SOLUTION: can someone please check this please? 3y/y^2+5y+4 plus 2y/y^2-1= 2y^2-1=(y-1)(y-1) y^2+5y+4=(y+1)(y+1)

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: can someone please check this please? 3y/y^2+5y+4 plus 2y/y^2-1= 2y^2-1=(y-1)(y-1) y^2+5y+4=(y+1)(y+1)      Log On


   

Question 77154: can someone please check this please?
3y/y^2+5y+4
plus
2y/y^2-1=
2y^2-1=(y-1)(y-1)
y^2+5y+4=(y+1)(y+1)
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
3y/y^2+5y+4
plus
2y/y^2-1=
------------
3y/[(y+4)(y+1)] + [2y/[(y+1)(y-1)]
Least common denominator is (y+4)(y+1)(y-1)
=3y(y-1)/lcd + 2y(y+4)/lcd
=[3y^2-3y+2y^2+8y]/lcd
=[5y^2+5y]/lcd
=[5y(y+1)]/lcd
=5y/[(y+4)(y-1)]
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Cheers,
Stan H.