Consider the following two sets of numbers:

$$
A=\left\{\frac{3 n-4}{5 n-3}: n \in \mathbb{Z}\right\} \quad \text { and } \quad B=\left\{\frac{4 k-3}{7 k-6}: k \in \mathbb{Z}\right\} \text {. }
$$

How many elements does the set $A \cap B$ have?
Solution. It is easy to check that if

$$
\frac{3 n-4}{5 n-3}=\frac{3 n^{\prime}-4}{5 n^{\prime}-3}
$$

then $n=n^{\prime}$, meaning that for different values of $n$, we get different elements of set $A$. A similar statement holds for set $B$.

Therefore, we need to determine how many pairs of integers $(n, k)$ satisfy

$$
\frac{3 n-4}{5 n-3}=\frac{4 k-3}{7 k-6}
$$

Since the denominators are never zero, the condition is equivalent to

$$
(3 n-4)(7 k-6)=(4 k-3)(5 n-3)
$$

Rearranging this, we get $(k-3) n=16 k-15$. Since $k=3$ does not provide a solution, we can divide by $(k-3)$, and thus we get $n=16+\frac{33}{k-3}$. Since $n$ is an integer, $k-3$ must be a divisor of 33. The number 33 has 4 positive and 4 negative divisors. Therefore, the number of suitable pairs $(n, k)$ is 8, which is the number of common elements between sets $A$ and $B$.