Question

What are the solutions to the system?

y = x^2 + 3x – 4
y = 2x + 2
answers
1 (–3, 6) and (2, –4)
2 (–3, –4) and (2, 6)
3 (–3, –4) and (–2, –2)
4 no solution

Answer 1

To find the solutions to the system, we can set the two equations equal to each other and solve for x:

x^2 + 3x – 4 = 2x + 2

Rearranging the equation, we get:

x^2 + x – 6 = 0

Factoring the quadratic equation, we have:

(x + 3)(x - 2) = 0

Setting each factor equal to zero, we find the possible values of x:

x + 3 = 0 => x = -3
x - 2 = 0 => x = 2

So the solutions to the system are x = -3 and x = 2. To find the corresponding y-values, we substitute these values into either of the original equations. Using the second equation, y = 2x + 2, we have:

For x = -3, y = 2(-3) + 2 = -6 + 2 = -4
For x = 2, y = 2(2) + 2 = 4 + 2 = 6

Therefore, the solutions to the system are (-3, -4) and (2, 6). Answer choice 2 is correct: (–3, –4) and (2, 6).