What Are The Solutions of The System?

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

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

Question

What Are The Solutions of The System?

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

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

anonymous · Answer

From this set up you can substitute y
y = x^2 + 3x - 4
y = 2x + 2

 2x + 2 = x^2 + 3x - 4

anonymous · Answer

Put everything on one side so we can have it equal zero then use the quadratic formula

anonymous · Answer

Actually, you can just factor it after bringing everything to one side.

anonymous · Answer

Yeah i did that but then i get confused >.< ... I was leaning toward d as the answer but idk if i'm right

anonymous · Answer

Once you factor it, you get (x + 3)(x - 2) = 0
x = -3, 2
ummm....I don't know what happened in you problem...I'm sorry. :(

anonymous · Answer

Oh >.< I'm Not very good at algebra. But which is the solution A B C or D

anonymous · Answer

Honestly, I would say no solution because none of the other choices make sense...