Last substitution problem, x^2 + 2xy + y^2 = 36, x^2 - xy = 0?
This is the final substitution question, it's much harder
x^2 + 2xy + y^2 = 36
x^2 - xy = 0
This is where i got to:
x^2 = xy
(xy) + 2xy + y^2 = 36
3xy + y^2 = 36
Where do i go from there? Thankyou!
- Nick SLv 61 decade agoFavorite Answer
Your first equation is better formulated as (x+y)^2 = 36, so
x+y is either 6 or -6.
Your second equation x(x-y)=0, which has solutions x=0 or x=y.
Hence all possible solutions are
(x,y) = (0,6), (0,-6), (3,3), (-3,-3).