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
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