$3y=\frac{x}{x-y}$
$x+4y=2x$
The top equation is the problem. But I think I've solved it correctly for x
$\frac{1}{3y} = \frac{x}{x}-\frac{y}{x}$
$\frac{1}{3y} = 1-\frac{y}{x}$
$\frac{y}{x} = 1-\frac{1}{3y}$
$\frac{X}{Y} = 1-3y$
$x = y-3y^2$
Put the X equal to eachother and solve for Y?
$y-3y^2 = 4y$
$-3y^2-3y=0$
That gives Y values $-1$ and $0$, which is not correct. So do I have to solve for Y first and find X-values or did I do a mistake somewhere?
http://www.wolframalpha.com/input/?i=3y%3Dx%2F%28x-y%29%2C+x%2B4y%3D2x