posted by James .
Solve algebraically and check
y = x^2 - 3
x + y = -1
y appears in both equations to the first degree. You can eliminate y by equating the value of y from each equation.
Solve the resulting quadratic equation for x.
Back-substitute the values of x into the equations to get the corresponding values of y.
Substitute x^2 - 3 for y in the second equation.
x + x^2 - 3 = -1
x^2 + x - 2 = 0
(x + 2)(x - 1) = 0
Find the two possible values for x and insert to find y. Insert both sets of values into second equation to check.