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please solve 3(n+1)/7 is equal to or greater than n+4/5
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\[\frac{{3}(n+1)}{7}\geq\frac{(n+4)}{5}\] Playing with the equation editor. Still there?
\[n \le -7\div5\]
Hmm. I'm getting a different answer. Check my work: \[\frac{5}{1}*\frac{7}{1}*\frac{{3}(n+1)}{7}\geq\frac{5}{1}*\frac{7}{1}*\frac{(n+4)}{5} \]
\[5(3n+3)\geq{7n+28}\Rightarrow15n+15\geq{7n+28}\]
\[8n\geq13\Rightarrow{{n}{\geq}\frac{13}{8}}\]
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