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

Algebra ->  Algebra  -> Distributive-associative-commutative-properties -> 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      Log On

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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(17075) About Me  (Show Source):
You can put this solution on YOUR website!
%286y%2F%28y%5E2%2B5y%2B4%29%29%2B%284y%2F%28y%5E2-1%29%29

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.

y%5E2%2B5y%2B4=%28y%2B4%29%28y%2B1%29.

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

%286y%2F%28%28y%2B4%29%28y%2B1%29%29%29%2B%284y%2F%28%28y-1%29%28y%2B1%29%29%29

Next you need a common denominator. In this case, you have one factor of y%2B4, one of y%2B1, and one of y-1, so:

%286y%28y-1%29%2B4y%28y%2B4%29%29%2F%28%28y%2B4%29%28y-1%29%28y%2B1%29%29%29
%286y%5E2-6y%2B4y%5E2%2B16y%29%2F%28%28y%2B4%29%28y-1%29%28y%2B1%29%29%29
%2810y%5E2%2B10y%29%2F%28%28y%2B4%29%28y-1%29%28y%2B1%29%29%29

Next, factor the numerator

%2810y%28y%2B1%29%2F%28%28y%2B4%29%28y-1%29%28y%2B1%29%29%29

Eliminate %28y%2B1%29%2F%28y%2B1%29

%2810y%2F%28%28y%2B4%29%28y-1%29%29%29

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:

%2810y%2F%28y%5E2%2B3y-4%29%29%29

Hope that helps,
John

Answer by edjones(7569) About Me  (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%2Across%28y%2B1%29%2F%28%28y%2B4%29cross%28y%2B1%29%28y-1%29%29
=10y/((y+4)(y-1)) Voila!
.
Ed