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OpenStudy (anonymous):
Solve for n.
n/9 < n/2 - 1/3
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OpenStudy (anonymous):
let us first try to simplify the right hand side
\[{n\over2}-{1\over 3}\]
OpenStudy (anonymous):
i would multiply by \(18\) to get rid of the annoying fractions
OpenStudy (anonymous):
this simplifies to \[{n\times3\over2\times3}-{1\times2\over3\times2}=\]
OpenStudy (anonymous):
I agree with satellitey
OpenStudy (anonymous):
Multiply both sides by 18, re-arrange to bring the n's on one side and then divide both sides so that n is left on one side and a number on the other.
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OpenStudy (anonymous):
both ways lead to rome
OpenStudy (anonymous):
i hate fractions !
\[\frac{n}{9}<\frac{n}{2}-\frac{1}{3}\]
\[18\left (\frac{n}{9}\right)<18\left (\frac{n}{2}-\frac{1}{3}\right)\]\[2n<9n-6\]
OpenStudy (anonymous):
subtract \(2n\) from both sides to get
\[0<7n-6\] add \(6\)\[6<7n\]divide by \(7\)
OpenStudy (anonymous):
6/7
OpenStudy (anonymous):
your answer should be an inequality, not a number
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OpenStudy (anonymous):
\[\frac{6}{7}<n\] or
\[n>\frac{6}{7}\]
OpenStudy (anonymous):
oh thank you
OpenStudy (anonymous):
yw
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