Question

Solve for $x$
$\frac{4}{x+1}\le 3\le \frac{6}{x+1},(x>0)$

Answer 1

Given : $\frac{4}{x+1}\le 3\le \frac{6}{x+1}$

$\Rightarrow \frac{4}{x+1}\le 3$ and $3\le \frac{6}{x+1}$

$\Rightarrow 4\le 3(x+1)$ and $3(x+1)\le 6$
$(\because x>0)$

$\Rightarrow 4\le 3x+3$ and $3x+3\le 6$

$\Rightarrow 3x\ge 1$ and $3x\le 3$

$\Rightarrow x\ge \frac{1}{3}$ and $x\le 1$


$\therefore x\in [\frac{1}{3},1]$

Hence, the required solution is

$x\in [\frac{1}{3},1]$