jmc

Solving $pq=\frac{9}{2}$ for $p$ we see that $p=\frac{9}{2q}$. Plugging this into $\frac{1}{p}+\frac{1}{q}=1$ we then get $$\frac{2q}{9}+\frac{1}{q}=1\Rightarrow 2q^2-9q+9=0.$$ Applying the quadratic equation we then see that $$q=\frac{9\pm \sqrt{81-72}}{4}=\frac{9\pm 3}{4}.$$ Now, the smaller root corresponds to $p$ and the larger to $q$, so we see that $\boxed{q=3}$.