<s>[INST] Solve the system $ x \lfloor y \rfloor = 7 $ and $ y \lfloor x \rfloor = 8 $. 
Solve the following system for $ x,y \in \mathbb{R} $: \begin{align} x \lfloor y \rfloor & = 7, \\ y \lfloor x \rfloor & = 8. \end{align}

It could be reducing to one variable, but it is not that simple.
Another idea is substitution, but it didnt lead me anywhere.
This was taken from a competition, and it’s tricky.
From what I noticed, systems with two or more variables involving "greatest integer" function are rarely seen, probably too difficult to analyze...

This was obtained with the help of Wolfram Alpha: (I swapped places of 7 and 8 by mistake, so it is not exactly the same system; anyway, the answer should not rely on software tools, of course)

 [/INST] Note: Here we use the following characterization of the floor function 
\begin{align*}
\lfloor x\rfloor = p\qquad \text{ with } \qquad p \leq x < p+1, \quad p\in \mathbb Z
\end{align*}
We use this relationship to transform the system of equations into a system of inequalities which can be easily analysed.

Let $x,y \in \mathbb{R}$. The system of equations
  \begin{align*}
x \lfloor y \rfloor & = 7 \tag{1}\\
y \lfloor x \rfloor & = 8
\end{align*}
  is equivalent with
  \begin{align*}
x p & = 7 \qquad \text{ and }\qquad p\leq y < p+1, \quad p\in \mathbb{Z}\\
y q & = 8 \qquad \text{ and }\qquad q\leq x < q+1, \quad q\in \mathbb{Z}
\end{align*}
  We substitute $x=\frac{7}{p}$ and $y=\frac{8}{q}$ and get the following system  of inequalities with $p,q\in \mathbb{Z}$
  \begin{align*}
p\leq \frac{8}{q} < p+1\tag{2}\\
q\leq \frac{7}{p} < q+1\tag{3}
\end{align*}
Note that the two inequalities are equivalent with the two equations in (1).

We are now looking for integer solutions of these inequalities. First we see from (1) that $x\neq 0$ and $y\neq 0$. Then we deduce from (2) that either both $p$ and $q$ are greater zero or both are less than zero.

First step: $p,q>0$

Since $q > 0$ we conclude from (3) that $1\leq p \leq 7$ and we obtain
\begin{align*}
p=1 \qquad \rightarrow \qquad q\leq &\frac{7}{1}<q+1 \qquad \rightarrow \qquad q=7\\
p=2 \qquad \rightarrow \qquad q\leq &\frac{7}{2}<q+1 \qquad \rightarrow \qquad q=3\\
p=3 \qquad \rightarrow \qquad q\leq &\frac{7}{3}<q+1 \qquad \rightarrow \qquad q=2\\
p=4 \qquad \rightarrow \qquad q\leq &\frac{7}{4}<q+1 \qquad \rightarrow \qquad q=1\\
&\ldots\\
p=7 \qquad \rightarrow \qquad q\leq &\frac{7}{7}<q+1 \qquad \rightarrow \qquad q=1\\
\end{align*}

Checking these pairs $(p,q)$ in inequality (2) we see, that only $p=1$ and $p=2$ are valid.
\begin{align*}
(p,q)=(1,7) \qquad \rightarrow \qquad 1\leq \frac{8}{7} < 2\tag{4}\\
(p,q)=(2,3) \qquad \rightarrow \qquad 2\leq \frac{8}{3} < 3\\
\end{align*}
Since $x=\frac{7}{p}$ and $y=\frac{8}{q}$ we get from (4) the solutions with positive $x,y$
  \begin{align*}
A=\left\{\left(7,\frac{8}{7}\right),\left(\frac{7}{2},\frac{8}{3}\right)\right\}
\end{align*}

$$ $$

Second step: $p,q<0$
First we note, that if $p<-7$ it follows from (3) that $q=-1$. Putting $q=-1$ into (2) we obtain $p\leq -8<p+1$ and $p=-8$ follows.
We conclude that in the case $p<-7$ only $(p,q)=(-8,-1)$ is a valid solution.

Next we consider $-7 \leq p \leq -1$ similar to above
\begin{align*}
p=-1 \qquad \rightarrow \qquad q\leq &-\frac{7}{1}<q+1 \qquad \rightarrow \qquad q=-7\\
p=-2 \qquad \rightarrow \qquad q\leq &-\frac{7}{2}<q+1 \qquad \rightarrow \qquad q=-4\\
p=-3 \qquad \rightarrow \qquad q\leq &-\frac{7}{3}<q+1 \qquad \rightarrow \qquad q=-3\\
p=-4 \qquad \rightarrow \qquad q\leq &-\frac{7}{4}<q+1 \qquad \rightarrow \qquad q=-2\\
&\ldots\\
p=-7 \qquad \rightarrow \qquad q\leq &-\frac{7}{7}<q+1 \qquad \rightarrow \qquad q=-1\\
\end{align*}

Checking these seven pairs $(p,q)$ in (2) shows that the following three pairs are valid
  \begin{align*}
(p,q)=(-2,-4)\qquad\Rightarrow\qquad -2\leq- \frac{8}{4}<-1\\
(p,q)=(-3,-3)\qquad\Rightarrow\qquad -3\leq- \frac{8}{3}<-2\\
(p,q)=(-4,-2)\qquad\Rightarrow\qquad -4\leq- \frac{8}{2}<-3\\
\end{align*}
Together with $(-8,-1)$ we derive from $x=\frac{7}{p}$ and $y=\frac{8}{q}$ the following $(x,y)$ pairs with negative solutions
  \begin{align*}
B=\left\{\left(-\frac{7}{2},-2\right),\left(-\frac{7}{3},-\frac{8}{3}\right),\left(-\frac{7}{4},-4\right),\left(-\frac{7}{8},-8\right)\right\}
\end{align*}

Note that $A\cup B$ coincides with the solution of @AlexR.
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