Question

Solve each of the following system of equations in R.
$\frac{4}{x+1}\le 3\le \frac{6}{x+1},x>0$

Answer 1

$$\begin{gathered} \frac{4}{x+1}\le 3\le \frac{6}{x+1},x>0 \\ \Rightarrow \frac{4}{x+1}\le 3\mathrm{and}3\le \frac{6}{x+1} \\ \mathrm{Now}, \\ \frac{4}{x+1}\le 3 \\ \Rightarrow \frac{4}{x+1}-3\le 0 \\ \Rightarrow \frac{4-3x-3}{x+1}\le 0 \\ \Rightarrow \frac{1-3x}{x+1}\le 0 \\ \Rightarrow \frac{3x-1}{x+1}\ge 0 \\ \Rightarrow x\in \left(-\infty ,-1\right)\cup [\frac{1}{3},\infty ) \end{gathered}$$



Thus, the solution set of the inequation is $\left(-\infty ,-1\right)\cup [\frac{1}{3},\infty )$.

$$\begin{gathered} \mathrm{And} \\ \frac{6}{x+1}\ge 3 \\ \Rightarrow \frac{6}{x+1}-3\ge 0 \\ \Rightarrow \frac{6-3x-3}{x+1}\ge 0 \\ \Rightarrow \frac{3-3x}{x+1}\ge 0 \\ \Rightarrow \frac{3x-3}{x+1}\le 0 \\ \Rightarrow x\in (-1,1] \end{gathered}$$



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 $\left[\frac{1}{3},1\right]$.