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# Help!

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

Apr 10, 2019

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