as you may know, we start off by doing a quick switcharoo on the variables, in order to get the inverse, and then solve for "y",
[tex]\bf \stackrel{f(x)}{y}=\cfrac{2x+3}{5}\qquad \qquad \stackrel{inverse}{\underline{x}=\cfrac{2\underline{y}+3}{5}}\qquad \implies 5x=2y+3
\\\\\\
5x-3=2y\implies \cfrac{5x-3}{2}=\stackrel{f^{-1}(x)}{y}
\\\\\\
\cfrac{5(3)-3}{2}=f^{-1}(3)[/tex]
