3. Determine all real numbers $x$ for which $\{x\}=\frac{x+[x]+\left[x+\frac{1}{2}\right]}{10}$, where $[x]$ is the greatest integer not greater than $x$ and $\{x\}=x-[x]$.
Solution. From $x=[x]+\{x\}$, it follows that the given equation is successively equivalent to the equations

$$
\begin{aligned}
& 10\{x\}=[x]+\{x\}+[x]+\left[x+\frac{1}{2}\right], \\
& 9\{x\}-2[x]-\left[x+\frac{1}{2}\right]=0 .
\end{aligned}
$$

Furthermore, $\{x\} \in[0,1)$, so we will consider two cases.

- If $\{x\} \in\left[0, \frac{1}{2}\right)$, then $\{x\}+\frac{1}{2} \in[0,1)$, so $\left[x+\frac{1}{2}\right]=\left[[x]+\{x\}+\frac{1}{2}\right]=[x]$. Substituting into (1), we get $9\{x\}-3[x]=0$, or $[x]=3\{x\} \in\left[0, \frac{3}{2}\right)$. According to this, we have two solutions $[x]_{1}=0$ or $[x]_{2}=1$. For $[x]_{1}=0$, we get $\{x\}_{1}=0$, from which we find $x_{1}=0$. For $[x]_{2}=1$, we get $\{x\}_{2}=\frac{1}{3}$, from which we find $x_{2}=\frac{4}{3}$.
- If $\{x\} \in\left[\frac{1}{2}, 1\right)$, then $\{x\}+\frac{1}{2} \in\left[1, \frac{3}{2}\right)$, so $\left[x+\frac{1}{2}\right]=[x]+1$. Substituting into (1), we get $9\{x\}-3[x]-1=0$, or $[x]=3\{x\}-\frac{1}{3} \in\left[\frac{7}{6}, \frac{8}{3}\right)$. From this, we get the solution $[x]_{3}=2$. For $[x]_{3}=2$, we get $\{x\}_{3}=\frac{7}{9}$, from which we find $x_{3}=\frac{25}{9}$.

Therefore, the solutions to the given equation are $x_{1}=0, x_{2}=\frac{4}{3}$, and $x_{3}=\frac{25}{9}$.