I need to solve for $x$ and $y$ in this system of equations, where $a$, $b$, and $n$ are constants:
$$\frac{-n}{x+y} + \frac{a}{x^2} = 0 $$
$$\frac{-n}{x+y} + \frac{b}{y^2} = 0 $$
It is really messy for me, but I tried getting this form $$x+y = \frac{nx^2}{a} = \frac{ny^2}{b}$$ and try replacing it into the system of equations, but I keep getting x =1, which should not be.
The answer should be $$x = \frac{a + \sqrt{ab}}{n} \qquad y = \frac{b + \sqrt{ab}}{n}$$
How do I fix my approach?
Cleary $x,y\neq0$
$$ {n\over x+y}={a\over x^2}={b\over y^2}={1\over k}\\ k^2={x^2y^2\over ab}\Rightarrow k=\pm {xy\over \sqrt{ab}} \\ \ \\ k^2={x^2+y^2+2xy\over n^2}={ak+bk\pm 2k\sqrt{ab}\over n^2}\\ \ \\ k((n^2\pm2\sqrt{ab})k-(a+b))=0 $$
Since $k\neq0$ $k={a+b\pm2\sqrt{ab}\over n^2}$
Now use this to solve the first equation and you get the required solutions