Example 6 The solution to the equation $6 x-3[x]+7=0$ is
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(21st "Hope Cup" Mathematics Invitational (Grade 8))
Solve the equation $6 x-3[x]+7=0$ which can be rewritten as $[x]=\frac{6 x+7}{3}$.
Substitute it into $x-1<[x] \leqslant x$, to get $x-1<\frac{6 x+7}{3} \leqslant x$.
Solving this system of inequalities yields $-\frac{10}{3}<x \leqslant-\frac{7}{3}$.
Thus, $[x]=-4$ or -3.
Therefore, $\frac{6 x+7}{3}=-4$ or -3.
Solving gives $x=-\frac{19}{6}$ or $-\frac{8}{3}$.
Upon verification, $x=-\frac{19}{6}$ and $x=-\frac{8}{3}$ are both solutions to the original equation.