# Solve for reals $x, y\in \mathbb R$ given system of two non-linear equations.

Solve for reals:-

\begin{align} 5x\left(1+\frac{1}{x^2+y^2}\right)& =12\\ 5y\left(1-\frac{1}{x^2+y^2}\right)&=4\end{align}

I got this relation

$$6x^{-1}+2y^{-1}=5$$ Now I substituted $x^{-1}=x_1$ and same for $y$ and got a four degree equation. Is there a short and elegant method for this?

• are you sure that ,the system has a real solution ? – Khosrotash Oct 21 '14 at 14:58
• @darya khosrotash yes $y=1$ and $x=2$ is one of the solutions. – Satvik Mashkaria Oct 21 '14 at 15:06

from yor equation we obtain $x=\frac{6y}{5y-2}$ plugging this in the first equation, simplifying and factorizing we get $- \left( y-1 \right) \left( 5\,y+1 \right) \left( 5\,{y}^{2}-4\,y+4 \right)=0$ from here you can compute all solutions.
The two given equations imply the equation $$2x+6y-5xy=0.$$ Since $x,y$ are different from zero, this is exactly the equation you have found. The rest is easy now. The equation implies that $5x-6\neq 0$, so that $y=\frac{2x}{5x-6}$. Substituting this into the given equations we obtain $$(5x^2 - 12x + 9)(5x - 2)(x - 2)=0.$$ The quadratic polynomial has no real roots, so that we obtain $(x,y)=(\frac{2}{5},\frac{-1}{5})$ and $(x,y)=(2,1)$.
There might be a more elegant solution, but this solution gives also another insight - the polynomial $2x+6y-5xy$ is an $S$-polynomial which appears from the Buchberger algorithm for polynomial equations.