Can you please help me with these algebra problems

For these two questions, find the solutions of the system.

1. 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 solution

2. y=x^2-2x-2
y=4x+5

a. (-1,1) and (-7,-23)
b. (-1,1) and (7,33)
c. (-1,33) and (7,1)
d. no solution

Please don't tell me to do my own work.
Thank you

1. C
2. D

Use substitution:
1) y= x^2 + 3x - 4
2x + 2 = x^2 +3x - 4
x^2 + 1x - 6 = 0 ............What two factors of 6 subtract and = 1?
(x - 2)(x + 3) = 0
x=2.....x= -3
y=2(2) + 2= 4+2 = 6 so (2, 6)
y= 2(- 3) + 2 = -6 + 2 = -4 so (-3, -4)

2)y= X^2 - 2x - 2
4x + 5 = x^2 - 2x - 2
x^2 - 6x - 7= 0.......Can you think of two factors of 7 that subtract and = -6?
(x +....)(x - ....) = 0.......finish up on your own.

Here is a video on the quadratic formula, use it if you can't factor

http://www.khanacademy.org/math/algebra/…

Set the two equations equal to each other: x^2 +3x-4 = 2x+2 yeilds: x^2+x-6 = 0, then factor (x-3)(x+2), so you know x= 3, -2, then plug those in to the original equations to find y.
Same process for the second one!

1).
y = x^2 + 3x - 4
y = 2x + 2
x^2 + 3x - 4 = 2x + 2
x^2 + x - 6 = 0
(x + 3)(x - 2) = 0
x = -3, y = -4
x = 2, y = 6
Answer:
b. (-3, -4) and (2, 6)

2).
y = x^2 - 2x - 2
y = 4x + 5
x^2 - 2x - 2 = 4x + 5
x^2 - 6x - 7 = 0
(x - 7)(x + 1) = 0
x = 7, y = 33
x = -1, y = 1
Answer:
b. (-1, 1) and (7, 33)

Do your own homework... ;) Just here for points