1. Find all $x$ for which $2[x]+\{3 x\}=\frac{7}{3}$, where $[x]$ is the integer part of $x$, and $\{x\}$ is the fractional part of $x$, that is, $\{x\}=x-[x]$.
Answer: $\left\{1 \frac{1}{9} ; 1 \frac{4}{9} ; 1 \frac{7}{9}\right\}$.

## Solution:

From the equation and definitions, it follows that $[x]=1$, and $\{3 x\}=\frac{1}{3}$.

Consider the equation $\{3 x\}=\frac{1}{3}$:

1) if $0 \leq\{x\}<\frac{1}{3}$, then $\{3 x\}=3\{x\} \Rightarrow\{x\}=\frac{1}{9}$;
2) if $\frac{1}{3} \leq\{x\}<1$, then $\left[\begin{array}{l}\{3 x\}=3\{x\}-1 \Rightarrow\{x\}=\frac{4}{9}, \\ \{3 x\}=3\{x\}-2 \Rightarrow\{x\}=\frac{7}{9}\end{array}\right.$.

Since $x=[x]+\{x\}$, the solutions to the original equation are

$$
x_{1}=1+\frac{1}{9}=1 \frac{1}{9}, x_{2}=1+\frac{4}{9}=1 \frac{4}{9}, x_{3}=1+\frac{7}{9}=1 \frac{7}{9}
$$