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July 29, 2014

July 29, 2014

Posted by **Mikayla** on Thursday, March 14, 2013 at 8:54pm.

- algebra 2 -
**drwls**, Friday, March 15, 2013 at 6:54amFirst factor the denominators.

3n/(n^2 -7n + 10) - 2n/(n^2-8n+15)

= 3n/[(n-5)(n-2)] -2n/[(n-5)(n-3)]

= [n/(n-5)] *{[3/(n-2)]-[2/(n-3)]}

= [n/(n-5)] *{[3(n-3)-2(n-2)/(n-2)(n-3)]}

= [n/(n-5)]*(n+5)/[(n-2)(n-3)]

This does not seem to be getting any simpler.

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