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

### 3 Answers | Add Yours

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)**

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

(x + 5 y) (x + y) = 0

so x = -y or x = -5y

For x = -y, we replace x = -y at 2xy +y^2 = 4

-2y^2 + y^2 = 4 => -y^2 = 4 => y = 2i or -2i.

We have two soltuions: **(-2i, 2i), and (2i, -2i)**

For x = -5y, plug it into 2xy +y^2 = 4

-10y^2 + y^2 = 4 ==> -9y^2 = 4

so y = 2/3 i or -2/3 i

another two solutions: **(-10/3i, 2/3 i), (10/3i, -2/3 i)**

We'll divide by y^2 the 2nd equation:

(x/y)^2 + 5 + 6(x/y) = 0

We'll replace x/y by t:

t^2 + 6t + 5 = 0

t1 = [-6+sqrt(36-20)]/2

t1 = (-6+4)/2

t1 = -1

t2 = -5

We'll solve the system for t = -1

x/y = -1 => x = -y

2xy+y^2=4 => -2y^2 + y^2 = 4 => -y^2 = 4 => y1 = sqrt-4

y1 = 2i and y2 = -2i

x1 = -2i and x2 = 2i

We'll solve the system for t = -5

x/y = -5 => x = -5y

2xy+y^2=4 => -10y^2 + y^2 = 4 => -9y^2 = 4 => y^2 = -4/9

y3 = 2i/3 and y4 = -2i/3

x3 = -10i/3 and x4 = 10i/3

**The complex solutions of the system are: (2i ; -2i) ; (-2i ; 2i) ; (10i/3 ; -2i/3) and (-10i/3 ; 2i/3).**