Question

What are the solutions of the system?
y=x^2+3x-4
y=2x+2

Answer 1

answers:
(–3, 6) and (2, –4)
(–3, –4) and (2, 6)
(–3, –4) and (–2, –2)
no solution

Answer 2

\[x ^{2}+3x-4=2x+2 => x ^{2}+x-6=(x-2) (x+3)=0\]
x=2 or -3 at x = 2, y=0, x=-3, y=0

Answer 3

@EarthCitizen Help him please. I am a little bit busy

Answer 4

Thanks

Answer 5

no solution

Answer 6

no p @Shayaan_Mustafa

Answer 7

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

if both equal y then they equal each other
x^2+3x-4= 2x+2
as posted above, this simplifies (add -2x-2 to both sides)
x^2 +x - 6=0
this factors to
(x+3)(x-2)=0
of course this gives two roots: x=2 and x=-3
to find y, use either of the original equations. Best to use the simple one:
y= 2x+2

if x= 2, y= 2*2+2= 6 so (2,6) is one of the solutions.
I'll leave it to you to find the 2nd solution (when x= -3)

Answer 8

kk, my bad.

Answer 9

so what is the answer??

Answer 10

@phi when either x is used to find y, does it give the same values for y ?

Answer 11

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

Answer 12

@EarthCitizen Yes. that is the whole idea.