Question 97682
3/(x^2 + 4x + 3)   -   1/(x^2 - 9) 

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Factor where you can to get:


= [3/(x+3)(x+1)] - [1/(x+3)(x-3)]

When you subtract or add fractions the fractions must
have the same denominator.  So you need a "least 
common denominator" or lcm.
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For your problem lcm = (x+3)(x-3)(x+1)
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Rewrite each fraction with the lcm as its denominator as follows:

=[3(x-3)/lcm] - [1(x+1)/lcm]

Since the denominators now the same combine the numerators over the lcm.

=[3x-9-x-1]/lcm

Simplify to get:

=[2x-10]/lcm

Rewrite:
[2(x-5)]/[(x+3)(x-3)(x+1)

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Cheers,
Stan H.