Solving $4\lfloor x \rfloor=x+\{x\}$. What is wrong in my solution? Can anyone tell? $\lfloor x \rfloor$ = Floor Function, and $\{x\}$ denotes fractional part function

Solve for $x$
$$4\lfloor x \rfloor= x + \{x\}$$

$ x = \lfloor x \rfloor + \{x\}$
$\implies x - \{x\} = \lfloor x \rfloor$
$\implies 4x - 4\{x\} = x + \{x\}$
$\implies 3x = 5\{x\}$
Note that ,
$\implies 0 \leq  \{x\}<1$
$\implies 0 \leq 5\{x\} < 5 \iff 0 \leq 3x <5 \iff\boxed{ 0 \leq x < \dfrac{5}{3}}$
But the answer is $ x = 0$

We have :
$$4 \lfloor x \rfloor = x + \{x\} = \lfloor x \rfloor + 2 \{x\}$$
then :
$$2 \{x\} = 3 \lfloor x \rfloor \in \mathbb{Z}$$
We deduce that :
$$\{x\} = 0 \text{ or } \{x\} = \dfrac{1}{2}$$

*

* If $\{x\} = 0$ then $3 \lfloor x \rfloor = 2 \{x\} = 0$. It means that $\lfloor x \rfloor = 0$ then $x = \lfloor x \rfloor + \{x\} = 0$.

* If $\{x\} = \dfrac{1}{2}$ then $3 \lfloor x \rfloor = 2 \{x\} = 1$.  It means that $\lfloor x \rfloor = \dfrac{1}{3}$ wich is impossible because $\lfloor x \rfloor \in \mathbb{Z}$.

The unique solution is $x = 0$.
