## Task A-2.1.

Determine all pairs $(x, y)$ of real numbers for which

$$
x+\frac{1}{y-x}=1 \quad \text { and } \quad y+\frac{1}{x-y}=2
$$
## Solution.

By subtracting the given equations, we obtain

$$
y-x-\frac{2}{y-x}=1
$$

Let $t=y-x$. From the previous equation, it follows that

$$
\begin{aligned}
t-\frac{2}{t} & =1 \\
t^{2}-t-2 & =0
\end{aligned}
$$

Solving the quadratic equation, we get $t=2$ or $t=-1$.

In the first case, from the first equation, we get $x=1-\frac{1}{t}=\frac{1}{2}$, and from the second equation, we get $y=2+\frac{1}{t}=\frac{5}{2}$.

In the second case, similarly, we get $x=2$ and $y=1$.

Therefore, all solutions to the system are $(2,1)$ and $\left(\frac{1}{2}, \frac{5}{2}\right)$.