Answer
$(a)f(x)+g(x)=\frac{5}{2}x -\frac{2}{5}$
$(b)f(x)-g(x)=-\frac{3}{2}x +\frac{4}{5}$
$(c)f(x)g(x)=x^2+\frac{1}{10}x-\frac{3}{25}$
$(d) \frac{f(x)}{g(x)}=\frac{ 5x+2 }{ 20x-6 }$
Work Step by Step
$f(x)=\frac{1}{2}x+\frac{1}{5}$
$g(x)=2x-\frac{3}{5}$
$(a)f(x)+g(x)=\frac{1}{2}x+\frac{1}{5}+2x-\frac{3}{5}$
$f(x)+g(x)=\frac{1}{2}x+2x +\frac{1}{5}-\frac{3}{5}$
$f(x)+g(x)=(\frac{1}{2}+2)x +\frac{1}{5}-\frac{3}{5}$
$f(x)+g(x)=\frac{5}{2}x -\frac{2}{5}$
$(b)f(x)-g(x)=\frac{1}{2}x+\frac{1}{5}-(2x-\frac{3}{5})$
$f(x)-g(x)=\frac{1}{2}x-2x +\frac{1}{5}+\frac{3}{5}$
$f(x)-g(x)=(\frac{1}{2}-2)x +\frac{1}{5}+\frac{3}{5}$
$f(x)-g(x)=-\frac{3}{2}x +\frac{4}{5}$
$(c)f(x)g(x)=(\frac{1}{2}x+\frac{1}{5})(2x-\frac{3}{5})$
$f(x)g(x)=\frac{1}{2}x(2x-\frac{3}{5})+\frac{1}{5}(2x-\frac{3}{5})$
$f(x)g(x)=\frac{1}{2}x(2x-\frac{3}{5})+\frac{1}{5}(2x-\frac{3}{5})$
$f(x)g(x)=\frac{1}{2}x\times2x-\frac{1}{2}x\times\frac{3}{5}+\frac{1}{5}\times2x-\frac{1}{5}\times\frac{3}{5}$
$f(x)g(x)=x^2-\frac{3}{10}x+\frac{2}{5}x-\frac{3}{25}$
$f(x)g(x)=x^2+\frac{1}{10}x-\frac{3}{25}$
$(d) \frac{f(x)}{g(x)}=\frac{ \frac{1}{2}x+\frac{1}{5} }{ 2x-\frac{3}{5} }$
$ \frac{f(x)}{g(x)}=\frac{ [\frac{1}{2}x+\frac{1}{5}] \times10 }{ [ 2x-\frac{3}{5}]\times10 }$
$ \frac{f(x)}{g(x)}=\frac{ [\frac{1}{2}\times10x+\frac{1}{5}\times10] }{ [ 2x\times10-\frac{3}{5}\times10 ] }$
$ \frac{f(x)}{g(x)}=\frac{ 5x+2 }{ 20x-6 }$
