Question

Solve:$\frac{2x-3}{4}+6\ge 4+\frac{4x}{3}$

Answer 1

$\frac{2x-3}{4}+6\ge 4+\frac{4x}{3}$

$\Rightarrow \frac{2x-3}{4}+6-6\ge 4+\frac{4x}{3}-6$

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

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

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

$\Rightarrow \frac{3x-8x}{6}\ge \frac{-8+3}{4}$

$\Rightarrow \frac{-5x}{6}\ge \frac{-5}{4}$

$\Rightarrow \frac{x}{3}\le \frac{1}{2}$

$\Rightarrow x\le \frac{3}{2}$

$\therefore x\in (-\infty ,\frac{3}{2}]$