<s>[INST] Solve $\lfloor \frac{2x-1}{3} \rfloor + \lfloor \frac{4x+1}{6} \rfloor=5x-4$ I came across this problem and it doesn't seem so tricky, but I didn't do it.

Solve the equation
  $$\lfloor {\frac{2x-1}{3}}\rfloor + \lfloor {\frac{4x+1}{6}}\rfloor=5x-4.$$

My thoughts so far:
Trying to use the inequality $k-1 < \lfloor k\rfloor \leq k$
$5x-4$ is integer, so $x=\frac{a}{5},$ where $a$ is from $\mathbb Z,$ then replace in the original equation, to get
$$\lfloor \frac{4a-10}{30}\rfloor + \lfloor \frac{4a+5}{30}\rfloor =a-4$$
Let $k= \frac{4a-10}{30}$
so $\lfloor k\rfloor +\lfloor k+\frac{1}{2}\rfloor =a-4$
but $\lfloor k\rfloor =\lfloor k+\frac{1}{2}\rfloor $ or $\lfloor k\rfloor +1=\lfloor k+\frac{1}{2}\rfloor $, so we have 2 cases
I strongly think that a shorter solution exists, I would appreciate a hint! 
 [/INST] 
The graph of the function $g(x) = \lfloor\frac{2x-1}{3}\rfloor + \lfloor\frac{4x+1}{6}\rfloor - (5x-4)$
Once we know $0 < x < 1$, we have $\lfloor\frac{2x-1}{3} \rfloor = 0$ and $\lfloor \frac{4x+1}{6} \rfloor = 0$. Thus $x = \frac{4}{5}$ is the only solution.
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