How to solve this system?

Question

x + [(3x-y)/(x^2+y^2)] = 3

y - [(x+3y)/(x^2+y^2)] = 0

Answer 1

from the first;
(3x - y)/(x^2 + y^2) = 3-x
x^2 + y^2 = (3-x)/(3x-y)

from the 2nd:
(x+3y)/x^2 + y^2) = y
x^2 + u^2 = (x+3y)/y

(3-x)/(3x-y) = (x+3y)/y

getting very nasty

Wolfram says:
http://www.wolframalpha.com/input/?i=solve+x+%2B+%5B%283x-y%29%2F%28x%5E2%2By%5E2%29%5D+%3D+3%2C+y+-+%5B%28x%2B3y%29%2F%28x%5E2%2By%5E2%29%5D+%3D+0