jmc

If we expand the left side, we have $$\begin{array}{rl}(x-p)(x-q)& =x(x-q)-p(x-q)\\ & =x^2-qx-px+pq\\ & =x^2-(p+q)x+pq.\end{array}$$ The other side of the equation is a constant, since there isn't an $x$ term. So, if we view the equation as a quadratic in $x$, the sum of the roots is $-[-(p+q)]=p+q$. We know that one of the roots is $r$, so if the other is $s$, we have $r+s=p+q$, so $s=\boxed{p+q-r}$.