How to solve $2x+3[x]-4\{-x\}=4$? I am beginner to greatest integer and fractional part functions, and I just came across this problem:
$$ 2x+3[x]-4\{-x\}=4,$$
where $[x]$ and $\{x\}$ are the greatest integer and the fractional part functions, respectively.
What I have tried is taken $[x] = x $ as $ x-1≤[x]≤x$ but no ideas for fractional part of $x$. However, I know that $ \{x\} = x-[x]$.
How do I proceed further?
 A: There are no integral solutions (clearly).
For non-integers $\{-x\}=1-\{x\}$.  Thus your expression can be rewritten as $$5\lfloor x \rfloor +6\{x\}=8$$
$\lfloor x \rfloor\in \mathbb Z$ then implies that $\{x\}=\frac a6$ for $a\in \{1,2,3,4,5\}$ (note:  here we do not assume that $\gcd (a,6)=1$).  Since $0<6\{x\}<6$ we see that $\lfloor x \rfloor =1$ and the problem now comes down to $$6\{x\}=3\implies \{x\}=\frac 12\implies x=\frac 32$$
A: By inspection, $x$ cannot be integer. Write $x:=i+f$ where $i$ is integer and $0<f<1$. The equation becomes
$$2(i+f)+3i-4(1-f)=4$$
or
$$5i+6f=8.$$
Eliminating $f$,
$$0<8-5i<6$$ and there is a single solution$$\color{green}{i=1,f=\frac12}.$$
A: Hint:
Your equation implies that $2x-4\{-x\}$ is an integer.
Use this, together with the equalities
$$x=[x]+\{x\}$$
and
$$\{-x\}=1-\{x\}$$
to show that $6\{x\}\in \Bbb Z$, and hence that $x=\frac n6$ for some $n\in\Bbb Z$.
A: If $x \in \mathbb{Z}$, then$$
4 = 2x + 3[x] - 4\{-x\} = 2x + 3x - 0 = 5x \Longrightarrow x = \frac{4}{5},
$$
a contradiction. Denote $a = [x]$, $b = \{x\}$, then $0 < b < 1$ and $\{-x\} = 1 - b$. Thus,$$
4 = 2x + 3[x] - 4\{-x\} = 2(a + b) + 3a - 4(1 - b) = 5a + 6b - 4,
$$
which implies $5a + 6b = 8$. Because $5a < 5a + 6b < 5a + 6$, thus $2 < 5a < 8$, which implies $a = 1$, and further $b = \dfrac{1}{2}$. Therefore, $x = \dfrac{3}{2}$.
A: $$2x  +3 \lfloor x \rfloor - 4 \{-x\} = 4$$
Let $x = n+\alpha$ where $n$ is an integer and $0 \le \alpha < 1$.
Then ${-x} =
\begin{cases}
          0 & \text{If $\alpha = 0$.} \\
   1-\alpha & \text{If $\alpha \ne 0$.}
\end{cases}$
So, when $\alpha = 0$
\begin{align}
   2x  +3 \lfloor x \rfloor - 4 \{-x\} &= 4 \\
   2n  +3n - 4(0) &= 4 \\
   5n &= 4
\end{align}
Which has no integer solution.
When $0 < \alpha < 1$, 
\begin{align}
   2x  +3 \lfloor x \rfloor - 4 \{-x\} &= 4 \\
   2(n + \alpha)  +3n - 4(1-\alpha) &= 4 \\
   5n+6\alpha &=8
\end{align}
The only possible solution is $n=1$ and $\alpha = \dfrac 12$
So $x=1 \frac 12$
Check:
\begin{align}
   \left \lfloor 1 \frac 12 \right \rfloor &= 1 \\
  \left\{ -1 \frac 12 \right\} &= \frac 12 \\
\end{align}
So
\begin{align}
   2x  +3 \lfloor x \rfloor - 4 \{-x\} 
   &= 2\left( 1 \frac 12 \right)   
      + 3 \left \lfloor 1 \frac 12 \right \rfloor 
      - 4 \left\{ -1 \frac 12 \right\} \\
   &= 3 + 3(1) - 4\left( \frac 12 \right) \\
   &= 3 + 3 - 2 \\
   &= 4
\end{align}
