Question

Solve the given inequality for real$x: \frac{x}{4} < \frac{(5x-2)}{3} -\frac{ (7x-3)}{5}$.

Answer 1

$\frac{x}{4} < \frac{(5x-2)}{3} -\frac{ (7x-3)}{5}$
$⇒ \frac{x}{4} < \frac{5(5x-2-2(7x-3))}{15}$
$⇒ \frac{x}{4} < \frac{25x-10-21x+9}{15}$
$⇒\frac{ x}{4} <\frac{ 4x-1}{15}$
$⇒ 15x < 4(4x - 1)$
$⇒ 15x < 16x - 4$
$⇒ 4 < 16x - 15x$
$⇒ 4 < x$
Thus, all real numbers x, which are greater than 4, are the solutions of the given inequality.
Hence, the solution set of the given inequality is (4, ∞).