Question

Solve the following system of equations in R.

$\frac{4}{x+1}\le 3\le \frac{6}{x+1},x>0$

Answer 1

Consider the first inequation,

$\frac{4}{x+1}\le 3\Rightarrow 4\le 3(x+1)\Rightarrow 4\le 3x+3\Rightarrow 4-3\le 3x\left[\text{subtracting 3 on both sides}\right]\Rightarrow 1\le 3x\therefore x\ge \frac{1}{3}...\left(i\right)$

and consider last two inequalities,

$3\le \frac{6}{x+1}\Rightarrow 3(x+1)\le 6\Rightarrow 3x+6\le 6\Rightarrow 3x\le 6-3\left[\text{subtracting 3 on both sides}\right]\Rightarrow 3x\le 3\therefore x\le 1\therefore Fromeqs.\left(i\right)and\left(ii\right)xϵ[\frac{1}{3},1]\frac{1}{3}\le x\le 1$