, 2008

We can rewrite the equations as
$$\begin{aligned} (x + 2y)(x^2 - 2xy + 4y^2) &= x + 2y, \\ 2xy(x + 2y) &= x + 2y. \end{aligned}$$
Obviously, every pair of numbers $x$ and $y$ that satisfies $x + 2y = 0$, solves the equations. Now, assume $x + 2y \neq 0$. 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 $x = 1$, $y = \frac{1}{2}$.