Question

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

Answer 1

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

$\Rightarrow \frac{2x-3}{4}+8-2-\frac{4x}{3}\ge 0$

$\Rightarrow \frac{2x-3}{4}-\frac{4x}{3}+6\ge 0$

$\Rightarrow \frac{6x-9-16x}{4}+6\ge 0$

$\Rightarrow \frac{-10x-9}{4}\ge -6$

$\Rightarrow \frac{10x+9}{4}\le 6$

$\Rightarrow 10x+9\le 24$

$\Rightarrow 10x\le 24-9=15$

$\Rightarrow x\le \frac{15}{10}=\frac{3}{2}$

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