4y/y^2 + 6y+ 5 plus 2y/ y^2 – 1?

Question

I need to add these and I am lost on even where to begin. please help!

Answer 1

Collect like terms.

Answer 2

First, group the like terms together:

(4y/y² + 6y + 5) + (2y/y² - 1) = (2y/y² + 4y/y²) + 6y + (5 - 1)

Then resolve the stuff in the parentheses:

2y/y² + 4y/y² = (2 + 4)*(y/y²) = 6y/y²
5 - 1 = 4

Answer:

6y/y² + 6y + 4

Answer 3

for 4y/(y^2 + 6y+ 5) plus 2y/( y^2 – 1)
factor the denominators of the fractions to get:

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

next find common denominator and add the fractions to get:

[4y(y-1) + (2y)(y+5)]/[(y+5)(y+1)(y-1)] the y+1 is used once

then simplify by multiplying out & combining terms to get

[4y^2-4y+2y^+10y]/[(y^2 + 6y+ 5)(y-1)]

[6y^2+6y]/(y^3+5y^2 + y-5)


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

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

Answer 4

I am assuming it is ((4y)/(y^2+6y+5))+((2y)/(y^2-1))
You need to get a common denominator, but this is made more complicated becase the left denominator isn't factorable and the right is but it isn't related at all. The most you can do is just do the typical cross multiply to get a very long common denominator:
(4y(y^2-1)+2y(y^2+6y+5))/((y^2+6y+5)(y^2-1))
Then it's just simplifying:
(4y^3-4y+2y^3+12y^2+10y)/((y^2+6y+5)(y^2-1))
(6y^3+12y^2+6y)/((y^2+6y+5)(y^2-1))
(6y(y^2+2y+1))/((y^2+6y+5)(y^2-1))
Factor what you can:
(6y(y+1)^2)/((y^2+6y+5)(y+1)(y-1))
Eleminate to get:
(6y(y+1))/((y^2+6y+5)(y-1))
redistribute:
(6y^2+6y)/(y^3+6y^2+5y-y^2-6y-5)
Simplify to get the final answer:
(6y^2+6y)/(y^3+5y^2-y-5)

Answer 5

4y / (y + 5)(y + 1) + 2y / (y - 1)(y + 1)
[ 4y(y - 1) + 2y(y + 5) ] / [ (y + 5)(y + 1)(y - 1) ]
[ 6y² + 6y ] / [ (y + 5)(y + 1)(y - 1) ]
6y (y + 1) / [ (y + 5)(y + 1)(y - 1) ]
6y / [ (y + 5) (y - 1) ]

Answer 6

4y/(y^2 + 6y+ 5) + 2y/( y^2 – 1)
=4y/(y+5)(y+1) + 2y/(y+1)(y-1)
={4y(y-1) + 2y(y+5)}/(y+1)(y-1)(y+5)
=(4y^2-4y+2y^2+10y)/(y+1)(y-1)(y+5)
=(6y^2+6y)/(y+1)(y-1)(y+5)
=6y(y+1)/(y+1)(y-1)(y+5)
=6y/(y-1)(y+5)

Answer 7

10y/y^2+2y/y^2+4 >> 12y/y^2+4 i think is as simplified as it can get..