SOLUTION: Please help me solve this equation If x^2 + 2xy = 40 and y^2 + 1/2xy = 15 find x^2 + y^2 x and y are positive integers thanks in advance for your help

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Question 994122: Please help me solve this equation
If x^2 + 2xy = 40 and y^2 + 1/2xy = 15 find x^2 + y^2
x and y are positive integers
thanks in advance for your help

Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.
x^2 + 2xy = 40,       (1)
y^2 + 1/2xy = 15.     (2)


Multiply the second equation by 4 (both sides). You will have an equivalent system


x^2 + 2xy = 40,       (3)
4y^2 + 2xy = 60.      (4)


Now add equations (3) and (4) of the last system. You will get


x%5E2+%2B+2xy+%2B+2xy+%2B+4y%5E2 = 100, or x%5E2+%2B+4xy+%2B+4y%5E2 = 100, or %28x+%2B+2y%29%5E2 = 100, or x + 2y = +/-10.      (5)


Next, distract the equation (4) from the equation (3). You will get


x%5E2+-+4y%5E2 = 20, or (x+2y)*(x-2y) = 20.                 (6)


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


system%28x+%2B+2y+=+10%2C%0D%0Ax-2y+=+2%29,        and     system%28x+%2B+2y+=+-10%2C%0D%0Ax-2y+=+-2%29.


First of these two systems has the solution x=6, y=2.

The second system has negative solutions x=-6, y=-2.


According to the condition, only the pair x=6, y=2 does suit.