If we expand the left side, we have \begin{align*} (x-p)(x-q) &=x(x-q) -p(x-q)\\ & = x^2 - qx - px +pq \\ &= x^2 -(p+q)x + pq.\end{align*} 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} $.