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If h(y) = (1 + y)/(2 - y), then what is the value of $h^{-1}(15)$? Express your answer in simplest form.

 Jul 31, 2021
 #1
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If

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

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

 

My attempt:

 

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

 

laugh

 Jul 31, 2021

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