So, either x + 2 = 0 or x - 1 = If x + 2 = 0, then x = -2. Substituting this back into the first equation -2 + y = y = So, one solution is x = -2 and y = 6.
If x - 1 = 0, then x = 1. Substituting this back into the first equation 1 + y = y = So, the other solution is x = 1 and y = 3.
Therefore, the solutions to the system of equations are x = -2, y = 6 and x = 1, y = 3.
We can solve this system of equations by substitution or elimination. Let's use substitution method here.
We have the equations
1) x + y =
2) x^2 - y = 2
From equation 1, we can write y = 4 - x. Now, substitute this value of y into equation 2:
x^2 - (4 - x) =
x^2 - 4 + x =
x^2 + x - 2 =
(x + 2)(x - 1) = 0
So, either x + 2 = 0 or x - 1 =
If x + 2 = 0, then x = -2. Substituting this back into the first equation
-2 + y =
y =
So, one solution is x = -2 and y = 6.
If x - 1 = 0, then x = 1. Substituting this back into the first equation
1 + y =
y =
So, the other solution is x = 1 and y = 3.
Therefore, the solutions to the system of equations are x = -2, y = 6 and x = 1, y = 3.