# Solve the system x+y+2xy=-11 2x^2y+2xy^2=-12 We have to solve

x+y+2xy=-11 ...(1)

2x^2y+2xy^2=-12 ...(2)

(2)

=> x^2y + xy^2 = -6

=> xy(x+y) = -6

(1)

=> x+ y + 2xy = -11

If we take the variables x+y and xy together as A and B, we get

A*B = -6

2A + B = -11

A( -11 - 2A) = -6

=> 11A + 2A^2 - 6 = 0

=> 2A^2 + 11A - 6 = 0

=> 2A^2 + 12A - A - 6 = 0

=> 2A ( a + 6) - 1(A + 6) = 0

=> (2A - 1)(A + 6) = 0

=> A = 1/2 and A = -6

=> B = -12 , 1

B = x+y = -12

A = xy = 1/2

From this we get the equation

x^2 + 12x - 1/2 = 0

=> 2x^2 + 24x - 1 = 0

x1 = [-24 + sqrt(576 + 8 )]/4

=> x1 = [-24 + sqrt 584]/4

=> x1 = -6 + sqrt 584 / 4

=> y1 = -6 - sqrt 584 / 4

x2 = -6 - sqrt 584 / 4

y2 = -6 + sqrt 584 / 4

Now, for the values -6 and 1 we have:

x^2 - x - 6 = 0

x^2 - 3x + 2x - 6 =0

=> x(x - 3) + 2( x - 3) =0

=> (x + 2)(x - 3) =0

x1 = -2

y1 = 3

x2 = 3

y2 = -2

So the solutions for x and y are (-2 , 3),( 3, 2), (-6 + sqrt 584 / 4, -6 - sqrt 584 / 4), ( -6 - sqrt 584 / 4 , -6 + sqrt 584 / 4)

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