SOLUTION: I need to find the sum or difference with unlike denominators
1. (y+2)/ (y^2+5y+6) + (2-y)/(y^2+y-6)
I did (y+2)/(y+3)(y+2) + (2-y) /(y+3)(y-2)
I crossed out the y+2 and the
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-> SOLUTION: I need to find the sum or difference with unlike denominators
1. (y+2)/ (y^2+5y+6) + (2-y)/(y^2+y-6)
I did (y+2)/(y+3)(y+2) + (2-y) /(y+3)(y-2)
I crossed out the y+2 and the
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Question 198040: I need to find the sum or difference with unlike denominators
1. (y+2)/ (y^2+5y+6) + (2-y)/(y^2+y-6)
I did (y+2)/(y+3)(y+2) + (2-y) /(y+3)(y-2)
I crossed out the y+2 and the y-2 and then left with 1/(y+3)(y+3) = 1/2y+6
2. (1)/(X) + (7)/(3x)
I times 1/x by 3/3 and got 1/x.3/3 + 7/3x.1/1 = 10/6x^2
Please help....I'm very confused
Thanks,
JC Found 2 solutions by user_dude2008, solver91311:Answer by user_dude2008(1862) (Show Source):
Once you have determined a common denominator, adding fractions is simply the sum of the numerators over that common denominator.
So far, so good. But realize that , so:
Now just add the numerators -- do not add the denominators
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You started with the correct strategy, i.e. you multiplied the left hand fraction by 3 over 3, but then you again added the denominators when you were adding the fractions. Your result should have been:
And, by the way, you also made an error when you added the denominators:
(
