- anonymous

how would i simplify this?

- chestercat

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- anonymous

##### 1 Attachment

- ZeHanz

First, try to factor both numerator and denominator.
If you're lucky, you can then divide out the common factors!

- ZeHanz

Can you factor y²+y-20?

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- anonymous

i woul eliminate the y^2 right?

- anonymous

y^3-20

- ZeHanz

No, write it as: y²+y-20=(y ...)(y ...)

- tyteen4a03

@andreadesirepen y^2 + y != y^3. Different powers.

- ZeHanz

On the dots you have to put numbers for which the sum is 1 and the product is -20.
Does that sound familiar?

- anonymous

yes

- ZeHanz

OK, then find these numbers!

- ZeHanz

(please)

- anonymous

is it 2x/y^2 ?

- ZeHanz

No, first you get: y²+y-20=(y-4)(y+5). That is just the numerator.

- ZeHanz

Now you have to do the same with the denominator:
y²-y-12=(y ...)(y ...)
On the dots are now numbers with sum -1 and product -12.
What are they?

- anonymous

what?

- anonymous

x^2/2y?

- ZeHanz

I get the feeling you don't understand what is going on.
If I want to factor y²-y-12, which is y² -1*y -12, I must look for two numbers that have sum -1 (the coefficient of y) and product -12 (the constant at the end).
These are: -4 and 3.
See: -4*3 = -12,
-4+3 = -1.
Why does this work?
Try it: (y -4)(y+3) can be expanded again, using FOIL:
First: y*y = y²
Outer: y*3 = 3y
Inner: -4*y = -4y
Last: -4*3 = -12
Together, this becomes: y² +3y-4y-12 = y²-y-12.
So you have to look at it like: y² +(a+b)y + ab = (y+a)(y+b)
That is why you say:
the sum is the number befor the y
the product is the number at the end.

- ZeHanz

So after all this factoring, we now have:\[\frac{ y^2+y-20 }{ y^2-y-12 }=\frac{ (y+5)(y-4) }{ (y+3)(y-4) }\]
Only now is the moment we can simplify!
Because only now we have a product in the numerator and also a product in the denominator.
Both products have x-4 as a common factor. That can therefore be divided out, leaving only:\[\frac{ y+5 }{ y+3 }\]

- ZeHanz

So, I'm sorry, but there is no quick way to just mess a bit with the numbers in your original fraction :(
What you have to do (always!) is:
1. write numerator as a product (=factor it)
2. write denominator as a product (=factor it)
3. Look for common factors. Divide them out.
4. See what is left. This is your answer!

- anonymous

it´ll take me a while to figure this. thanks though!

- ZeHanz

yw!

- anonymous

do you know this one ? http://learn.flvs.net/webdav/assessment_images/educator_algebra2_v10/09_14_31.gif

- anonymous

THIS one is tuff

- tyteen4a03

@andreadesirepen You need to download the file and reupload it on OpenStudy.

- anonymous

##### 1 Attachment

- anonymous

- tyteen4a03

@andreadesirepen Same drill. Remember that this expression can also be written as \(\frac{12x}{x^2+6x+9} \times \frac{x^2-9}{4x^2}\).

- anonymous

stay, change, flip?

- anonymous

or do we cross multiply?

- anonymous

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