помогите решить систему уравнений

Cложим 
xy+x^2+xy+y^2=3+6
x^2+2xy+y^2=9
(x+y)^2=9
получаем
1, x+y=3
2, x+y=-3
вычитаем 
xy+x^2-xy-y^2=6-3
x^2-y^2=3
(x-y)(x+y)=3
подставляем
 1. x+y=3
3(x-y)=3
x-y=1
2x=4
x=2
y=1
2. x+y=-3
-3(x-y)=3
x-y=-1
2x=-4
x=-2
y=-1
Ответ (-2 -1) (2 1)

Дано:
xy + x^2 = 6
xy + y^2 = 3

Решение:
x(y + x) = 6
y(x + y) = 3

x/y = 2

x = 2y

2y y + y^2 = 3

y = 1
y = 1

x 1 + x^2 = 6
x ( 1) + x^2 = 6

x = 2
x = 3
x = 3
x = 2

(x1 ; y1) = (2 ; 1)
(x2 ; y2) = ( 3 ; 1)
(x3 ; y3) = (3 ; 1)
(x4 ; y4) = ( 2 ; 1)

2 1 + 2^2 = 6
2 1 + 1^2 = 3

3 1 + ( 3)^2 = 6
3 1 + 1^2 = 3

3 ( 1) + 3^2 = 6
3 ( 1) + ( 1)^2 = 3

2 ( 1) + ( 2)^2 = 6
2 ( 1) + ( 1)^2 = 3


6 = 6
3 = 3

6 = 6
2 = 3

6 = 6
2 = 3

6 = 6
3 = 3

(x1 ; y1) = (2 ; 1)
(x2 ; y2) = ( 2 ; 1)