Solve for $x$: $x\lfloor x + 2 \rfloor +\lfloor 2x - 2 \rfloor +3x =12$ Solve for $x$: $$x\lfloor x + 2 \rfloor +\lfloor 2x - 2 \rfloor +3x =12$$
My attempt: I have changed this equation into the fractional part function. so that we know $0 \leq \{x\}<1$. I have final equation $x^2 +7x-x\{x+2\}-\{2x-2\}=14$. How to proceed now?
 A: First, with a bit of simplification, we get to $x(\lfloor x \rfloor+5) +\lfloor 2x \rfloor =14$.  
Case: $x> 0$:
Clearly the LHS is an increasing function in $x$, so if there is a solution, it is unique. It is also not hard to quickly narrow down to $x \in (\frac32, 2)$, and then the solution is easy $x = \frac{11}6$.   
Case $x< 0$:
While the function is no longer monotone, it is still piecewise linear, with jump discontinuities when $2x$ is an integer.  We may once again find a suitable interval to narrow the study, but perhaps being more systematic will help. Using $ x-1 < \lfloor x \rfloor \le x$,  we find bounds: $$ x^2 +5x+2(x-1) < x \lfloor x \rfloor + 5x + \lfloor 2x \rfloor < x(x-1)+5x+2x $$
$$ \implies  x^2 +7x -2 < x \lfloor x \rfloor + 5x + \lfloor 2x \rfloor < x^2+6x$$
The bounds imply that any negative root if it exists, $ \in (-\frac{7+\sqrt{113}}2, -3-\sqrt{23}) \approx (-8.8, -7.8)$.  It remains to show by considering the intervals $(-9, -8.5), [-8.5, -8), [-8, -7.5)$, where the function is linear, there is no solution.  In these intervals, the function is $-4x-18, -4x-17, -3x-16$ respectively, and is never $14$.  Hence there is no negative root.
A: Since
$$\lfloor x+2\rfloor=\lfloor x\rfloor+2,\quad \lfloor 2x-2\rfloor=\lfloor 2x\rfloor-2$$
the equation can be written as
$$x(\lfloor x\rfloor+2)+\lfloor 2x\rfloor-2+3x=12,$$
i.e.
$$x\lfloor x\rfloor+5x+\lfloor 2x\rfloor=14\tag 1$$
Now let us separate it into cases :
Case 1 : $x=m+\alpha$ where $m\in\mathbb Z$ and $0\le\alpha\lt \frac 12$.
$$\begin{align}(1)&\Rightarrow (m+\alpha)m+5m+5\alpha+2m=14\\&\Rightarrow (m+5)\alpha=-m^2-7m+14\\&\Rightarrow \alpha=\frac{-m^2-7m+14}{m+5}\qquad (\text{since $m+5\not=0$})\\&\Rightarrow 0\le \frac{-m^2-7m+14}{m+5}\lt \frac 12\end{align}$$
There is no such $m$.
Case 2 : $x=m+\alpha$ where $m\in\mathbb Z$ and $\frac 12\le\alpha \lt 1$.
$$\begin{align}(1)&\Rightarrow (m+\alpha)m+5m+5\alpha+2m+1=14\\&\Rightarrow (m+5)\alpha=-m^2-7m+13\\&\Rightarrow \alpha=\frac{-m^2-7m+13}{m+5}\qquad (\text{since $m+5\not=0$})\\&\Rightarrow \frac 12\le \frac{-m^2-7m+13}{m+5}\lt 1\\&\Rightarrow m=1,\alpha=\frac 56\end{align}$$
Hence, $\color{red}{x=\frac{11}{6}}$ is the only solution.
A: I would let $x=y+z$, with integer $y$ and $0\leq z<1,z=\{x\}$.  Then either
$$(y+z)(y+2)+2y-2+3y+3z=12$$
or 
$$(y+z)(y+2)+2y-1+3y+3z=12$$
So, for example, $z(y+5)$ is an integer, so only some fractions are allowed.
Also, 12 is between $y(y+2)+2y-2+3y$ and $(y+1)(y+2)+2y-2+3y+3$, for the first case, or something similar for the second case; and $y$ is an integer.
