Solving a system of non-linear equations

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
Luke53
Posts: 54
Joined: Sun Mar 13, 2011 9:46 am
Contact:

Solving a system of non-linear equations

Hello, I need help on this one, tried out about everything, but still no use, can't solve the following system:
y² - x y - 4x = 0
x² - x y + 4 y - 12 = 0
I can't isolate x or y in either of the two equations for substituting in the other.
So what's the trick to solve this?

Luke.

nona.m.nona
Posts: 288
Joined: Sun Dec 14, 2008 11:07 pm
Contact:

Re: Solving a system of non-linear equations

y² - x y - 4x = 0
x² - x y + 4 y - 12 = 0
One method might be to solve the first equation for $y.$ Substitute the result into the second equation. Multiply through to clear the denominators, and expand.

Quite a few terms will drop out, leaving you with a quadratic which solves nicely with the Quadratic Formula.

Luke53
Posts: 54
Joined: Sun Mar 13, 2011 9:46 am
Contact:

Re: Solving a system of non-linear equations

y² - x y - 4x = 0
x² - x y + 4 y - 12 = 0
x=y²/(y+4) (didn't see that I could factor x out in the first eqn.)
Substituting this in the second eqn and solving gives: 20y²-32y-192=0
y=4 and y= -12/5
subst. y=4 into the system gives x=2
Thanks.
Luke.