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?

Hint :

$$ \lfloor a \rfloor + \lfloor a + \frac{1}{2} \rfloor = \lfloor 2a \rfloor $$

This follows from Hermite's identity.

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 $-5 <k < 27$.

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.