Question

let {x} and [x] denote the fractional and integral part of real number x respectively. Solve 4 {x} = x + [x] . Three times the non integral solution is

Answer 1

$4\left\{x\right\}=x+\left[x\right]$
$\Rightarrow 4(x-[x\left]\right)=x+\left[x\right]$
$\Rightarrow 5\left[x\right]-3x=0$ ....(1)
Case I : If $x\in I$
$\left[x\right]=x$
Using this in (1),
$\Rightarrow 2x=0$
$\Rightarrow x=0$
Case II: When $x\in I$
$\left[x\right]=n;$
Using this in (1), we get
$\Rightarrow 3x=5n$
$\Rightarrow x=\frac{5n}{3};n\in I$
For $n=1$
$x=\frac{5}{3}$
So, three times non-integral solution is 5