Match List-I with List-II

List-I	List-II
(Inequality)	(Solution Set)
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(A) $2x - 3 < x + 2 \le 3x + 5, x \in \mathbb{R}$	(I) $x \in (-1, \infty)$
(B) $\vert 2x + 3\vert < 7, x \in \mathbb{R}$	(II) $x \in (-\infty, 120]$
(C) $\frac{1}{2}\left(\frac{3}{5}x + 4\right) \ge \frac{1}{3}(x - 6), x \in \mathbb{R}$	(III) $x \in (-5, 2)$
(D) $\frac{\vert x + 1\vert }{x + 1} > 0, x \in \mathbb{R} - \{-1\}$	(IV) $x \in \left[-\frac{3}{2}, 5\right)$

Choose the correct answer from the options given below:

Splitting $2x - 3 < x + 2 \leq 3x + 5$ into two parts:

$2x - 3 < x + 2$

$x < 5$

 

$x + 2 \leq 3x + 5$

$-2x \leq 3$

$x \geq -\dfrac{3}{2}$

Both must hold simultaneously:

$x \in \left[-\dfrac{3}{2},\ 5\right) \Rightarrow$ (IV)

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Using $|A| < k \implies -k < A < k$:

$|2x + 3| < 7$

$-7 < 2x + 3 < 7$

$-10 < 2x < 4$

$-5 < x < 2$

$x \in (-5,\ 2) \Rightarrow$ (III)

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$\dfrac{1}{2}\left(\dfrac{3}{5}x + 4\right) \ge \dfrac{1}{3}(x - 6)$

Clearing fractions by multiplying by $6$:

$3\left(\dfrac{3}{5}x + 4\right) \ge 2(x - 6)$

$\dfrac{9}{5}x + 12 \ge 2x - 12$

Multiplying by $5$:

$9x + 60 \ge 10x - 60$

$120 \ge x$

$x \in (-\infty,\ 120] \Rightarrow$ (II)

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$\dfrac{|x+1|}{x+1} > 0$, where $x \neq -1$

Since $|x+1| > 0$ always, the sign depends entirely on the denominator.

The fraction is positive only when $x + 1 > 0$, i.e. $x > -1$.

$x \in (-1,\ \infty) \Rightarrow$ (I)

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Therefore, the correct answer is (A)-(IV), (B)-(III), (C)-(II), (D)-(I).