x^2 + 3xy + (2y)^2 = x^2 - xy
2x + y = 12 (1)
The first equation simplifies to
4xy + 4y^2 = 0 divide both sides by 4
xy + y^2 = 0 (2)
(1) can be transformed into y = 12 - 2x
Subbing this into (2), we have
x (12 - 2x) + (12 - 2x)^2 = 0 simplify
12x - 2x^2 + 4x^2 - 48x + 144 = 0
2x^2 - 36x + 144 = 0 divide through by 2
x^2 - 18x + 72 = 0 factor
(x - 6) (x - 12) = 0
Setting each factor to 0, x = 6 or x = 12
When x = 6, y = 12 -2(6) = 0
When x = 12, y = 12 - 2(12) = -12
So.....the solutions are (6, 0) and (12, -12)
These two solutions are verified by WolframAlpha, here : http://www.wolframalpha.com/input/?i=xy+%2B+y^2+%3D+0+,+2x+%2B+y+%3D+12
