Number of real solutions of $\begin{array}{r} {\left[\frac{2 x+1}{3}\right]+\left[\frac{4 x+5}{6}\right]} =\frac{3 x-1}{2} \end{array}$ Solve for $x \in \mathbb{R}$ $$\begin{array}{r}
{\left[\frac{2 x+1}{3}\right]+\left[\frac{4 x+5}{6}\right]} 
=\frac{3 x-1}{2}
\end{array}$$ where $[x]$ denotes greatest integer less than or equal to $x$.
My try:
Letting $a=\frac{2x+1}{3}$ we get
$$\left[a\right]+\left[a+\frac{1}{2}\right]=\frac{9 a-5}{4} \tag{1}$$
Now knowing that:
$$a-1<[a]\leq a$$ and $$a-\frac{1}{2}<\left[a+\frac{1}{2}\right]\leq a+\frac{1}{2}$$
Adding both the above inequalities and using $(1)$ we get
$$2a-\frac{3}{2}<\frac{9a-5}{4}\leq 2a+\frac{1}{2}$$
So we get $$a \in (-1, 7]$$
Any help here?
 A: Hint :
$$ \lfloor a \rfloor + \lfloor a + \frac{1}{2} \rfloor = \lfloor 2a \rfloor $$
This follows from Hermite's identity.
A: HINT: The RHS $\frac{3x-1}{2}$ of the original equation must be an integer, which implies that $3x$ has to be an odd integer [typo noted via comment below]. Your bounds for $a$ leave only a few values to check for $x$, in particular $-2<x<9$;  this an $3x$ an odd integer leaves only to check each $x$ satisfying $x= \frac{k}{3}$;  $k$ is an odd integer satisfying $-5 <k < 27$.
A: 
Letting $\displaystyle a=\frac{2x+1}{3}$ we get
$$\left[a\right]+\left[a+\frac{1}{2}\right]=\frac{9 a-5}{4} \tag{1}$$

The solution will involve identifying all satisfying values of $a$, and then translating these values into all satisfying values of $x$ via $\displaystyle a=\frac{2x+1}{3}$.
The problem may be attacked analytically, from scratch.
Let $a$ be represented by $P + r ~: P \in \Bbb{Z}, ~0 \leq r < 1.$ 
This implies that $[a] = P$.
From (1) above, the from scratch procedure is to consider the mutually exclusive cases:

*

*$\displaystyle 0 \leq r < \frac{1}{2}.$


*$\displaystyle \frac{1}{2} \leq r < 1.$

$\underline{\textbf{Case 1:} ~\displaystyle 0 \leq r < \frac{1}{2}}$
The LHS of (1) above is $2P.$
Therefore $\displaystyle 2P = \frac{9[P + r] - 5}{4} \implies $ 
$8P = 9P + 9r - 5 \implies $
$$ 5 = P + 9r.\tag2 $$
Combining the Case 1 constraint with (2) above, you have that $(P,r)$ must be an element in
$$\left\{(5,0), \left(4,\frac{1}{9}\right), \left(3,\frac{2}{9}\right), \left(2,\frac{3}{9}\right), \left(1 + \frac{4}{9}\right)\right\}.\tag3 $$
This means that the Case 1 candidate values for $a$ are
$$\left\{(5), \left(4 + \frac{1}{9}\right), \left(3 + \frac{2}{9}\right), \left(2 + \frac{3}{9}\right), \left(1 + \frac{4}{9}\right)\right\}.\tag4 $$
In fact, all five of the candidate values for $(a)$ shown in (4) above satisfy (1) above.

$\underline{\textbf{Case 2:} ~\displaystyle \frac{1}{2} \leq r < 1}$
The LHS of (1) above is $2P + 1.$
Therefore $\displaystyle (2P + 1) = \frac{9[P + r] - 5}{4} \implies $ 
$8P + 4 = 9P + 9r - 5 \implies $
$$ 9 = P + 9r.\tag5 $$
Combining the Case 2 constraint with (5) above, you have that $(P,r)$ must be an element in
$$\left\{\left(4,\frac{5}{9}\right), \left(3,\frac{6}{9}\right), \left(2,\frac{7}{9}\right), \left(1 + \frac{8}{9}\right)\right\}.\tag6 $$
This means that the Case 2 candidate values for $a$ are
$$\left\{\left(4 + \frac{5}{9}\right), \left(3 + \frac{6}{9}\right), \left(2 + \frac{7}{9}\right), \left(1 + \frac{8}{9}\right)\right\}.\tag7 $$
In fact, all four of the candidate values for $(a)$ shown in (7) above satisfy (1) above.

