=\frac{ 2x+1 }{ 15x-9 }$
Work Step by Step
$ f(x)=\frac{2}{3}x+\frac{1}{3}=\frac{1}{3}(2x+1)$
$g(x)=5x-3$
$(a) f(x)+g(x)=\frac{1}{3}(2x+1)+5x-3$
$ f(x)+g(x)=\frac{2}{3}x+\frac{1}{3}+5x-3$
$ f(x)+g(x)=(\frac{2}{3}+5)x+\frac{1}{3}-3$
$ f(x)+g(x)=\frac{17}{3}x-\frac{8}{3}$
$(b) f(x)-g(x)=\frac{2}{3}x+\frac{1}{3} -(5x-3)$
$ f(x)-g(x)=\frac{2}{3}x+\frac{1}{3}-5x+3$
$ f(x)-g(x)=(\frac{2}{3}-5)x+\frac{1}{3}+3$
$ f(x)-g(x)=-\frac{13}{3}x+\frac{10}{3}$
$(c)f(x)g(x)=\frac{1}{3}(2x+1)(5x-3)$
$f(x)g(x)=\frac{1}{3}(2x(5x-3)+1(5x-3))$
$f(x)g(x)=\frac{1}{3}( 10x^2-6x+5x-3)$
$f(x)g(x)=\frac{1}{3}( 10x^2-x-3)$
$(d)\frac{f(x)}{g(x)}=\frac{ \frac{1}{3}(2x+1) }{ 5x-3 }$
$\frac{f(x)}{g(x)}=\frac{ [\frac{1}{3}(2x+1) ] \times3 }{ (5x-3 ) \times 3 }$
$\frac{f(x)}{g(x)}=\frac{ 2x+1 }{ 15x-9 }$
