**Question:**

Solve the following systems of linear in equations:

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

**Solution:**

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

When,

$\frac{4}{x+1} \leq 3$

Subtracting 3 from both the sides

$\frac{4}{x+1}-3 \leq 3-3$

$\frac{4-3(x+1)}{x+1} \leq 0$

$\frac{4-3 x-3}{x+1} \leq 0$

$\frac{1-3 x}{x+1} \leq 0$

Signs of $1-3 x$ :

$1-3 x=0 \rightarrow x=\frac{1}{3}$

(Subtract 1 from both the sides and then divide both sides by -3)

$1-3 x>0 \rightarrow x<\frac{1}{3}$

(Subtract 1 from both the sides, then multiply by -1 on both sides and then divide both sides by 3)

$1-3 x<0 \rightarrow x>\frac{1}{3}$

(Subtract 1 from both the sides, then multiply by -1 on both sides and then divide both sides by 3)

Interval satisfying the required condition ≤ 0 , x > 0

$\mathrm{X}=\frac{1}{3}$ or $\mathrm{X}>\frac{1}{3}$

Or

$x \geq \frac{1}{3}$

Now when,

$3 \leq \frac{6}{x+1}$

Subtracting 3 from both the sides

$3-3 \leq \frac{6}{x+1}-3$

$0 \leq \frac{6-3(x+1)}{x+1}$

$0 \leq \frac{6-3 x-3}{x+1}$

$0 \leq \frac{3-3 x}{x+1}$

Dividing both sides by 3

$0 \leq \frac{1-x}{x+1}$

Multiplying by -1 on both sides

$0 \geq \frac{x-1}{x+1}$

Signs of $x-1$ :

$x-1=0 \rightarrow x=1$ (Adding 1 to both the sides)

$x-1>0 \rightarrow x>1$ (Adding 1 to both the sides)

$x-1<0 \rightarrow x<1$ (Adding 1 to both the sides)

Interval satisfying the required condition: ≤ 0

x ≤ 1

Combining the intervals:

$\frac{1}{3} \leq x<1$ such that $x>0$