Question 994121
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x^2 + xy = 12,       (1)
xy + y^2 =  6.       (2)


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


{{{x^2 + 2xy + y^2}}} = {{{18}}}, or {{{(x + y)^2}}} = {{{18}}}, or x + y = +/- {{{3sqrt(2)}}}.      (3)


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


{{{x^2 - y^2}}} = {{{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(x + y = 3sqrt(2),
x-y = 6/(3sqrt(2)))}}},        and     {{{system(x + y = -3sqrt(2),
x-y = -6/(3sqrt(2)))}}}.


Simplify right sides:


{{{system(x + y = 3sqrt(2),
x-y = sqrt(2))}}},        and     {{{system(x + y = -3sqrt(2),
x-y = -sqrt(2))}}}.


First of these two systems has the solution x = {{{2sqrt(2)}}}, y = {{{sqrt(2)}}}.

The second system has negative solutions x = -{{{2sqrt(2)}}}, y = -{{{sqrt(2)}}}.


According to the condition, only the pair x = {{{2sqrt(2)}}}, y = {{{sqrt(2)}}} does suit. 
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