Help Please Algebra?

Question

4y/(y^2+6+5)+2y/y^2-1)

Is this the right answer??? 3y/y(y-2)

Answer 1

First of all there is a typo in your question: The 1st denominator should be y^2+6y+5.

Anyway, Here are the steps:
1. Factor the denominators.
y^2+6y+5 = (y+5)(y+1)
y^2-1 = (y+1)(y-1)
2. Figure out the common denominator: (y+5)(y+1)(y-1)
3. Multiply the first fraction by (y-1) on both numerator and denominator
4y(y-1)/(y+5)(y+1)(y-1)
4. Multiply the second fraction by (y+5)
2y(y+5)/(y+5)(y+1)(y-1)
5. Simplify the numerators, add them and factor... then you should have 6y(y+1)/(y+5)(y+1)(y-1)
6. All you need to do now is to cancel out (y+1)
So your final answer is 6y/(y+5)(y-1)

Hope this is helpful!

Answer 2

I assumed that the second term is not written correctly, that the proper statement is:

4y/(y^2+6+5)+2y/(y^2-1)

The common denominator is

(y^2+11)(y^2-1)

so (4y(y^2-1)+2y(y^2+11))/(y^2+11)(y^2-1)

2y(2y^2-1+y^2+11)/(y^2+11)(y^2-1)
2y(3y^2+10)/(y^2+11)(y^2-1)


If the first denominator is written (y^2+6y+5)
then the common denominator is
(y+1)(y+5)(y-1)
giving the first term as
4y(y-1) over the denominator
the second term is
2y(y+5) over the denominator

solving the numerator
2y(2y-2+y+5)
gives
2y(3y+3)
or
6y(y+1)
cancel (y+1) from the demoniator and you get
6y/((y+5)(y-1))
or
6y/(y^2+4y-6)

Answer 3

Quite obviously , there is a misprint in the den of 1st term: it should be y2+6y+5, whichi s = (y+1)(y+5)
What you have is then
[4y/(y+1)(y+5)]+[2y/(y+1)(y-1)]

=(2y/(y+1))[2/(y+5).+1/(y-1)}
Consider the num of term in the square bracket:
It is
=[2(y-1)+(y+5)}=3(y+1)
The full term is then 3(y+1) /((y+5)(y+1)
Combining it with the 1st term cancels out the factor (y+1) giving finally
6y/(y+5)(y+1)//
it is easier to do the problem , or follow the steps,on paper than on the computer screen.

Answer 4

I don't get that either. For one, your answer is reducible to 3/(y-2) and is that your intent to reduce the equation or solve the equation for the condition Eq. f(y) = 0? I get as an answer:
4 / (y+2+10/y). Also your ()'s are NOT balanced. It makes a difference where the missing left ( is.... In my case, I assumed it was at the left most position of the denominator.

Answer 5

Hmmm ... not what I get

I get 6y(y^2+3) / (y^4 + 10y^2 - 11)

Or is that first fraction 4y/(y^2 + 6y + 5) cos that'd change it a lot. In THIS case my answer becomes:

6y/(y^2+4y-5)

Answer 6

no....factor the bottoms first into (y+1)(y+5) and (y+1)(y-1)
Multiply the tops by the missing pieces...... 4y(y-1) and 2y(y+5)
combine like terms and factor out yhe 6y(y+1) on top ..... cancel the y+1 one top and bottom and you are left with 6y/(y-1)(y+5)

Answer 7

4y/(y+5)(y+1) +2y/(y+1)(y-1)
LCD is (y+5)(y+1)(y-1)
4y+2y(y+5) /(y+5)(y+1)(y-1)
2y^2+14y /(y+5)(y+1)(y-1)
2y(y+7) /(y+5)(y+1)(y-1)

Answer 8

No it isn't - also the problem makes more sense if it were 6y not 6 in the first denominator, could you check it?

Answer 9

there's an extra bracket at the end.. so what's the actual question?