How to solve $20(x-\lfloor x\rfloor)=x+\lfloor x\rfloor+\left\lfloor x+\frac{1}{2}\right\rfloor$ analytically? How to solve this analytically?
$$20(x-\lfloor x\rfloor)=x+\lfloor x\rfloor+\left\lfloor x+\frac{1}{2}\right\rfloor$$
where $\lfloor .\rfloor$ is the floor function.
I attempted to solve this equation numerically for the first 7 numbers, after which no solution exists. However, solving numerically is a pain.
How can I solve this analytically?
My working (an example for the first 3-4 reals )
$$19x-21\lfloor x\rfloor =\left\lfloor x+\frac{1}{2}\right\rfloor$$

*

*When $x$ is between $0$ and $0.5$, there is a single solution at $x=0$.


*When $x$ is between $0.5$ and $1$, we have $19x=1$, which means no solution in the given interval


*Next, solving on $[1,1.5]$, we have $19x-21=1$, which gives us another solution in the given interval.
And so on, until no solutions occur for two or 3 tries, at which point all solutions have been obtained.
I haven't yet found all the elements as it would obviously take forever. (I know there are 7 solutions as I graphed these on Desmos to confirm my idea.)
Any suggestions?
 A: Write $x = \lfloor x \rfloor + (x - \lfloor x \rfloor)$ and then split it into the cases where $0 \leq x - \lfloor x \rfloor < 1/2$ and $1/2 \leq x - \lfloor x \rfloor < 1$. Now the term $\lfloor x + 1/2 \rfloor$ evaluates to $\lfloor x \rfloor$ in the first case and $\lfloor x \rfloor + 1$ in the second.
In both cases, $x$ can be expressed in a linear equation in terms of $\lfloor x \rfloor$; the scaling factor in front of $\lfloor x \rfloor$ ensures there are only a finite number of solutions. Since $\lfloor x \rfloor$ is an integer you can just test all candidates.
A: Denote $\{x\}$ the fractional part of $x$, then
$$\Longleftrightarrow 19\{x\} = 3[x] +\left[\{x\} +\frac{1}{2} \right] $$
We deduce that
$$-1<3[x]<19  \Longleftrightarrow 0\le [x] \le6$$
Case 1: If $\{x\} < \frac{1}{2}$, then
$$19\{x\} = 3[x]  \Longleftrightarrow  [x] < \frac{1}{3}\cdot\frac{19}{2} \Longleftrightarrow [x] \le 3$$
then for $[x] =n \in \{0,1,2,3 \}$, we have
$$ \{x\} = \frac{3n}{19}\Longleftrightarrow x = n +\frac{3n}{19}  \qquad \text{for } n= 0,1,2,3 \tag{1}$$
Case 2: If $\{x\} \ge \frac{1}{2}$, then
$$19\{x\} = 3[x] +1 \Longleftrightarrow  [x] \ge \frac{1}{3}\left(\frac{19}{2}-1\right) \Longleftrightarrow [x] \ge 3$$
then for $[x] =n \in \{3,4,5,6 \}$, we have
$$ \{x\} = \frac{3n+1}{19}\Longleftrightarrow x = n +\frac{3n+1}{19}  \qquad \text{for } n= 3,4,5 \tag{2}$$
Attention: We need to remove the case $n = 6$ as in this case, $\{x\} = \frac{3*6+1}{19} = 1$ that cannot occur ($\{x\}$ must be in $[0,1)$).
From $(1),(2)$, we have 7 solutions
$$x \in \left\{0, \frac{22}{19}, \frac{44}{19}, \frac{66}{19}, \frac{67}{19}, \frac{89}{19}, \frac{111}{19}  \right\}$$
