Question

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

Answer 1

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

$\Rightarrow \frac{2x}{3}-\frac{1}{3}\ge \frac{3x}{4}-\frac{2}{4}-\frac{2}{5}+\frac{x}{5}$

$\Rightarrow \frac{2x}{3}-\frac{3x}{4}-\frac{x}{5}\ge \frac{-2}{4}-\frac{2}{5}+\frac{1}{3}$

$\Rightarrow \frac{40x-45x-12x}{60}\ge \frac{-39-24+20}{60}$

$\Rightarrow \frac{-17x}{60}\ge \frac{-34}{60}$

Multiplying both sides by 60, we have

$-17x\ge -34$

Dividing both sides by -17, we have

$\frac{-17x}{-17}\le \frac{-34}{-17}\Rightarrow x\le 2$

Thus the solution set is $(-\infty ,2]$.