Question

$\frac{1}{2}(\frac{3x}{5}+4)\ge \frac{1}{3}(x-6)$.

Answer 1

Here $\frac{1}{2}(\frac{3x}{5}+4)\ge \frac{1}{3}(x-6)$

$\Rightarrow \frac{3x}{10}+2\ge \frac{x}{3}-2\Rightarrow \frac{3x}{10}-\frac{x}{3}\ge -2-2$

$\Rightarrow \frac{9x-10x}{30}\ge -4\Rightarrow \frac{-x}{30}\ge -4$

Multiplying both sides by 30, we have

$-x\ge -120$

Dividing both sides by -1, we have

$x\le 120$

Thus the solution set is $[-\infty ,120]$.