Use the elimination to solve the linear-quadratic system y = x^2 + 10, y = -7x - 2.
A. (3, -23) and (4, -30)
B. (-4, 26)
C. (-3, 19) and (-4, 26)
D. (-3, 19)
First, we rearrange the quadratic equation to be in standard form: x^2 - y + 10 = 0.
Now, we can set both equations equal to each other: x^2 + 10 = -7x - 2.
Rearranging this equation, we get: x^2 + 7x + 12 = 0.
Now, we can factor this equation: (x + 3)(x + 4) = 0.
Setting each factor equal to zero gives us two possible values for x: x = -3 or x = -4.
Substituting these values back into either equation, we can solve for the corresponding y-values.
For x = -3, we have y = (-3)^2 + 10 = 19.
For x = -4, we have y = (-4)^2 + 10 = 26.
Therefore, the solution to the system is (-3, 19) and (-4, 26).
So, the correct answer is C.