Please help me add 4y/y^2+6y +5 plus 2y/y^2 -1?

Question

Please help me add 4y/y^2+6y +5 plus 2y/y^2 -1? show me how to do this step by step so I can learn how it works

Answer 1

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

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

Multiply everything by (y + 5)(y - 1)(y + 1)

((4y * (y - 1)) + (2y * (y + 5)))/((y + 5)(y - 1)(y + 1))

(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))

take out the (y + 1)s

(6y)/((y + 5)(y - 1))

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

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

Answer 2

Firstly, I'm making the assumption that you don't really mean the following (if so, then my answer is not correct and I appologize):

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

And I'm making the assumption that you mean the following (like typing it into a TI graphing calculator):

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

Firstly, the y/y^2 can be reduced to 1/y (because y/y = 1):

4/y+6y+5+2/y-1

It may help to rewrite 1/y as y^-1 (a property of exponents):

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

It may help to regroup with the commutative property and associative property:

[4y^(-1)+2y^(-1)]+6y+(5-1)

Now combine like terms:

6y^(-1)+6y+4

That's the answer. It can also be rewritten like the following:

6/y+6y+4

Verification: 4y/y^2+6y+5+2y/y^2-1=6/y+6y+4

Answer 3

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

First, I would get rid of the denominators by multiplying them through. Below you will see me multiplying (y^2 + 6y + 5) on each side, which cancels it out of the first part to leave only 4y:

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

Then, multiply through by (y^2 - 1) to get:

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

Then, I would multiply the numbers outside of the parenthesis through:

4y^3 - 4y + 2y^3 + 12y^2 + 10y

Then, you add similar terms to get:

6y^3 + 12y^2 + 6y

=)