Question

Solve the inequality

$\frac{2x}{x + 3} \leqslant 3$
The solution set is

Answer 1

$Domain:x\neq-3\\\\\dfrac{2x}{x+3}\leq3\qquad\text{subtract 3 from both sides}\\\\\dfrac{2x}{x+3}-3\leq0\\\\\dfrac{2x}{x+3}-\dfrac{3(x+3)}{x+3}\leq0\\\\\dfrac{2x}{x+3}-\dfrac{3x+9}{x+3}\leq0\\\\\dfrac{2x-(3x+9)}{x+3}\leq0\\\\\dfrac{2x-3x-9}{x+3}\leq0\\\\\dfrac{-x-9}{x+3}\leq0\Rightarrow(x+3)(-x-9)\leq0\\\\x=-3,\ x=-9\\\\x\in(-\infty,\ -9]\ \cup\ (-3,\ \infty)$