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Question 193826: (3b+5) + (-b+5)
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(b^2+4b+3) (b^2+2b-3)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! (3b+5)/(b^2+4b+3) + (-b+5)/(b^2+2b-3)
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Factor where you can:
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(3b+5)/(b^2+4b+3) + (-b+5)/(b^2+2b-3)
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(3b+5)/(b+3)(b+1) - (b-5)/(b+3)(b-1)
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lcd = (b+3)(b-1)(b+1)
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Rewrite the two fractions:
[(3b+5)(b-1)]/lcd - [(b-5)(b+1)]/lcd
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Combine the numerators over the lcd:
[3b^2+ 2b -5 -(b^2 - 4b -5)]/lcd
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[3b^2+ 2b -5 -(b^2 - 4b -5)]/lcd
(2b^2 + 6b]/[(b+3)(b-1)(b+1)]
Factor:
[2b(b+3)]/[(b+3)(b-1)(b+1)]
= 2b/[b^2-1]
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Cheers,
Stan H.
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