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): You can put this solution on YOUR website!
(y+2)/ (y^2+5y+6) + (2-y)/(y^2+y-6)
(y+2)/((y+3)(y+2)) + (2-y)/((y+3)(y-2))
(y+2)/((y+3)(y+2)) - (y-2)/((y+3)(y-2))
1/(y+3) - 1/(y+3)
0
Ans: (y+2)/ (y^2+5y+6) + (2-y)/(y^2+y-6) = 0
------------------------------------------------------------------
1/x+7/(3x)
3/(3x)+7/(3x)
(3+7)/(3x)
10/(3x)
Ans: 1/x+7/(3x) = 10/(3x)
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Once you have determined a common denominator, adding fractions is simply the sum of the numerators over that common denominator.
(y + 2)} + \frac{2 - y}{(y + 3)(y - 2)})
So far, so good. But realize that )
, so:
Now just add the numerators -- do not add the denominators
===============================================
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:
rather 
Remember adding fractions in elementary school:
not
John
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