<s>[INST] Find the domain of x $\left \lfloor x \right \rfloor + \left \lfloor x+\frac{1}{2} \right \rfloor + \left \lfloor x-\frac{1}{3} \right \rfloor =8$ Find the domain of x
$$\left  \lfloor x \right \rfloor + \left  \lfloor x+\frac{1}{2}  \right \rfloor   + \left  \lfloor x-\frac{1}{3}  \right \rfloor    =8$$ 
My approach
When x is an integer $x+x+x-1=8$ or x=3
But for case when x is not an integer i am not able to substitute, manually x is $\frac{10}{3}$ which i did by trial method.
My final answer is $3\le x <\frac{10}{3} $
 [/INST] Let $x = k + y$, where $k$ is an integer part of $x$ ($k=\left  \lfloor x \right \rfloor$) and $y$ is its fractional part. Consider $y$-s belonging to the following intervals 
$\left[0, \frac{1}{3}\right)$, $\left[\frac{1}{3}, \frac{1}{2}\right)$ and $\left[\frac{1}{2}, 1\right)$.
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