Solve the system of equations 2xy+y^2=4 and x^2+5y^2+6xy=0.

We have to solve the system of
equations:

2xy + y^2 = 4
...(1)

x^2 + 5y^2 + 6xy = 0
...(2)

x^2 + 5y^2 + 6xy =
0

2xy + y^2 = 4

=> x =
(4 - y^2)/2y

substitute in
(2)

=> [(4 - y^2)/2y]^2 + 5y^2 + 12 - 3y^2 =
0

=> (4 - y^2)^2/4y^2 + 5y^2 + 12 - 3y^2 =
0

=> (4 - y^2)^2/4y^2 + 5y^2 + 12 - 3y^2 =
0

let y^2 = z

=> (4 -
z)^2/4z + 2z + 12 = 0

=> 16 + z^2 - 8z + 8z^2 + 48z
= 0

=> 9z^2 + 40z + 16 =
0

=> 9z^2 + 36z + 4z + 16 =
0

=> 9z(z + 4) + 4(z + 4) =
0

=> (9z + 4)(z + 4) =
0

=> z = -4 and z =
-4/9

y^2 = 2i, -2i and y = 2i/3,
-2i/3

x = (4 - y^2)/2y

for y =
2i

x = 4/2i = -2i

for y =
-2i

x = 4/-2i = 2i

for y =
2i/3

x = (4 + 4/9)/(4i/3) =
-10i/3

for y = -2i/3

x = (4 +
4/9)/(4i/3) = 10i/3

The solution of the
equations are (-2i, 2i), (2i, -2i), (-10i/3, 2i/3) and (10i/3,
-2i/3)