Question

A.


`x + y = 0 \\ y = {x}^(2) + 2xy`
can any one tell this answer with dafination and all steps​

Answer 1

The system has two solutions:

(1, -1) and (0, 0)

Step-by-step 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:

`y = (-(-1) \pm √((-1)^2 - 4*1*0) )/(2*1) = (1 \pm 1)/(2)`

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)