<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] 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.
$\begin{array}{|l|ccccccccc|}\hline
r & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\\hline
p & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\\hline
n & 9 & 10 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\\hline
x & \frac{19}3 & \frac{21}3 & \frac{5}3 & \frac{7}3 & \frac{9}3 & \frac{11}3 & \frac{13}3 & \frac{15}3 & \frac{17}3\\\hline
\end{array}$
</s>