SOLUTION: if x^2 + xy= 12 and xy + y^2 = 6 find x and y note : x>0

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Question 994121: if x^2 + xy= 12 and xy + y^2 = 6 find x and y
note : x>0
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
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x^2 + xy = 12,       (1)
xy + y^2 =  6.       (2)


Add two equations (1) and (2). You will get


x%5E2+%2B+2xy+%2B+y%5E2 = 18, or %28x+%2B+y%29%5E2 = 18, or x + y = +/- 3sqrt%282%29.      (3)


Next, distract the equation (2) from the equation (1). You will get


x%5E2+-+y%5E2 = 6, or (x+y)*(x-y) = 6.                                (4)


By combining (5) and (6), you have two linear systems of two equations in two unknowns


system%28x+%2B+y+=+3sqrt%282%29%2C%0D%0Ax-y+=+6%2F%283sqrt%282%29%29%29,        and     system%28x+%2B+y+=+-3sqrt%282%29%2C%0D%0Ax-y+=+-6%2F%283sqrt%282%29%29%29.


Simplify right sides:


system%28x+%2B+y+=+3sqrt%282%29%2C%0D%0Ax-y+=+sqrt%282%29%29,        and     system%28x+%2B+y+=+-3sqrt%282%29%2C%0D%0Ax-y+=+-sqrt%282%29%29.


First of these two systems has the solution x = 2sqrt%282%29, y = sqrt%282%29.

The second system has negative solutions x = -2sqrt%282%29, y = -sqrt%282%29.


According to the condition, only the pair x = 2sqrt%282%29, y = sqrt%282%29 does suit.