What are the solutions of the system?

y=x^2+3x-4

y=2x+2

a) (-3,6) and (2,-4)**
b) (-3,-4) and (2,6)
c) (-3,-4) and (-2,-2)
d) no solutions

2x+2 = x^2 + 3x -4

x^2 + x - 6 = 0
(x-2)(x+3) = 0
x = +2 ---> y = 6 or (2,6)
x = -3 ---> y = -4 or (-3,-4)

so I disagree with you

You got x = -3,2 correct, but your final answer is wrong.

My secon guess was C

guess? No need to guess !!!!

So am I right or wrong

I suck at this

which one is

(-3,-4) and (2,6) ?????

To find the solutions of the system, we need to find the values of x and y that satisfy both equations simultaneously.

Let's start by setting the two equations equal to each other:

x^2 + 3x - 4 = 2x + 2

Next, we can simplify the equation by moving all the terms to one side:

x^2 + 3x - 2x - 4 - 2 = 0

x^2 + x - 6 = 0

Now, we can factor the quadratic equation:

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

To find the values of x, we set each factor equal to zero and solve for x:

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

x - 2 = 0 --> x = 2

So, we have two possible values for x: -3 and 2.

To find the corresponding y-values, we can substitute these x-values into either of the original equations. Let's use the second equation:

y = 2x + 2

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

So, one solution is (-3, -4).

For x = 2:
y = 2(2) + 2
y = 4 + 2
y = 6

So, another solution is (2, 6).

Therefore, the correct answer is a) (-3,6) and (2,-4).