This leads to the following chart of solutions (in terms of $x$):
\begin{array}{| r | r |}
  \hline                       
  a & x = \displaystyle \frac{3a - 1}{2} \\
  \hline                       
  5 & 7 \\
  \hline                       
  \left(4 + \frac{1}{9}\right) & \left(5 + \frac{2}{3}\right) \\
  \hline      
  \left(3 + \frac{2}{9}\right) & \left(4 + \frac{1}{3}\right) \\
  \hline      
  \left(2 + \frac{3}{9}\right) & \left(3\right) \\
  \hline      
  \left(1 + \frac{4}{9}\right) & \left(1 + \frac{2}{3}\right) \\
  \hline 
  \left(4 + \frac{5}{9}\right) & \left(6 + \frac{1}{3}\right) \\
  \hline      
  \left(3 + \frac{6}{9}\right) & \left(5\right) \\
  \hline      
  \left(2 + \frac{7}{9}\right) & \left(3 + \frac{2}{3}\right) \\
  \hline      
  \left(1 + \frac{8}{9}\right) & \left(2 + \frac{1}{3}\right) \\
  \hline           
\end{array}
Edit
Actually, there is a shortcut.  Because the relationship between $a$ and $x$ is linear, each distinct value of $a$ gives a distinct value in $x$.
Further, the satisfying solutions for $a$ are all of the form $\displaystyle P + \frac{s}{9}$, where $s$ runs through each of the elements in $\{0,1,2,\cdots, 8\}$ exactly once.
Therefore, if all that is desired is a count of the distinct satisfying values of $x$, you can conclude, looking only at the fractional part of the confirmed Case 1 and Case 2 satisfying values for $a$, that there are exactly $9$ solutions.
A: Thanks for hermites identity that @mymolecules showed...
$\left[\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{3}}\right]+\left[\frac{\mathrm{4}{x}+\mathrm{5}}{\mathrm{6}}\right]=\frac{\mathrm{3}{x}−\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\: \\ $
$\left[\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{3}}\right]+\left[\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}\right]=\frac{\mathrm{3}{x}−\mathrm{1}}{\mathrm{2}} \\ $
$\left[{p}\right]=\left[\frac{\mathrm{4}{x}+\mathrm{2}}{\mathrm{3}}\right]=\frac{\mathrm{3}{x}−\mathrm{1}}{\mathrm{2}}={n}\in{Z}\:\:\:\:\:\:\:\:\: \\ $
${x}=\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{3}}\Rightarrow{p}=\frac{\mathrm{8}{n}+\mathrm{10}}{\mathrm{9}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $
$\frac{\mathrm{8}{n}+\mathrm{10}}{\mathrm{9}}−{n}=\frac{\mathrm{10}−{n}}{\mathrm{9}}=\left\{{p}\right\}\in\left[\mathrm{0},\mathrm{1}\right)\:\:\: \\ $
${n}\in\left\{\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9},\mathrm{10}\right\}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $
${x}\in\left\{\frac{\mathrm{5}}{\mathrm{3}},\frac{\mathrm{7}}{\mathrm{3}},\frac{\mathrm{9}}{\mathrm{3}},\frac{\mathrm{11}}{\mathrm{3}},\frac{\mathrm{13}}{\mathrm{3}},\frac{\mathrm{15}}{\mathrm{3}},\frac{\mathrm{17}}{\mathrm{3}},\frac{\mathrm{19}}{\mathrm{3}},\frac{\mathrm{21}}{\mathrm{3}}\right\} \\ $
A: The LHS of this equation is an integer $\,n=\overbrace{\left\lfloor\dfrac{2x+1}3\right\rfloor}^\text{integer}+\overbrace{\left\lfloor\dfrac{4x+5}6\right\rfloor}^\text{integer}=\dfrac{3x-1}2$
We can express $x=\dfrac{2n+1}{3}$ and report in the equation to get $\dfrac 49n+\cdots$ inside the floor values.
This motivates us to set $\,n=9p+r\,$ with $p\in\mathbb Z$ and $r\in\{0,1,2,3,4,5,6,7,8\}$.
Reporting in the equation gives $\quad 9p+r=\left\lfloor 4p+\frac{4r+5}9\right\rfloor+\left\lfloor 4p+\frac{8r+19}{18}\right\rfloor$
And we can get the integer $4p$ out of the floor() function, after simplification we get:
$$p=\left\lfloor \frac{4r+5}9\right\rfloor+\left\lfloor \frac{8r+19}{18}\right\rfloor-r$$
Pluging the $9$ possible values for $r$ gives you as many solutions.
$\begin{array}{|l|ccccccccc|}\hline
r & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\\hline
p & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\\hline
n & 9 & 10 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\\hline
x & \frac{19}3 & \frac{21}3 & \frac{5}3 & \frac{7}3 & \frac{9}3 & \frac{11}3 & \frac{13}3 & \frac{15}3 & \frac{17}3\\\hline
\end{array}$
