SOLUTION: Please help me with this problem. I have similar problems in my homework but it does not show how to solve this type of problem. You help is appreciated. Problem: 6y over y^2+5y+4

Question 117515: Please help me with this problem. I have similar problems in my homework but it does not show how to solve this type of problem. You help is appreciated.
Problem: 6y over y^2+5y+4 add to 4y over y^2-1. I tried to factor the first denominator but that just got me more confused. I don't know where to start.
Found 2 solutions by solver91311, edjones:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

Factoring both denominators is the key to doing this problem just as you suspected. Since 4 X 1 = 4 and 4 + 1 = 5, this trinomial factors rather neatly.

.

Now, you can write your problem with fully factored denominators, thus:

Next you need a common denominator. In this case, you have one factor of , one of , and one of , so:

Next, factor the numerator

Eliminate

Sometimes it might be more convenient to leave your answer just like that. At least the domain of the function is readily apparent (all Reals except -4 and 1). But if need be, multiply the binomial, thus:

Hope that helps,
John
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
6y/(y^2+5y+4)+4y/(y^2-1)
=6y/((y+4)(y+1))+4y/((y+1)(y-1)) Factor the denominators.
=(6y(y-1)+4y(y+4))/((y+4)(y+1)(y-1)) LCD is the denominator
=6y^2-6y+4y^2+16y/((y+4)(y+1)(y-1)) Expand the numerator.
=10y^2+10y/((y+4)(y+1)(y-1)) Combine terms.
=
=10y/((y+4)(y-1)) Voila!
.
Ed
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