Question

Solve:$\frac{2x-3}{4}+8\ge 2+\frac{4x}{3}$

Answer 1

$\frac{2x-3}{4}+8\ge 2+\frac{4x}{3}$

$\Rightarrow \frac{2x-3+32}{8}\ge \frac{6+4x}{3}$

$\Rightarrow \frac{2x+29}{8}\ge \frac{6+4x}{3}$

$\Rightarrow 6x+87\ge 48+32x$

$\Rightarrow 6x+87-87\ge 32x+48-87$

$\Rightarrow 6x\ge 32x-39$

$\Rightarrow 6x-32x\ge -39$

$\Rightarrow -26x\ge -39$

$\Rightarrow 26x\le 39$

$\Rightarrow x\le \frac{39}{26}=\frac{3}{2}$

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