i need some help?

Question

add and express in simplest form.



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

Answer 1

The first thing is to factor each of the denominators... using the reverse FOIL method.

If you factor (y^2+6y+5)... you need to numbers that will multiply to give you +5, and those same to numbers must also add up to +6... The two factors are +1, and +5.... So, now you know (y^2 + 6y + 5) = (y+1)(y+5)

Okay... now look at the denominator of the 2nd term... how can (y^2 - 1) be written? Well, you will notice that this polynomial is the difference of two squares. So... your factors must multiply to 1.... BUT they will have opposite signs.... so (y^2 - 1) can also be written as (y+1)(y-1).

Okay.... Now rewrite the given equation, substituting with what you found....

So...

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

is the same as....

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

BUT.... in order to add these two fractions together... you have to have a common denominator... As you can see, both of the denominators are not the same...

What factor is the first fraction missing? It's missing (y-1) which is found in the 2nd denominator. So in the 1st fraction, you will need to multiply by (y-1)/(y-1), which is essentially "1". (Remember, since you are going to multiply the denominator by "y-1".... what you do to the bottom, you also have to do to the numerator... meaning... you will also have to multiply the numberator by "y-1")

Now, what factor is the 2nd fraction missing? It's missing (y+5) which is found in the 1st denominator. So in the 2nd fraction, you will need to multiply by (y+5)/(y+5), which is essentially "1". (Remember, since you are going to multiply the denominator by "y+5".... what you do to the bottom, you also have to do to the numerator... meaning... you will also have to multiply the numberator by "y+5")

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

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

See what I just did?

** now it's just math ... distribute the 4y... and the 2y... **

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

** now combine "like" terms... **

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

** factor "6y" out of the numerator **

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

Do you notice anything? You should notice that you have (y+1) in the numerator that will cancel out with the (y+1) in the denominator.... leaving you with...

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

This is in simplest form because you can't reduce the fraction down any further.

Hope this helps!!!

Answer 2

34

Answer 3

shooot me

Answer 4

what do I do again ?