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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?

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

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  • $\begingroup$ I have done this only but I am searching for a shorter and better solution. $\endgroup$ – Satvik Mashkaria Oct 21 '14 at 14:59
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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.

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