# Solving a non-linear system, urgent reply needed!

• Jul 15th 2008, 08:35 PM
mankvill
Solving a non-linear system, urgent reply needed!
http://img228.imageshack.us/img228/9513/math1zq5.jpg

Not sure where to begin on this one. If you tell me how to do this one, it'd help me with the others. :D
• Jul 15th 2008, 08:52 PM
Isomorphism
Quote:

Originally Posted by mankvill
http://img228.imageshack.us/img228/9513/math1zq5.jpg

Not sure where to begin on this one. If you tell me how to do this one, it'd help me with the others. :D

Substitute for y in the second equation to get a quadratic equation. Now solve this quadratic and try both roots to get values for y.

Try it (Wink)
• Jul 15th 2008, 08:57 PM
mankvill
Oh yeah, I always forget to do that!

Now I'm having trouble with this! (Headbang)

(x-2)(x-2) doesn't work.
(x+2)(x-2) doesn't work...

(Doh)
• Jul 15th 2008, 08:58 PM
mankvill
Wait.

Is there no solution?
• Jul 15th 2008, 09:08 PM
Isomorphism
Sigh...

x^2+ 4x + 1 = - 3

x^2+ 4x + 4 = 0

(x+2)^2 = 0

x = -2

Substiute this value in y = x^2 + 1 = (-2)^2 + 1 = 4+1 = 5
• Jul 15th 2008, 09:10 PM
mankvill
Quote:

Originally Posted by Isomorphism
Sigh...

x^2+ 4x + 1 = - 3

x^2+ 4x + 4 = 0

(x+2)^2 = 0

x = -2

Substiute this value in y = x^2 + 1 = (-2)^2 + 1 = 4+1 = 5

oh, okay. yeah, i'm not good with that.

So is there 1 answer: -2,5?
• Jul 15th 2008, 09:13 PM
Isomorphism
Quote:

Originally Posted by mankvill
oh, okay. yeah, i'm not good with that.

So is there 1 answer: -2,5?

Yes, after all that working out... you should not be this doubtful :p
• Jul 15th 2008, 09:14 PM
mankvill
I know, I'm just taking a test right now! Gotta be sure.

Thanks!