SOLUTION: Could I get some help on this problem? 3/x^2 + 4x + 3 - 1/x^2 - 9 I do not understand how to solve any help is appreciated. Thank you.

Question 97682: Could I get some help on this problem?
3/x^2 + 4x + 3 - 1/x^2 - 9
I do not understand how to solve any help is appreciated. Thank you.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
3/(x^2 + 4x + 3) - 1/(x^2 - 9)
-----------------
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.
------------
For your problem lcm = (x+3)(x-3)(x+1)
-------------
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)
==================
Cheers,
Stan H.
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