# Solve the system x+y+2xy=-11 2x^2y+2xy^2=-12

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### 2 Answers

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)**

We'll note xy = p and x + y = s.

We'll re-write the 2nd equation:

2xy (x + y) = -12

We'll divide by 2:

xy (x + y) = -6

Now, we'll re-write the system in s and p:

s + 2p = -11

s = -2p - 11 (3)

p*s = -6 (4)

We'll substitute (3) in (4):

-(2p + 11)*p = -6

We'll multiply by -1 and we'll remove the brackets:

2p^2 + 11p = 6

We'll subtract 6:

2p^2 + 11p - 6 = 0

We'll apply quadratic formula:

p1 = [-11+sqrt(121 + 48)]/4

p1 = (-11+13)/4

p1 = 1/2

p2 = -6

s1 = -2p1 - 11

s1 = -1 - 11

s1 = -12

s2 = -2p2 - 11

s2 = 12 - 11

s2 = 1

x+y = -12

xy = 1/2

We'll form the quadratic to determine x and y:

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

2x^2 + 24x - 1 = 0

x1 = [-24 + sqrt(576 + 8 )]/4

x1 = [-24 + sqrt(584 )]/4

x1 = (-12 + sqrt146)/2

x2 = (-12 - sqrt146)/2

**So, the solutions of the symmetric system are: {(-12 + sqrt146)/2 ; (-12 - sqrt146)/2} and {(-12 - sqrt146)/2 ; (-12 + sqrt146)/2}.**

Now, we'll calculate the solutions of the system for s2 and p2:

x^2 - x - 6 = 0

x1 = [1 + sqrt(1 + 24)]/2

x1 = (1+5)/2

x1 = 3

x2 = -2

**The solutions of the symmetric system are: {3 ; 2} and {-2 ; 3}.**