Answer:
The system has two solutions:
(1, -1) and (0, 0)
Explanation:
We have the system of equations:
x + y = 0
x = x^2 + 2*x*y
To solve this, the first step is to isolate one of the variables in one of the equations, I will isolate x on the first one.
x = -y
Now we can replace this on the other equation, to get:
x = x^2 + 2*x*y
(-y) = (-y)^2 + 2*(-y)*y
Now we can solve this equation for y.
-y = y^2 - 2*y^2
-y = -y^2
y^2 - y = 0
We can solve this using the Bhaskara's formula:
The solutions are then:
[tex]y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4*1*0} }{2*1} = \frac{1 \pm 1}{2}[/tex]
Then the two possible solutions are:
y = (1 + 1)/2 = 1
and
y = (1 - 1)/2 = 0
Suppose that we take the first one, y = 1.
Then the solution for x is given by "x = -y"
Then:
x = -1
This means that one solution of the system is (-1, 1)
If we take the other solution for y, y = 0
The value of x will be:
x = -y = -0 = 0
Then another solution of the system is (0, 0)
