3. Solve the equation $\left[\frac{25 x-2}{4}\right]=\frac{13 x+4}{3}$ ( $[a]=z$ if $z \in \mathbb{Z}$ and $z \leq a<z+1$ ).
Solution. Let $\frac{13 x+4}{3}=y$. From there, $x=\frac{3 y-4}{13}$. If we substitute in the equation, we get $\left[\frac{75 y-126}{52}\right]=y$. It follows that $y \leq \frac{75 y-126}{52}<y+1$, from which $126 \leq 23 y \leq 178$ or $\frac{126}{23} \leq y<\frac{178}{23}$. Since $y$ is an integer, it follows that $y_{1}=6$ or $y_{2}=7$. Therefore, $x_{1}=\frac{14}{13}$ and $x_{2}=\frac{17}{13}$ are the solutions to the given equation.