Solve for $x$: $\dfrac{4}{x+1} \le 3 \le \dfrac{6}{x+1}$, $(x 0)$.

Question

Solve for $x$: $\dfrac{4}{x+1} \le 3 \le \dfrac{6}{x+1}$, $(x > 0)$.

Accepted Answer

> [!TLDR]
> Splitting the double inequality (with $x+1>0$ since $x>0$) gives $\dfrac13 \le x \le 1$.

Since $x > 0$, we have $x + 1 > 0$, so multiplying by $x+1$ preserves the inequality directions. Split the compound inequality into two parts.

**Part 1:** $\dfrac{4}{x+1} \le 3$
$$4 \le 3(x+1) \implies 4 \le 3x + 3 \implies 1 \le 3x \implies x \ge 	frac13.$$

**Part 2:** $3 \le \dfrac{6}{x+1}$
$$3(x+1) \le 6 \implies x + 1 \le 2 \implies x \le 1.$$

Taking the intersection of $x \ge 	frac13$ and $x \le 1$ (both consistent with $x>0$):
$$	frac13 \le x \le 1.$$

> [!ANSWER]
> The solution set is $\left[\dfrac13,\ 1ight]$, i.e. $\dfrac13 \le x \le 1$.

Inequality	Meaning	Number line representation
$x > a$	All numbers strictly greater than $a$	Open circle at $a$ , shade right
$x \ge a$	All numbers greater than or equal to $a$	Closed (filled) circle at $a$ , shade right
$x < a$	All numbers strictly less than $a$	Open circle at $a$ , shade left
$x \le a$	All numbers less than or equal to $a$	Closed circle at $a$ , shade left

Q.Solve for $x$ : $\dfrac{4}{x+1} \le 3 \le \dfrac{6}{x+1}$ , $(x > 0)$ .