Question

Solve the inequality for real $x$.

$\frac{(2x-1)}{3}\ge \frac{(3x-2)}{4}-\frac{(2-x)}{5}$

Answer 1

Given, $\frac{(2x-1)}{3}\ge \frac{(3x-2)}{4}-\frac{(2-x)}{5}$

$\Rightarrow \frac{(2x-1)}{3}\ge \frac{5(3x-2)-4(2-x)}{20}$

$\Rightarrow \frac{(2x-1)}{3}\ge \frac{15x-10-8+4x}{20}$

$\Rightarrow \frac{(2x-1)}{3}\ge \frac{19x-18}{20}$

$\Rightarrow 20(2x-1)\ge 3(19x-18)$

$\Rightarrow 40x-20\ge 57x-54$

$\Rightarrow -20+54\ge 57x-40x$

$\Rightarrow 34\ge 17x$

$\Rightarrow x\le 2\text{or}x\in (-\infty ,2]$