The solution set of the linear inequation $|4x-3|≤\frac{3}{4}$ is

$[\frac{9}{16},\frac{15}{16}]$

The correct answer is Option (3) → $[\frac{9}{16},\frac{15}{16}]$ **

Given:

$|4x - 3| \le \frac{3}{4}$

Write as a double inequality:

$-\frac{3}{4} \le 4x - 3 \le \frac{3}{4}$

Add 3 to all sides:

$3 - \frac{3}{4} \le 4x \le 3 + \frac{3}{4}$

$\frac{12}{4} - \frac{3}{4} \le 4x \le \frac{12}{4} + \frac{3}{4}$

$\frac{9}{4} \le 4x \le \frac{15}{4}$

Divide by 4:

$\frac{9}{16} \le x \le \frac{15}{16}$

Solution set: $\left[\frac{9}{16},\ \frac{15}{16}\right]$