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Need help with simplifying this: (3m + n) / (n - 1) * (n^2 - n)/5
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\[\frac{3m+n}{n-1} \cdot \frac{n^2-n}{5}\] I would suggest factoring the (n^2-n)
after you factor n^2-n you should see a common factor and top and bottom
n * (n - 1)?
\(\bf \cfrac{3m+n}{n-1} \cdot \cfrac{n^2-n}{5}\implies \cfrac{3m+n}{\cancel{n-1}} \cdot \cfrac{n\cancel{(n-1)}}{5}\)
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Ah, okay. So, the final result is 3m+n/5. Thank you, freckles and jdoe.
Thanks for giving freckles a medal, jdoe. :-)
well \(\bf \cfrac{3m+n}{\cancel{n-1}} \cdot \cfrac{n\cancel{(n-1)}}{5}\implies \cfrac{(3m+n)n}{5}\implies \cfrac{3mn+n^2}{5}\)
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