Question

Solve that: $\frac{1}{3}(2x-1)\ge \frac{1}{4}(3x-2)-\frac{1}{5}(2-x)$ for $xϵN$.

Answer 1

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

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

$⟹\frac{3x}{4}+\frac{x}{5}-\frac{2x}{3}\le \frac{2}{4}+\frac{2}{5}-\frac{1}{3}$ (on rearranging)

$⟹\frac{45x+12x-40x}{60}\le \frac{30+24-20}{60}$

$⟹17x\le 34$

$⟹x\le 2$

This is the required solution.