1. Find all pairs of real numbers $x$ and $y$ that satisfy the equations

$$
\begin{aligned}
& x+\frac{1}{y-x}=1 \\
& y+\frac{1}{x-y}=2
\end{aligned}
$$
II/1. 1. way. We add the equations and get $x+y=3$. From this, we express $y=3-x$ and substitute it into the first equation to get $x+\frac{1}{3-2 x}=1$. We eliminate the fractions and rearrange to get $2 x^{2}-5 x+2=0$. The left side of this equation can be factored as $2(x-2)\left(x-\frac{1}{2}\right)=0$. This gives us $x=2$ or $x=\frac{1}{2}$. In the first case, $y=1$, and in the second case, $y=\frac{5}{2}$. Since in both cases $x-y \neq 0$, $(2,1)$ and $\left(\frac{1}{2}, \frac{5}{2}\right)$ are indeed solutions to the given system of equations.

2. way. We subtract the equations and get $y-x-\frac{2}{y-x}=1$. By introducing the variable $a=y-x$, we rearrange to get $a^{2}-a-2=0$. The left side of this equation can be factored as $(a+1)(a-2)=0$. This gives us $a=-1$ and $a=2$. For each case of $a$, we substitute back into the first equation and get $x=2$ and $x=\frac{1}{2}$. In the first case, from $a=y-x$ we get $y=1$, and in the second case, $y=\frac{5}{2}$. Since in both cases $y-x \neq 0$, $(2,1)$ and $\left(\frac{1}{2}, \frac{5}{2}\right)$ are indeed solutions to the given system of equations.

## 1. way:

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-10.jpg?height=55&width=1631&top_left_y=852&top_left_x=218)

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-10.jpg?height=52&width=1631&top_left_y=905&top_left_x=224)

Factorization or use of the quadratic formula ................... 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-10.jpg?height=52&width=1633&top_left_y=1005&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-10.jpg?height=63&width=1633&top_left_y=1051&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-10.jpg?height=125&width=1653&top_left_y=1154&top_left_x=204)

The obtained quadratic equation $a^{2}-a-2=0 ; a=y-x \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \ldots . \ldots$ points

Factorization or use of the quadratic formula .................. 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-10.jpg?height=51&width=1633&top_left_y=1365&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-10.jpg?height=68&width=1633&top_left_y=1411&top_left_x=220)

A contestant who solves the problem but does not verify that the solutions satisfy the condition $x-y \neq 0$ will lose 1 point.