3n/n^2-49 - 2/3n+21

3n/(n^2-49) - 2/(3n+21) <==Is this what you want?

LCD = 3(n+7)(n-7)

3n/(n^2-49) - 2/(3n+21)

= (9n - 2(n-7))/LCD

= (7n + 14)/LCD

= 7(n+2)/[3(n+7)(n-7)] <==Answer

3n/n^2-49 - 2/3n+21

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3n/(n^2-49) - 2/(3n+21) <==Is this what you want?

LCD = 3(n+7)(n-7)

3n/(n^2-49) - 2/(3n+21)

= (9n - 2(n-7))/LCD

= (7n + 14)/LCD

= 7(n+2)/[3(n+7)(n-7)] <==Answer

Let me be sure about the format of this expression before I give you a solution, because it will definitely make a difference! May I take this expression literally, as it is written? Or is it actually a difference of two fractions... in other words, with 3n being the numerator of the first fraction, n-squared minus 49 the denominator of the first fraction; 2 being the numerator of the second fraction, and 3n+21 the denominator of the second fraction? That would make it: 3n/(n^2-49) - 2/(3n+21).

If you are studying order of operations, your original format makes sense. If you are studying factoring, the alternate format I described would make more sense.

Note: I didn't notice earlier that Robert J. has answered, making the same assumption that I have... your solution is there.

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