Have fun solving the equation

$\begin{aligned} \frac{\lfloor 2x \rfloor - \lfloor x + 1\rfloor}{x} & = 2 \\ \Rightarrow \color{#3D99F6}{\lfloor 2x \rfloor} - \lfloor x + 1\rfloor & = 2\color{#D61F06}{x} \\ \color{#3D99F6}{2 \lfloor x \rfloor + \lfloor 2\{x\} \rfloor} - \lfloor x \rfloor - 1 & = 2 \color{#D61F06}{\lfloor x \rfloor} + 2 \color{#D61F06}{\{ x \}} \quad \quad \small \color{#3D99F6}{0 \le \{ x \} < 1 \text{ is the fractional part of }x} \\ \lfloor x \rfloor + 1 & = \lfloor 2\{x\} \rfloor - 2 \{ x \} \end{aligned}$

Since LHS is integral, RHS must also be integral and there are only two cases when $2\{x\}$ in the RHS is an integer.

$\begin{aligned} \text{When } \{x\} = 0: \quad \Rightarrow \lfloor x \rfloor + 1 & = 0 \\ \lfloor x \rfloor & = -1 \\ \lfloor x \rfloor + \{x\} & = -1 + 0 \\ \Rightarrow x & = -1 \end{aligned}$

$\begin{aligned} \text{When } \{x\} = \frac{1}{2}: \quad \Rightarrow \lfloor x \rfloor + 1 & = 0 \\ \lfloor x \rfloor & = -1 \\ \lfloor x \rfloor + \{x\} & = -1 + \frac{1}{2} \\ \Rightarrow x & = -\frac{1}{2} \end{aligned}$

Therefore the sum of $x$ satisfying the equation is $-1 - \frac{1}{2} = \boxed{-1.5}$ .

From Hermite's Identity; $\left\lfloor 2x \right\rfloor = \left\lfloor x \right\rfloor + \left\lfloor x + \displaystyle \frac{1}{2} \right\rfloor$ , we get

$\left\lfloor x \right\rfloor + \left\lfloor x + \displaystyle\frac{1}{2} \right\rfloor - \left\lfloor x+1 \right\rfloor = 2x$

From $\left\lfloor x \right\rfloor - \left\lfloor x+1 \right\rfloor = -1$ which is easily proved.

Therefore, $\left\lfloor x + \displaystyle\frac{1}{2} \right\rfloor = 2x+1 = 2\left(x+\displaystyle\frac{1}{2}\right) = 2\left\lfloor x+\displaystyle\frac{1}{2}\right\rfloor + 2\left\{ x + \displaystyle\frac{1}{2}\right\}$ .

Which we get $\left\lfloor x + \displaystyle\frac{1}{2} \right\rfloor = -2\left\{ x + \displaystyle\frac{1}{2}\right\}$ .

Since we get that $-2 < RHS \leq 0$ , therefore, $\left\lfloor x + \displaystyle\frac{1}{2} \right\rfloor = 0,-1$ .

And from $\left\lfloor x + \displaystyle\frac{1}{2} \right\rfloor = 2x+1$ , we know that $\{x\} = 0, \displaystyle\frac{1}{2}$ because $LHS$ is integer.

If $\left\lfloor x + \displaystyle\frac{1}{2} \right\rfloor = 0$ we get $\boxed{x = \displaystyle -\frac{1}{2}}$ .

If $\left\lfloor x + \displaystyle\frac{1}{2} \right\rfloor = -1$ we get $\boxed{x = -1}$ .

My first two steps were similar to Chew-Seong's solution and found that 2x has to be an integer (RHS=LHS).

Then, I also had similar two cases (if 2x is an integer, then it is either odd or even).

Case 1: if 2x is odd, then 2x=2k+1 and x= (2k+1)/2

2k+1-(k+1)=2k+1

k=-1

x=-0.5

Case 2: if 2x is even, 2x=2k and x=k

2k - (k+1) = 2k

k=-1

x=-1

Finally, I determined their sum for the answer: -0.5 + -1 = -1.5.

(And this way, avoided the use of the fractional parts.)

$\begin{array}{c}\text{ }\text{we get that}\quad [2x]-[x+1]=2x\\ \text{let frac(x) denote fractional part of x.we find 2 cases.}\\ \text{case 1:}frac(x)<.5\Longrightarrow [2x]=2[x],[x+1]=[x]+1\\ [2x]-[x+1]=2x\Longrightarrow [x]-1=2x\\ \text{note that [x] is always an integer, which means [x]-1 is an integer and so is 2x. this implies}\\ frac(x)=.5(\text{rejected since it contradiicts frac(x)<.5)} \quad or\quad 0\\ [x]=x\Longrightarrow x-1=2x\Longrightarrow x=-1\\ \text{case 2:} frac(x)\geq .5\Longrightarrow [2x]=2[x]+1,[x+1]=[x]+1\\ 2[x]+1-[x]-1=[x]=2x\\ \text{using same logic as before,:} frac(x)=.5\quad or\quad 0(rejected)\\ [x]=x-.5\Longrightarrow x-.5=2x\Longrightarrow x=-.5\\ -1-.5=\boxed{-1.5}\end{array}$

Anyone with graphical approach...I did by this:(/