Quadratic Systems

y=3x^2-5
x-y=3

Could you tell me the answer.I want to make sure I'm right and could you show work

2 answers

It's better if you show your work first, but here goes.

Since y = x-3,
x-3 = 3x^2-5
3x^2 - x - 2 = 0
(3x+2)(x-1) = 0
x = 1 or -2/3
and you can now evaluate y for each x.

as shown at

http://www.wolframalpha.com/input/?i=solve+y%3D3x^2-5%2C+x-y%3D3
Thank you so much it help a lot