[tex]\begin{gathered} f(x)=\frac{2x-3}{5} \\ y=\frac{2x-3}{5} \\ x=\frac{2y-3}{5} \\ cross\text{ multiply} \\ 5x=2y-3 \\ 2y=5x+3 \\ divide\text{ through by 2} \\ y=\frac{5}{2}x+\frac{3}{2} \\ f^{-1}(x)=\frac{5}{2}x+\frac{3}{2} \end{gathered}[/tex]
The answer is B.
