<s>[INST] Solve the System of Equations in $x$ and $y$ \begin{equation}
x+\frac{3\,x-y}{x^2+y^2}=3 \tag{1}
\end{equation}
\begin{equation}
y=\frac{x+3\,y}{x^2+y^2} \tag{2}
\end{equation}
 [/INST] We have:
\begin{equation}
x\left(1+\frac{3}{x^2+y^2}\right)=3+\frac{y}{x^2+y^2} \tag{3}
\end{equation}
\begin{equation}
y\left(1-\frac{3}{x^2+y^2}\right)=\frac{x}{x^2+y^2} \tag{4}
\end{equation}
Multiplying Eqn $3$ with $y$ and Eqn $4$ with $x$ we get
\begin{equation}
x\,y\left(1+\frac{3}{x^2+y^2}\right)=3\,y+\frac{y^2}{x^2+y^2} \tag{5}
\end{equation}
\begin{equation}
x\,y\left(1-\frac{3}{x^2+y^2}\right)=\frac{x^2}{x^2+y^2} \tag{6}
\end{equation} Adding Eqns $5$ and $6$ we get
$$y=\frac{1}{2\,x-3}$$ Using $y=\frac{1}{2\,x-3}$ in Eqn $2$ we get Bi Quadratic in $x$ as:
$$4x^4-24x^3+57x^2-63x+26=0$$
Can i get a Hint without involvement of BiQuadratic...
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