Question 37558
Okay, from 4y/(y^2+6y+5) + 2y/(y^2-1),  we can find the lowest common denominator (after factoring both denoms.) as (y + 5)(y + 1)(y - 1)...then we'll need to convert each fraction to that denominator...so we have
4y/(y^2+6y+5) + 2y/(y^2-1) = 
4y(y - 1)/[(y + 5)(y + 1)(y - 1)] + 2y(y + 5)/[(y + 5)(y + 1)(y - 1)] =
and expanding and combining the numerator's terms, we get
(4y^2 - 4y + 2y^2 + 10y) / [(y + 5)(y + 1)(y - 1)] =
(6y^2 + 6y) / [(y + 5)(y + 1)(y - 1)] =
6y(y + 1) / [(y + 5)(y + 1)(y - 1)] =
notice here the (y + 1)'s cancel...
6y / [(y + 5)(y - 1)]
and we're done...