If\( h(y)=\dfrac{1+y}{2-y}\), then what is the value of \(h^{-1}(5)\)? Express your answer in simplest form.

\(h(y) = \dfrac{1+y}{2-y}\\ x = \dfrac{1+y}{2-y}\\ (2-y)x = 1+y\\ 2x-1=(1+x)y\\ y = \dfrac{2x-1}{x+1}\\ \text{now substitute }x \to y\\ h^{-1}(y) = \dfrac{2y-1}{y+1}\\ h^{-1}(5) = \dfrac{2(5)-1}{5+1}=\dfrac{9}{6}=\dfrac 3 2\)