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Answer:
the first one
Step-by-step explanation:
Express the left side as a single fraction
3 + [tex]\frac{4}{x}[/tex]
= [tex]\frac{3x+4}{x}[/tex], hence
[tex]\frac{3x+4}{x}[/tex] ≥ [tex]\frac{x+2}{x}[/tex]
Subtract [tex]\frac{3x+4}{x}[/tex] from both sides
0 ≥ [tex]\frac{x+2}{x}[/tex] - [tex]\frac{3x+4}{x}[/tex]
0 ≥ [tex]\frac{-2x-2}{x}[/tex]
Multiply both sides by - 1, remembering to reverse the inequality symbol as a consequence
0 ≤ [tex]\frac{2x+2}{x}[/tex], hence
[tex]\frac{2x+2}{x}[/tex] ≥ 0
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Rewritten by : Barada
Answer:
The first alternative is correct
Step-by-step explanation:
We move all the expressions to the left hand side of the inequality then combine like terms using lcm;
[tex]\frac{3}{1}+\frac{4}{x}-\frac{x+2}{x}\geq0\\\\\frac{3x+4-(x+2)}{x}\geq0\\\frac{2x+2}{x}\geq0[/tex]
