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?
 A: As $2x-\dfrac{1}{2}$ is an integer, $x$ is either $n+\dfrac{1}{4}$ or $n+\dfrac{3}{4}$, $n \in \mathbb{Z}$. (Recall any real $x$ can be written as sum of an integer and a positive fraction).
Substitute this in original equation to solve for $n$.

$$n=-1 \Rightarrow x=-1+\dfrac{1}{4}$$
$$n=-2 \Rightarrow x=-2+\dfrac{3}{4}$$

A: $y=3x+1 \implies [y] = (4y-7)/6$.
Let $[y]=n, \{y\}=r$ then $n=(4n+4r-7)/6, 2n=4r-7$.
Since $0\le r < 1$, $0 \le 4r < 4$, we must have $4r=1,3$, so either $r=\frac 14, n=-3$ or $r=\frac 34, n=-2$.
Finally, $3x+1 = -\frac{11}{4}$ or $-\frac 54$, $x=-\frac 54$ or $-\frac 34$.
(You don't need $y$ but it simplifies things a bit.)
A: Let us find  a necessary condition.
$$\lfloor 3x+1\rfloor=2x-\frac 12 \implies$$
$$\lfloor 3x\rfloor=2x-\frac 32 \implies $$
$$3x-\lfloor 3x\rfloor =x+\frac 32\implies$$
$$0\le x+\frac 32<1 \implies$$
$$-\frac 32\le x<-\frac 12\implies $$
$$-\frac 72\le 2x-\frac 12<-\frac 32\implies$$
$$2x-\frac 12\in \{-3,-2\}\implies$$
$$x\in\{-\frac 54,-\frac 34\}$$
We check this condition is sufficient.
the sum of the roots is
$$S=-\frac 54-\frac 34=-2$$
A: 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}
