Solve each of the following system of equations in R.
$\frac{4}{x+1} \leq 3 \leq \frac{6}{x+1}, x>0$
$\frac{4}{x+1} \leq 3 \leq \frac{6}{x+1}, x>0$
$\Rightarrow \frac{4}{x+1} \leq 3$ and $3 \leq \frac{6}{x+1}$
Now,
$\frac{4}{x+1} \leq 3$
$\Rightarrow \frac{4}{x+1}-3 \leq 0$
$\Rightarrow \frac{4-3 x-3}{x+1} \leq 0$
$\Rightarrow \frac{1-3 x}{x+1} \leq 0$
$\Rightarrow \frac{3 x-1}{x+1} \geq 0$
$\Rightarrow x \in(-\infty,-1) \cup\left[\frac{1}{3}, \infty\right)$
Thus, the solution set of the inequation is $(-\infty,-1) \cup\left[\frac{1}{3}, \infty\right)$.
And,
$\frac{6}{x+1} \geq 3$
$\Rightarrow \frac{6}{x+1}-3 \geq 0$
$\Rightarrow \frac{6-3 x-3}{x+1} \geq 0$
$\Rightarrow \frac{3-3 x}{x+1} \geq 0$
$\Rightarrow \frac{3 x-3}{x+1} \leq 0$
$\Rightarrow x \in(-1,1]$
Thus, the solution set of the inequation is $(-1,1]$.
The common values of $x$ in both the inequation is $\left[\frac{1}{3}, 1\right]$.
Hence, the solution set of both the inequation is $\left[\frac{1}{3}, 1\right]$.
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