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).
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
1 answer