Example 6 Solve the equation $\left[\frac{5+6 x}{8}\right]=\frac{15 x-7}{5}$.
Let $\frac{15 x-7}{5}=n(n \in Z)$, then $x=\frac{5 n+7}{15}$, substituting into the original equation and rearranging gives $\left[\frac{10 n+39}{40}\right]=n$. By the definition of the floor function, we have $0 \leqslant \frac{10 n+39}{40}-n<1$, solving this yields $-\frac{1}{30}<n \leqslant \frac{13}{10}$, thus $n=0$ or $n=1$.
If $n=0$, then $x=\frac{7}{15}$; if $n=1$, then $x=\frac{4}{5}$.