How do you solve a system of equations with rational functions?

[anemone]

Here is this week's POTW:

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Find all reals $x$ and $y$ that satisfy the system $\dfrac{3x-y}{x^2+y^2}=3-x$ and $\dfrac{x+3y}{x^2+y^2}=y$.

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[anemone]

Congratulations to Opalg for his correct solution(Cool), which you can find below:

Spoiler

Let $r = x^2+y^2$ (and note that $r\ne0$). The equations become $$3x-y = r(3-x), \qquad x+3y = ry.$$ Write these as $$\begin{aligned}(3+r)x - y &= 3r\\ x + (3-r)y &= 0.\end{aligned}$$ Solve those simultaneous equations for $x$ and $y$, to get $$x = \frac{3r(r-3)}{r^2-10},\qquad y = \frac{3r}{r^2-10}.$$ But $x^2 + y^2 = r$. Therefore $$\frac{9r^2(r-3)^2}{(r^2+10)^2} + \frac{9r^2}{(r^2+10)^2} = r,$$ $$9r^2\bigl((r-3)^2 + 1\bigr) = (r^2-10)^2r,$$ $$9r(r^2 - 6r + 10) = r^4 - 20r^2 + 100,\qquad \dagger$$ $$r^4 - 9r^3 + 34 r^2 - 90r + 100 = 0$$ (in the line labelled $\dagger$ it's okay to divide through by $r$ because $r\ne0$).

A bit of experimentation shows that $r=2$ and $r=5$ are roots of this equation, which becomes $$(r-2)(r-5)(r^2-2r+10) = 0$$ (the other two roots are non-real complex numbers).

If $r=2$ then the original equations become $$3x-y = 2(3-x),\qquad x+3y = 2y,$$ with the solution $(x,y) = (1,-1)$.

If $r=5$ then they become
$$3x-y = 5(3-x),\qquad x+3y = 5y,$$ with the solution $(x,y) = (2,1)$.

Those are the only solutions.