Answer
$(a)f(x)+g(x)=\frac{22}{5}x-\frac{32}{5}$
$(b)f(x_-g(x)=-\frac{18}{5}x+\frac{38}{5}$
$(c)f(x)g(x)=\frac{1}{5}(8x^2-2x-21)$
Work Step by Step
$f(x)=\frac{2}{5}x+\frac{3}{5}$
$g(x)=4x-7$
$(a)f(x)+g(x)$
$=\frac{2}{5}x+\frac{3}{5}+4x-7$
$=\frac{2}{5}x+4x+\frac{3}{5}-7$
$=\frac{22}{5}x-\frac{32}{5}$
$(b)f(x)-g(x)$
$=\frac{2}{5}x+\frac{3}{5}-(4x-7)$
$=\frac{2}{5}x+\frac{3}{5}-4x+7$
$=\frac{2}{5}x-4x+\frac{3}{5}+7$
$=-\frac{18}{5}x+\frac{38}{5}$
$(c)f(x)g(x)$
$=(\frac{2}{5}x+\frac{3}{5})(4x-7)$
$=\frac{1}{5}(2x+3)(4x-7)$
$=\frac{1}{5}(8x^2-14x+12x-21)$
$=\frac{1}{5}(8x^2-2x-21)$
