Biby98 Biby98
  • 29-11-2018
  • Mathematics
contestada

Solve the inequality

[tex] \frac{2x}{x + 3} \leqslant 3[/tex]
The solution set is

Respuesta :

gmany
gmany gmany
  • 01-12-2018

[tex]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)[/tex]

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