<s>[INST] Solve, $[3x + 1] = 2x - \frac{1}{2}$ , and find the sum of all roots. 
Solve, $[3x + 1] = 2x - \frac{1}{2}$ , and find the sum of all roots.

What I Tried: I have :-
$$\rightarrow [3x + 1] = \frac{4x - 1}{2}$$
Now, the LHS is an integer, so $2|(4x-1)$ .
I can see that if $x$ is an integer, then $(4x - 1)$ is odd, so there are no integer solutions. I am not sure about fractions, as $x = \frac{1}{2}$ still makes it odd, but $x = \frac{1}{4}$ makes it even, although $2$ does not divided $0$.
This is where I am stuck, how should I proceed next?
Can anyone help me?
 [/INST] I'm leaving this here, but it's wrong. I added a correct answer below.
\begin{align}
   \lfloor 3x + 1 \rfloor &= 2x - \frac{1}{2} \\
   \lfloor 3x \rfloor + 1 &= 2x - \frac{1}{2} \\
   \lfloor 3x \rfloor &= 2x - \frac 32
\end{align}
So, for some integer, $n$,
\begin{array}{ccccc}
                       &&\lfloor 3x \rfloor = n \\
\hline
      n           &\le  &3x  &< &n+1 \\
      \frac n3    &\le  &x   &< &\frac n3 + \frac 13 \\
\hline
      \frac {2n}3            &\le  &2x                   &<   &\frac {2n}3 + \frac 23 \\
      \frac {2n}3 - \frac 32 &\le  &2x - \frac 32        &<   &\frac {2n}3 - \frac 56 \\
      \frac {2n}3 - \frac 32 &\le  &n                    &<   &\frac {2n}3 - \frac 56 \\
      -\frac 32              &\le  &\frac n3             &<   &-\frac 56 \\
      -\frac 92              &\le  & n                   &<   &-\frac 52 \\
      -4                     &\le  & n                   &\le &-3 \\
\hline
      -4                     &\le  & \lfloor 3x \rfloor  &\le &-3 \\
      -4                     &\le  & 2x - \frac 12       &\le &-3
\end{array}
Hence $x \in \left\{ -\dfrac 74, -\dfrac 54 \right\}$
The sum of the roots is $-3$.
This is a correct answer.
Let $x = n + \epsilon$ where $n \in \mathbb R$ and $\epsilon \in [0,1)$.
\begin{align}
   \lfloor 3x \rfloor &= 2x - \frac 32 \\
   3n + \lfloor 3\epsilon \rfloor &= 2n + 2\epsilon - \frac 32 \\
   6n + 2\lfloor 3\epsilon \rfloor &= 4n + 4\epsilon - 3 \\
   2n + 2\lfloor 3\epsilon \rfloor + 3 &= 4\epsilon \\
\end{align}
Since the LHS is an integer, then so too must $4\epsilon$ be an integer.
It follows that $4\epsilon \in \{0,1,2,3\}$
If $\epsilon = 0$.
\begin{align}
   2n + 2\lfloor 3\epsilon \rfloor + 3 &= 4\epsilon \\
   2n + 3 &= 0 \\
   &\text{No solution.}
\end{align}
If $\epsilon =  \dfrac 14$.
\begin{align}
   2n + 2\lfloor 3\epsilon \rfloor + 3 &= 4\epsilon \\
   2n + 3 &= 1 \\
   n &= -1 \\
   x &= -\dfrac 34
\end{align}
If $\epsilon =  \dfrac 12$.
\begin{align}
   2n + 2\lfloor 3\epsilon \rfloor + 3 &= 4\epsilon \\
   2n + 5 &= 2 \\
   &\text{No solution.}
\end{align}
If $\epsilon =  \dfrac 34$.
\begin{align}
   2n + 2\lfloor 3\epsilon \rfloor + 3 &= 4\epsilon \\
   2n + 7 &= 3 \\
   n &= -2 \\
   x &= -\dfrac 54
\end{align}
</s>