Determine all possible real pairs (p, q) satisfying the following system of equations:
                     4p(p+q-5) = 3q
                      q(p+q-4)  = 16p

Multiply both equations to get 4pq(p+q-5)(p+q-4)= 48pq. Assuming p and q are not zero (p=q=0 is a solution, BTW) simplify to get (p+q-5)(p+q-4)=12. Let p+q=x, so (x-5)(x-4)=12, which means x=8 or x=1.

If p+q=8, the first equation becomes 4p.(8-5)=3q, or 12p=3(8-p) so p=8/5, and q=32/5.

If p+q=1, the same method produces 4p(1-5)=3(1-p) so p=-3/13 and q=16/13.

Posted by Federico Kereki on 2007-07-20 08:17:25