### Instruction: If $h(y)=\dfrac{1+y}{2-y}$, then what is the value of $h^{-1}(5)$? Express your answer in simplest form.### Response: I want to find the inverse function of $h(y)$, which means that I need to switch the roles of $y$ and $x$ and then solve for $y$.
So, I start by writing $x=\dfrac{1+y}{2-y}$ and then I cross-multiply to get $2x-xy=1+y$.
Next, I want to isolate $y$ on one side of the equation, so I subtract $2x$ from both sides and get $-xy=1+y-2x$.
Then, I factor out $y$ from the left side and get $y(-x-1)=1-2x$.
To solve for $y$, I divide both sides by $-x-1$ and get $y=\dfrac{1-2x}{-x-1}$.
This is the expression for the inverse function of $h(y)$.
Now, I just need to plug in $x=5$ and simplify.