Solution of the inequality $\frac{2x+3}{4x-5}≥ 0$ is

$\left(-∞,-\frac{3}{2}\right]\left(\frac{5}{4},∞\right)$

The correct answer is Option (4) → $\left(-∞,-\frac{3}{2}\right]\left(\frac{5}{4},∞\right)$ **

Given inequality:

$\frac{2x+3}{4x-5} \ge 0$

Critical points from numerator and denominator:

$2x+3 = 0 \Rightarrow x = -\frac{3}{2}$

$4x-5 = 0 \Rightarrow x = \frac{5}{4}$ (undefined point)

Check sign intervals:

1. $x < -\frac{3}{2}$ → numerator −, denominator − → fraction + → included.

2. $x = -\frac{3}{2}$ → numerator 0 → fraction = 0 → included.

3. $-\frac{3}{2} < x < \frac{5}{4}$ → numerator +, denominator − → fraction − → excluded.

4. $x > \frac{5}{4}$ → numerator +, denominator + → fraction + → included.

Final solution:

$x \le -\frac{3}{2}$ or $x > \frac{5}{4}$