$\lfloor x \rfloor$ = 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$