1. Let $A$ and $B$ be subsets of $\mathbb{R}^{2}=\mathbb{R} \times \mathbb{R}$ defined by:

$$
\begin{aligned}
& A=\left\{(x, y) \mid \text { for every } a \in \mathbb{R}, y \neq \frac{x-a}{a x-1}\right\} \\
& B=\{(x, y) \mid \text { there exists } a \in \mathbb{R}, a(1+x y)=x+y\}
\end{aligned}
$$

Find $A \cap B$, and then represent it graphically.
Solution. Let $(x, y) \in A \cap B$. Since $(x, y) \in B$, there exists $a \in \mathbb{R}$ such that $a(1+x y)=x+y$, i.e., $y(a x-1)=x-a$. If $a x-1 \neq 0$, then $y=\frac{x-a}{a x-1}$, which contradicts $(x, y) \in A$. Therefore, $a x=1$ and $x=a$, from which it follows that $x^{2}=1$. For $x=1$, we have $-1=\frac{1-a}{a-1}$ and for $x=-1, 1=\frac{-1-a}{-a-1}$, which means that $x y \neq-1$. Consequently,

$$
A \cap B=\left\{(x, y) \mid x^{2}=1, x y \neq 1\right\}
$$

The graphical representation is given on the right-

![](https://cdn.mathpix.com/cropped/2024_06_05_777e4c6fa99d478c3e87g-7.jpg?height=349&width=492&top_left_y=1611&top_left_x=975)