Solve for x, the inequalities in
Question:

Solve for x, the inequalities in

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

Solution:

According to the question,

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

Multiplying each term by (x + 1)

⇒ 4 ≤ 3(x + 1) ≤ 6

⇒ 4 ≤ 3x + 3 ≤ 6

Subtracting each term by 3, we get,

⇒ 1 ≤ 3x ≤ 3

Dividing each term by 3, we get,

⇒ (1/3) ≤ x ≤ 1