We can solve this system of equations by substitution or elimination method.
Let's solve it by substitution:
From the second equation, we can solve for y:
y = 2 - x
Substituting this value of y into the first equation, we have:
2 - x - x^2 + 2x = -4
Now, simplify and rearrange the equation:
-x^2 + 3x + 2 = -4
-x^2 + 3x + 6 = 0
Rearranging again, we have a quadratic equation:
x^2 - 3x - 6 = 0
We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula.
By factoring:
(x - 2)(x + 3) = 0
Setting each factor equal to zero, we get:
x - 2 = 0 → x = 2
x + 3 = 0 → x = -3
Now, substitute these values of x back into the second equation to find the corresponding y-values:
For x = 2, y + 2 = 2 → y = 0
For x = -3, y - 3 = 2 → y = 5
Therefore, the solution to this system of equations is:
(x, y) = (2, 0) and (-3, 5)