Have You Tried Adding Floors?

Since the LHS is an integer, the RHS must also be an integer $n$ , $\Rightarrow \dfrac{3x-1}{2} = n$ and $x = \dfrac{2n+1}{3}$ . And that:

$\begin{aligned} \left \lfloor \frac{2x+1}{3} \right \rfloor + \left \lfloor \frac{4x+5}{6} \right \rfloor & = \frac{3x-1}{2} \\ \left \lfloor \frac{2\left(\frac{2n+1}{3}\right)+1}{3} \right \rfloor + \left \lfloor \frac{4\left(\frac{2n+1}{3}\right)+5}{6} \right \rfloor & = n \\ \left \lfloor \frac{4n+5}{9} \right \rfloor + \left \lfloor \frac{8n+19}{18} \right \rfloor & = n \\ \color{#3D99F6}{\left \lfloor \frac{4n+5}{9} \right \rfloor} + \color{#D61F06}{\left \lfloor \frac{8n+1}{18} \right \rfloor} + 1 & = n \end{aligned}$

We note that the LHS increases with $n$ and LHS $>$ RHS for $n \le 1$ and LHS $<$ RHS for $n \ge 11$ . And when:

$\begin{cases} n = 2, & \Rightarrow \color{#3D99F6}{1} + \color{#D61F06}{0} + 1 = 2, & \text{acceptable} \\ n = 3, & \Rightarrow \color{#3D99F6}{1} + \color{#D61F06}{1} + 1 = 3, & \text{acceptable} \\ n = 4, & \Rightarrow \color{#3D99F6}{2} + \color{#D61F06}{1} + 1 = 4, & \text{acceptable} \\ n = 5, & \Rightarrow \color{#3D99F6}{2} + \color{#D61F06}{2} + 1 = 5, & \text{acceptable} \\ n = 6, & \Rightarrow \color{#3D99F6}{3} + \color{#D61F06}{2} + 1 = 6, & \text{acceptable} \\ n = 7, & \Rightarrow \color{#3D99F6}{3} + \color{#D61F06}{3} + 1 = 7, & \text{acceptable} \\ n = 8, & \Rightarrow \color{#3D99F6}{4} + \color{#D61F06}{3} + 1 = 8, & \text{acceptable} \\ n = 9, & \Rightarrow \color{#3D99F6}{4} + \color{#D61F06}{4} + 1 = 9, & \text{acceptable} \\ n = 10, & \Rightarrow \color{#3D99F6}{5} + \color{#D61F06}{4} + 1 = 10, & \text{acceptable} \\ \end{cases}$

There are $\boxed{9}$ real values of $x$ satisfying the equation.