problem: If $h(y)=\dfrac{1+y}{2-y}$, then what is the value of $h^{-1}(5)$? Express your answer in simplest form.
solution: $h^{-1}(5)$ is defined as the number $y$ such that $h(y)=5$. Thus, we solve the equation $$\frac{1+y}{2-y} = 5.$$Multiplying both sides by $2-y$, we have $$1+y = 5(2-y).$$Expanding gives $$1+y = 10-5y,$$then adding $5y-1$ to both sides gives $$6y = 9.$$Finally, we divide both sides by $6$ and simplify to get $y=\boxed{\dfrac{3}{2}}$.
  
  Note that we can check our work by plugging $\dfrac{3}{2}$ into the formula for $h$: $$\dfrac{1+\frac32}{2-\frac32} = \dfrac{\left(\frac52\right)}{\left(\frac12\right)} = 5,$$which is what we expected.