Question

Please help me solve the system of equations.

Answer 1

Answer: (-2, 0) and (0, -2)

Explanation:

This system is:

y + x = -2

y = (x + 1)^2 - 3

To solve this we first need to isolate one of the variables in one fo the equations, in the second equation we have already isolated the variable y, so we can just replace it in the first equation:

(x + 1)^2 - 3 + x = -2

Now we can solve this for x.

x^2 + 2*x + 1 - 3 = -2

x^2 + 2*x + 1 -3 + 2 = 0

x^2 + 2*x + 0 = 0

The solutions of this equation are given by the Bhaskara's formula, then the solutions are:

`x = (-2 +- √(2^2 -4*1*0) )/(2*1) = (-2 +- 2)/(2)`

The two solutions are:

x = (-2 - 2)/2 = -2

In this case, we replace this value of x in the first equation and get:

y - 2 = -2

y = -2 + 2 = 4

This solution is x = -2, y = 0, or (-2, 0)

The other solution for x is:

x = (-2 + 2)/2 = 0

If we replace this in the first equation we get:

y + 0 = -2

y = -2

This solution is x = 0, y = -2, or (0, -2)