How to solve $2x+3[x]-4\{-x\}=4$?

Question

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?

Answer 1

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$$\color{green}{i=1,f=\frac12}.$$

Answer 2

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 $n\in\Bbb Z$.

Answer 3

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 $$6\{x\}=3\implies \{x\}=\frac 12\implies x=\frac 32$$

Answer 4

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, $x = \dfrac{3}{2}$.

Answer 5

$$2x +3 \lfloor x \rfloor - 4 \{-x\} = 4$$

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}