anonymous
  • anonymous
how would i simplify this?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
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ZeHanz
  • ZeHanz
First, try to factor both numerator and denominator. If you're lucky, you can then divide out the common factors!
ZeHanz
  • ZeHanz
Can you factor y²+y-20?

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anonymous
  • anonymous
i woul eliminate the y^2 right?
anonymous
  • anonymous
y^3-20
ZeHanz
  • ZeHanz
No, write it as: y²+y-20=(y ...)(y ...)
tyteen4a03
  • tyteen4a03
@andreadesirepen y^2 + y != y^3. Different powers.
ZeHanz
  • 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
  • anonymous
yes
ZeHanz
  • ZeHanz
OK, then find these numbers!
ZeHanz
  • ZeHanz
(please)
anonymous
  • anonymous
is it 2x/y^2 ?
ZeHanz
  • ZeHanz
No, first you get: y²+y-20=(y-4)(y+5). That is just the numerator.
ZeHanz
  • 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
  • anonymous
what?
anonymous
  • anonymous
x^2/2y?
ZeHanz
  • 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
  • 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
  • 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
  • anonymous
it´ll take me a while to figure this. thanks though!
ZeHanz
  • ZeHanz
yw!
anonymous
  • anonymous
do you know this one ? http://learn.flvs.net/webdav/assessment_images/educator_algebra2_v10/09_14_31.gif
anonymous
  • anonymous
THIS one is tuff
tyteen4a03
  • tyteen4a03
@andreadesirepen You need to download the file and reupload it on OpenStudy.
anonymous
  • anonymous
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anonymous
  • anonymous
tyteen4a03
  • 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
  • anonymous
stay, change, flip?
anonymous
  • anonymous
or do we cross multiply?
anonymous
  • anonymous

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