Question 994122
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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^2 + 2xy + 2xy + 4y^2}}} = {{{100}}}, or {{{x^2 + 4xy + 4y^2}}} = {{{100}}}, or {{{(x + 2y)^2}}} = {{{100}}}, or x + 2y = +/-10.      (5)


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


{{{x^2 - 4y^2}}} = {{{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(x + 2y = 10,
x-2y = 2)}}},        and     {{{system(x + 2y = -10,
x-2y = -2)}}}.


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. 
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