gobathegreat wrote: Find all real numbers $(x,y)$ satisfying the following: $$x+\frac{3x-y}{x^2+y^2}=3$$$$y-\frac{x+3y}{x^2+y^2}=0$$ Setting $x=r\cos t$ and $y=r\sin t$ (with $r\ne 0$), second equation becomes $\cos t=(r^2-3)\sin t$ Plugging this in first equation, we get $r^4\ne 10$ and $\sin t=\frac {3r}{r^4-10}$ and so $\cos t=\frac{3r(r^2-3)}{r^4-10}$ Writing then $\sin^2 t+\cos^2=1$, we get $r^8-9r^6+34r^4-90r^2+100=0$ Which is $(r^2-2)(r^2-5)(r^4-2r^2+10)=0$ $r^2=2$ implies $x=r\cos t=\frac{3r^2(r^2-3)}{r^4-10}=1$ and $y=r\sin t=\frac {3r^2}{r^4-10}=-1$ $r^2=5$ implies $x=r\cos t=\frac{3r^2(r^2-3)}{r^4-10}=2$ and $y=r\sin t=\frac {3r^2}{r^4-10}=1$ Hence the conclusion $\boxed{(x,y)\in\{(1,-1),(2,1)\}}$ which indeed both are solutions

What's the advantage of substitution $r \sin t / \cos t$ and when shall we consider to do it

AlastorMoody wrote: What's the advantage of substitution $r \sin t / \cos t$ and when shall we consider to do it I can not give you a clever answer to this question. I saw the bump on this problem and concluded that direct algebra approach was heavy and difficult (since no answer) and I even did not try it. Presence of $x^2+y^2$ was for me an invitation to try $x=r\cos t$ and $y=r\sin t$ I just tried it and it worked. Maybe direct algebra with some clever simplifications would be simpler. I dont know.

if $x=0$ or $y=0 \to x=y=0$(unsatisfactory) if $x,y \ne 0$ then: $xy+\frac{3xy-y^2}{x^2+y^2}=3y$ and $xy-\frac{x^2+3xy}{x^2+y^2}=0$ $\to 2xy-1=3y \to y=\frac{1}{2x-3}$ $\to x+\frac{3x-\frac{1}{2x-3}}{x^2+(\frac{1}{2x-3})^2}=3$ $\iff x=0$(unsatisfactory); $x=1$;$x=2$ Retry and ...