SOLUTION: Here is another problem with addition just in case the steps are different than subtraction. (4y/y^2+6y+5)+(2y/y^2-1)

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Question 56144This question is from textbook Beginning Algebra
: Here is another problem with addition just in case the steps are different than subtraction.
(4y/y^2+6y+5)+(2y/y^2-1) This question is from textbook Beginning Algebra

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
(4y/y^2+6y+5)+(2y/y^2-1)
Factor where you can to get:
4y/[(y+5)(y+1)] + 2y/[(y-1)(y+1)]
The LCD is (y+5)(y+1)(y-1)]
Rewrite each fraction with LCD as its denominator:
(4y)(y-1)/LCD + 2y(y+5)/LCD
Combine the numerators over the LCD
[4y^2-4y+2y^2+10y]/LCD
[6y^2+6y]/LCD
[6y(y+1)]/LCD
Cancel the common (y+1) factors to get:
6y/[(y+5)(y-1)]
Cheers,
Stan H.