Inverse Functions

If

$h(y) = \dfrac{1 + y}{2 - y}$,

then what is the value of $h^{-1}(15)$?

My attempt:

$\begin{array}{|rcll|} \hline h^{-1}\Big(h(y)\Big) &=& y \\\\ &&\boxed{h(y) = 15 \\ \frac{1+y}{2-y}=15\\ 1+y = 15(2-y) \\ 1+y=30-15y\\ 16y = 29 \\ \mathbf{y=\frac{29}{16}} } \\\\ \mathbf{h^{-1}(15)} &=& \mathbf{\dfrac{29}{16}} \\ \hline \end{array}$