Slovenija 2008

We can rewrite the equations as $$\begin{array}{rl}(x+2y)(x^2-2xy+4y^2)& =x+2y,\\ 2xy(x+2y)& =x+2y.\end{array}$$ Obviously, every pair of numbers $x$ and $y$ that satisfies $x+2y=0$, solves the equations. Now, assume $x+2y\ne 0$. We can then divide by $x+2y$ and get $x^2-2xy+4y^2=1$ and $2xy=1$, so $$(x+2y{)}^2=x^2+4xy+4y^2=(x^2-2xy+4y^2)+6xy=4.$$ We consider two cases: $x+2y=2$ or $x+2y=-2$. In the first case we have $1=2xy=2(2-2y)y=4y-4y^2$, so $0=4y^2-4y+1=(2y-1{)}^2$. This implies $y=\frac{1}{2}$ and $x=2-2y=1$. In the second case we have $x=-2-2y$, so $1=2xy=2(-2-2y)y=-4y-4y^2$ and $0=(2y+1{)}^2$. This implies $y=-\frac{1}{2}$ and $x=-1$.

Solutions are all pairs of real numbers $x$ and $y$ such that $x+2y=0$, as well as the pairs $x=-1$, $y=-\frac{1}{2}$ and $x=1$, $y=\frac{1}{2}$.