The solution set of the inequality $\frac{7x-4}{3}>\frac{9x-3}{7}+\frac{x-5}{9}, x\in R$ is :

$\left(\frac{22}{59}, ∞\right)$

The correct answer is Option (3) → $\left(\frac{22}{59}, ∞\right)$

$\frac{7x-4}{3}>\frac{9x-3}{7}+\frac{x-5}{9}$

$\frac{7x-4}{3}>\frac{9(9x-3)+7(x-5)}{63}$

$\frac{7x-4}{3}>\frac{81x-27+7x-35}{63}$

$\frac{7x-4}{3}>\frac{88x-62}{63}$

$21(7x-4)>88x-62$

$147x-84>88x-62$

$59x>22$

$x>\frac{22}{59}$

$⇒\left(\frac{22}{59}, ∞\right)$