jmc

If $(7,-22)$ lies on both $y=f(x)$ and the graph of its inverse, then $f(7)=-22$ and $f(-22)=7.$ Hence, $$\begin{array}{rl}\frac{-7p-3}{-7q+3}& =-22,\\ \frac{22p-3}{22q+3}& =7.\end{array}$$Then $-7p-3=-22(-7q+3)=154q-66$ and $22p-3=7(22q+3)=154q+21.$ Solving, we find $p=3$ and $q=\frac{3}{11},$ so $p+q=3+\frac{3}{11}=\boxed{\frac{36}{11}}.$