Thank you for reporting, we will resolve it shortly
Q.
The solution set of the equation $4\{x\}=x+[x]$, where $\{x\}$ and $[ x ]$ denote the fractional and integral parts of a real number $'x'$ respectively, is
Linear Inequalities
Solution:
Let $x=[x]+\{x\}$,
the equation becomes
$4\{x\}=[x]+\{x\}+[x] $
$\Rightarrow 3\{x\}=2[x]$
$\Rightarrow \{x\}=\frac{2}{3}[x] \ldots$ (1)
$\because 0 \leq\{x\} < 1$
$ \Rightarrow 0 \leq \frac{2}{3}[x] < 1$
$ \Rightarrow 0 \leq[x] < \frac{3}{2}$
and $[ x ]$ in integer
$\therefore [ x ]=0$ or 1, from (1)
$\{x\}=0$ or $\frac{2}{3}$
$\therefore x =0+0$ or $1+\frac{2}{3}$
$ \Rightarrow x =0$ or $\frac{5}{3} .$
The solution set is
$x \in\left\{0, \frac{5}{3}\right\}$.