Solve the equation $10 x-5=9[x]$ in the set of real numbers (where $[x]$ denotes the integer part of $x$).
Solution. First, consider the case when $x$ is an integer. In this case, $[x]=x$, so the equation becomes: $10 x-5=9 x$, from which $x=5$. If $x$ is not an integer, we can break it down into the sum of two numbers: $x=a+b$, where $a=[x]$. Then $b$ will be the fractional part of $x$. Since $x$ is not an integer, $0<b<1$.

In this case, the equation takes the following form: $10(a+b)-5=9 a$. This can be rearranged to $a+10 b=5$. Since $a$ is an integer, $10 b$ must also be an integer for their sum to be an integer. Since $0<b<1$, it follows that $0<10 b<10$, so $10 b \in\{1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 7 ; 8 ; 9\}$.

Let's examine all cases:

| $b$ | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 | 0.7 | 0.8 | 0.9 |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $a=5-10 b$ | 4 | 3 | 2 | 1 | 0 | -1 | -2 | -3 | -4 |
| $x=a+b$ | 4.1 | 3.2 | 2.3 | 1.4 | 0.5 | -0.4 | -1.3 | -2.2 | -3.1 |

Therefore, the solutions to the equation are: $x_{1}=5 ; x_{2}=4.1 ; x_{3}=3.2 ; x_{4}=2.3 ; x_{5}=1.4 ; x_{6}=0.5 ; x_{7}=-0.4 ; x_{8}=-1.3$; $x_{9}=-2.2 ; x_{10}=-3.1$.