<s>[INST] 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?
 [/INST] 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.
</s>