<s>[INST] 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?
 [/INST] 
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.
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