46. The solution to the equation $3 x+5[x]-2017=0$ is $\qquad$. (For any real number $x$, let $[x]$ denote the greatest integer less than or equal to $x$)
Solution: Let $x=[x]+r, 0 \leqslant r<1$, substitute it into the original equation,

we get
$$
3[x]+3 r+5[x]-2017=0,
$$

so
$$
r=\frac{2017-8[x]}{3} \text {. }
$$

Thus, from $0 \leqslant \frac{2017-8[x]}{3}<1$, i.e., $0 \leqslant 2017-8[x]<3$, or equivalently $-3 \leqslant 8[x]-2017 \leqslant 0$,
we obtain
$$
\frac{2014}{8} \leqslant[x] \leqslant \frac{2017}{8},
$$

hence
$$
[x]=\frac{2016}{8}=252 \text {. }
$$

Therefore,
$$
r=\frac{2017-8[x]}{3}=\frac{2017-2016}{3}=\frac{1}{3} \text {, }
$$

so
$$
x=[x]+r=252 \frac{1}{3} \text {. }
$$