Answer:
[tex]\frac{2x+2}{x}\ge \:0[/tex]
Step-by-step explanation:
[tex]3+\frac{4}{x}\ge \frac{x+2}{x}[/tex]
Subtract [tex]\frac{x+2}{x}[/tex] from both sides:
[tex]3+\frac{4}{x}-\frac{x+2}{x}\ge \frac{x+2}{x}-\frac{x+2}{x}[/tex]
[tex]3+\frac{4}{x}-\frac{x+2}{x}\ge \:0[/tex]
[tex]\frac{3x}{x} +\frac{-x+2}{x}[/tex]
[tex]\frac{2x+2}{x}\ge \:0[/tex]
