## Task 3 - 181213

Determine all pairs $(x, y)$ of real numbers that satisfy the following system of equations:

$$
x+\frac{1}{y}+\frac{y}{x}=3, \quad y+\frac{1}{x}+\frac{x}{y}=3
$$
Assuming $(x, y)$ is a solution to the given system of equations. Then $x \neq 0$ (and $y \neq 0$), and it holds that

$$
\begin{aligned}
& x^{2} y + x + y^{2} = 3 x y \\
& x y^{2} + y + x^{2} = 3 x y
\end{aligned}
$$

By subtracting one from the other, we obtain

$$
(x-y)(x y + 1 - x - y) = 0 \quad \text{thus} \quad (x-y)(x-1)(y-1) = 0
$$

This implies that at least one of the equations $x=y, x=1, y=1$ holds.

From $x=y$ and (1), we get $x^{3} - 2 x^{2} + x = 0, x(x-1)^{2} = 0$, and since $x=y$, we have $x=1, y=1$.

From $x=1$ and (1), we get $y^{2} - 2 y + 1 = 0, (y-1)^{2} = 0$, thus $y=1$.

From $y=1$ and (1), we get $x^{2} - 2 x + 1 = 0, (x-1)^{2} = 0$, thus $x=1$.

Therefore, the only pair that can be a solution to the system of equations is $(1,1)$. Indeed, it satisfies both equations of the system because $1 + \frac{1}{1} + \frac{1}{1} = 3$.

Adapted from [5]