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# 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!

### 1 Answer

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- 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).

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