The solution set of the inequation, $\frac{4 x+5}{7}-6<\frac{5(x+3)}{4}, x \in R$ is:

$\left(-\frac{253}{19}, \infty\right)$

The correct answer is Option (2) → $\left(-\frac{253}{19}, \infty\right)$

$\frac{4x+5}{7} - 6 < \frac{5(x+3)}{4}$

$\frac{4x+5 - 42}{7} < \frac{5x+15}{4}$

$\frac{4x - 37}{7} < \frac{5x+15}{4}$

$4(4x - 37) < 7(5x + 15)$

$16x - 148 < 35x + 105$

$-148 - 105 < 35x - 16x$

$-253 < 19x$

$x > -\frac{253}{19}$

$\text{Solution set} = \left(-\frac{253}{19}, \infty\right)$