The solution set of the inequality $\frac{7x-4}{3}>\frac{9x-3}{7}+\frac{x-5}{9}, x\in R$ is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\left(\frac{22}{59}, ∞\right)$
$\frac{7x-4}{3}>\frac{9x-3}{7}+\frac{x-5}{9}$
$\frac{7x-4}{3}>\frac{9(9x-3)+7(x-5)}{63}$
$\frac{7x-4}{3}>\frac{81x-27+7x-35}{63}$
$\frac{7x-4}{3}>\frac{88x-62}{63}$
$21(7x-4)>88x-62$
$147x-84>88x-62$
$59x>22$
$x>\frac{22}{59}$
$⇒\left(\frac{22}{59}, ∞\right)$