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! 
 A: $$\lfloor{\frac{2x-1}{3}}\rfloor + \lfloor{\frac{4x+1}{6}}\rfloor=5x-4.$$
As you have been mentioned; 
$5x-4$ must be an integer, 
so $x=\frac{a}{5},$ 
where $a$ is from $\mathbb Z,$ 
then replace in the original equation, to get
$$\lfloor\frac{4a-10}{30}\rfloor + \lfloor\frac{4a+5}{30}\rfloor=a-4.$$ 

Notice that by Euclid's algorithm
there are integers $n,t \in \mathbb{Z}$; 
such that $a=15n+t$, with $0 \leq t \leq 14$,
by replacing we get that: 
$$ 
\lfloor\frac{60n+4t-10}{30}] + \lfloor\frac{60n+4t+5}{30}]=15n+t-4 
\Longrightarrow 
\\ 
2n+\lfloor\frac{4t-10}{30}] + 2n+\lfloor\frac{4t+5}{30}]=15n+t-4 
\Longrightarrow 
\\ 
\lfloor\frac{4t-10}{30}] + \lfloor\frac{4t+5}{30}] -t +4 =11n 
\ 
; 
\ \ \ \ \ 
\color{Blue}{\star} 
$$ 

but notice that $0 \leq t \leq 14$; implies the following inequalities: 
$$ 
-10 \leq 4t-10 \leq 46 
\ \ 
%%\text{and} 
; 
\ \ \ 
  5 \leq 4t+5  \leq 61 
\ \ 
%%\text{and} 
; 
\ \ \ 
-10 \leq -t+4 \leq 4 
\ \ 
\Longrightarrow 
\\ 
 \color{Red}{-1} \leq \lfloor\frac{4t-10}{30}\rfloor \leq  \color{Red}{1} 
\ \ 
%%\text{and} 
; 
\ \ \ 
  \color{Red}{0} \leq \lfloor\frac{4t+5}{30}\rfloor  \leq  \color{Red}{2} 
\ \ 
%%\text{and} 
; 
\ \ \ 
-10 \leq -t+4 \leq 4 
\ \ 
\Longrightarrow 
\\ 
 -1+0-10 
\leq 
\lfloor\frac{4t-10}{30}\rfloor + 
\lfloor\frac{4t+ 5}{30}\rfloor 
-t +4 
\leq  
1 + 2 + 4
\ \ 
\color{Green}{\star} 
; 
\overset{\color{Blue}{\star}}{\Longrightarrow} 
\\ 
-11 
\leq 
11n 
\leq  
7; 
$$

so we have only two choices for $n$ ; 


*

*$n=-1$, gives: 
$$ -1+0-10 = -11 = 
\lfloor\frac{4t-10}{30}\rfloor + 
\lfloor\frac{4t+ 5}{30}\rfloor 
-t +4 
; 
$$ 
which implies that: 
$$ 
4t-10 < 0 
\ \ 
; 
\ \ \ 
4t +5 < 30 
\ \ 
; 
\ \ \ 
-t+4 = -10 
; 
$$ 
so there is no possibilities for $t$ in this case. 

*$n=0$, gives: 
$$ 0 = 
\lfloor\frac{4t-10}{30}\rfloor + 
\lfloor\frac{4t+ 5}{30}\rfloor 
-t +4 
; 
$$ 
which implies that: 
$$ 
-3=-(\color{Red}{1+2}) 
\leq 
-t+4 
\leq 
-(\color{Red}{-1+0})=1 
; 
$$ 
so the only possibilities for $t$ is: 
$t=3, 4, 5, 6, 7$.
Now you can check by-hand that only $\color{Green}{t=4}$ gives an answer.
A: Let $\lfloor{\frac{2x-1}{3}}\rfloor + \lfloor{\frac{4x+1}{6}}\rfloor=n$, where $n\in\mathbb Z$.
Hence, $5x-4=n$, which gives $x=\frac{n+4}{5}$.
Thus,
$$\lfloor{\frac{2n+13}{15}}\rfloor + \lfloor{\frac{4x+21}{30}}\rfloor=n,\tag{1}$$
which gives
$$n\leq\frac{2n+13}{15}+ \frac{4x+21}{30}<n+2,\tag{2}$$ which gives
$0\leq n\leq2$.
Now, for $n=0$ we get $x=\frac{4}{5}$ and a substitution in the original equation gives $1=1$, 
which says that $1$ is the root.
$n=1$ gives $0=1$, $n=2$ gives $0=2$, 
which says that the answer is
$$\left\{\frac{4}{5}\right\}.$$
A: 
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.
A: Use $\lfloor x \rfloor + \lfloor x + 1/2 \rfloor = \lfloor 2x \rfloor$ identity . So $$\lfloor \frac{4a-10}{15} \rfloor = a-4 $$ Try $a = 0 , 1,2,3, \dots$ The only solution is $a = 4$ .
