New idea to solve this equation I was teaching $\left \lfloor x  \right \rfloor$ function properties and equation . I solved this equation in my class . My works are show below. Some students ask me for new Idea...,and now I am looking for various method to solve this (like this)  equation .$$\left \lfloor  x\right \rfloor+2x=1$$
1st: $$x=n+p \\n \in\mathbb{Z} , 0 \leq p<1 \to \left \lfloor x \right \rfloor=n ,p=x-n\\  $$ 
and
 $$ \left \lfloor  x\right \rfloor+2x=1\\n+2(n+p)=1 \to 3n+2p=1 \\3n=0 ,\pm3 ,\pm 6,\pm9,... $$in this case $$3n=0 \to 2p=1 \\n=0 , p=\frac{1}{2}  \to x=n+p=0+\frac{1}{2} $$
2nd:
$$\left \lfloor  x\right \rfloor+2x=1 \to \left \lfloor  x\right \rfloor=1-2x$$ like $f(x)=g(x)$ by drawing both of them obtain the answer

With respect to the picture ,it suffice to solve $0=1-2x$s o the answer is $x=\frac{1}{2}$
3rd :
we know $$\left \lfloor  x\right \rfloor =k \in \mathbb{Z} \to k \leq x <k+1 $$ so $$\left \lfloor  x\right \rfloor=1-2x \to 1-2x=k \in  \mathbb{Z}\\x=\frac{1-k}{2} \to \left \lfloor  \frac{1-k}{2}\right \rfloor =k$$ so we have 
$$k \leq \frac{1-k}{2} <k+1 \to \\\left\{\begin{matrix}
 k\leq \frac{1-k}{2} \to &  2k \leq 1-k \to & k \leq \frac{1}{3} \to k=\left \{ 0,-1,-2,-3,... \right \}\\ 
 \frac{1-k}{2}<k \to &  1-k<2k \to  &k> -\frac{1}{3} \to k=\left \{ 0,1,2,3,... \right \}
\end{matrix}\right.\\
\left \{ ...,-3,-2,-1,0 \right \}\bigcap \left \{ 0,1,2,3,... \right \}=\left \{ 0 \right \}\rightarrow  k=0  \\\rightarrow  x=\frac{1-k}{2}=\frac{1}{2}$$
 A: Note that the function $f(x) =\lfloor x \rfloor + 2x$ is strictly increasing. 
It is always between $3x-1$ and $3x$. 
Solving $3x-1=1$ and $3x=1$ gives that $x$ is between $\frac13$ and $\frac23$, because $f$ is strictly increasing.
So $\lfloor x \rfloor = 0$, since $\frac13 < x < \frac23$. So we solve $2x=1$ so $x=\frac12$. 
A: Rewrite the given equation as 
$$\lfloor x \rfloor = 1-2x$$
Note that $1-2x$ must be an integer. So $x=\frac{k+1}{2}$. Now simplification is much easier by considering two cases (1)$k$ is even (2)$k$ is odd.
OR alternatively 
you can do as follows for $x > 1/2$ the right side is negative but left side is non-negative so no equality. Likewise for $x < 1/2$ the situation reverses so still no equality. The only value left is when both sides are $0$.
A: $x-1<\lfloor x\rfloor \leq x$, so $3x-1<\lfloor x\rfloor +2x=1\leq 3x$, this implies $x\in [1/3, 2/3)$, so $\lfloor x\rfloor=0$, then $2x=1$, $x=1/2$.
A: Note that for any integer $n$, $\;\lfloor x+n\rfloor=\lfloor x\rfloor+n $,and that $$\lfloor 2x\rfloor=\begin{cases}2\lfloor x\rfloor&\text{if}\enspace 0\le x-\lfloor x\rfloor<\dfrac12,\\2\lfloor x\rfloor+1&\text{if}\enspace \dfrac12\le x-\lfloor x\rfloor<1.\end{cases} $$
The given equation implies $\;1=\bigl\lfloor\lfloor x\rfloor+ 2x\bigr\rfloor=\lfloor x\rfloor+\lfloor 2x\rfloor=\begin{cases}3\lfloor x\rfloor&\text{if}\enspace 0\le x-\lfloor x\rfloor<\dfrac12,\\3\lfloor x\rfloor+1 &\text{if}\enspace\dfrac12\le x-\lfloor x\rfloor<1 \end{cases}.$
The first case cannot happen, so necessarily $\lfloor x\rfloor=0\;$  and $\;\dfrac12\le x-\lfloor x\rfloor<1$. The equation becomes $\;2x=1$, whence 
$$x=\smash{\frac12}.$$
A: Write $x=\lfloor x\rfloor+\{x\}$. Then we have to solve
$$3\lfloor x\rfloor+2\{x\}=1\ .$$
There is no solution with $\lfloor x\rfloor<0$ or $\lfloor x\rfloor\geq1$. When $\lfloor x\rfloor=0$ we need $2\{x\}=1$, which implies $x={1\over2}$. It is easy to check that this is indeed a solution.
