4. Find all pairs of real numbers $x$, $y$ that satisfy the system of equations

$$
\begin{aligned}
& x \cdot\lfloor y\rfloor=7, \\
& y \cdot\lfloor x\rfloor=8 .
\end{aligned}
$$

(The notation $\lfloor a\rfloor$ denotes the floor of $a$, i.e., the greatest integer less than or equal to $a$.)
SOLUTION. From the second equation, it follows that $\lfloor x\rfloor \neq 0$, and $y=8 /\lfloor x\rfloor$ is therefore a non-zero number with an absolute value of at most 8. Substituting into the first equation, we get the equation

$$
x \cdot\left\lfloor\frac{8}{\lfloor x\rfloor}\right\rfloor=7
$$

which is actually equivalent to the given system of equations in the following sense: if we assign to any solution $x$ of equation (1) the value $y=8 /\lfloor x\rfloor$, the pair $(x, y)$ will clearly be a solution to the original system.

We will therefore sequentially find the solutions to equation (1) for individual values of the integers $a=\lfloor 8 /\lfloor x\rfloor\rfloor \in -8, \ldots,-1,1, \ldots, 8\}$ by calculating $x=7 / a, y=8 /\lfloor x\rfloor$ and verifying whether $\lfloor y\rfloor=a$. Moreover, given the inequality $\lfloor x\rfloor \neq 0$ from equation (1), it is clear that $a$ cannot be 8.

For $a=-8$, $x=-\frac{7}{8},\lfloor x\rfloor=-1$ and $y=-8$, so $\lfloor y\rfloor=a$.

For $a=-7$, $x=-1=\lfloor x\rfloor$ and $y=-8$, so $\lfloor y\rfloor \neq a$.

For $a=-5$, $x=-\frac{7}{5},\lfloor x\rfloor=-2$ and $y=-4$, so $\lfloor y\rfloor > a$.

For $a=-4$, $x=-\frac{7}{4},\lfloor x\rfloor=-2$ and $y=-4$, so $\lfloor y\rfloor=a$.

For $a=-3$, $x=-\frac{7}{3},\lfloor x\rfloor=-3$ and $y=-\frac{8}{3}$, so $\lfloor y\rfloor=a$.

For $a=-2$, $x=-\frac{7}{2},\lfloor x\rfloor=-4$ and $y=-2$, so $\lfloor y\rfloor=a$.

For $a=-1$, $x=-7=\lfloor x\rfloor$ and $y=-\frac{8}{7}$, so $\lfloor y\rfloor \neq a$.

For $a \in\{4,5,6,7\}$, $\lfloor x\rfloor=1$ and $y=8$, so $\lfloor y\rfloor > a$.

Conclusion. The system of equations has 6 solutions, namely $\left[-\frac{7}{8},-8\right],\left[-\frac{7}{4},-4\right],\left[-\frac{7}{3},-\frac{8}{3}\right],\left[-\frac{7}{2},-2\right]$, $\left[7, \frac{8}{7}\right]$ and $\left[\frac{7}{2}, \frac{8}{3}\right]$.

GUIDING AND SUPPLEMENTARY PROBLEMS: