The first equation can be rearranged to obtain p=10+q.
Substituting this into the second equation,
[tex]q(2(10+q)-10)=1000 \\ \\ q(10+2q)=1000 \\ \\ q(5+q)=500 \\ \\ q^2 + 5q-500=0 \\ \\ (q+25)(q-20)=0 \\ \\ q=-25, 20 \\ \\ q=-25 \implies p=-15 \\ \\ q=20 \implies p=30 \\ \\ \therefore \boxed{(p,q)=\{(-15, -25), (30, 20)\}}[/tex]