<s>[INST] Solving $\Bigl\{\begin{smallmatrix}x+\frac{3x-y}{x^2+y^2}=3\\y-\frac{x+3y}{x^2+y^2}=0\end{smallmatrix}$ in $\mathbb R$ 
$$\begin{cases}
x+\dfrac{3x-y}{x^2+y^2}=3 \\
y-\dfrac{x+3y}{x^2+y^2}=0
\end{cases}$$
  Solve in the set of real numbers.

The furthest I have got is summing the equations, and I got
$$x^3+(y-3)x^2+(y^2+2)x+y^3-3y^2-4y=0.$$
But I have no idea how to solve this problem. How can I solve it?
 [/INST] Sketch of a solution rewriting the given system in terms of linear algebra:


*

*Set $\boxed{a = \binom{x}{y}}$. Then the given system looks as follows:
$$\begin{cases}
x+\dfrac{3x-y}{x^2+y^2}=3 \\
y-\dfrac{x+3y}{x^2+y^2}=0
\end{cases} \Longleftrightarrow \boxed{a +\frac{1}{||a||^2}\begin{pmatrix}3 & -1 \\ -1 & -3 \end{pmatrix}a =  \binom{3}{0}}$$

*Set $\boxed{t = \frac{1}{ ||a||^2}}$ and $\boxed{A = \begin{pmatrix}3 & -1 \\ -1 & -3 \end{pmatrix}} \Rightarrow  (I + tA)a = \binom{3}{0} \Rightarrow \boxed{a = (I + tA)^{-1}\binom{3}{0}}$.

*So, calculate $a$ by inverting $(I + tA)$: 
$$\boxed{a =} (I + tA)^{-1}\binom{3}{0} = \begin{pmatrix}1+3t & -t \\ -t & 1-3t \end{pmatrix}^{-1}\binom{3}{0} = \ldots = \boxed{\frac{3}{1-10t^2}\binom{1-3t}{t} }$$

*Find $t >0 $ such that $t = \frac{1}{ ||a||^2} \Rightarrow \boxed{t= \frac{1}{5}, \frac{1}{2}}$

*Plug these values into $a = \frac{3}{1-10t^2}\binom{1-3t}{t}$:
$$\Rightarrow \begin{cases} t= \frac{1}{2} \Rightarrow \boxed{a= \binom{1}{-1}} \\ t= \frac{1}{5} \Rightarrow \boxed{a= \binom{2}{1}} \end{cases}$$
</s>