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ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.1First, try to factor both numerator and denominator. If you're lucky, you can then divide out the common factors!

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.1Can you factor y²+y20?

andreadesirepen
 one year ago
Best ResponseYou've already chosen the best response.0i woul eliminate the y^2 right?

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.1No, write it as: y²+y20=(y ...)(y ...)

tyteen4a03
 one year ago
Best ResponseYou've already chosen the best response.0@andreadesirepen y^2 + y != y^3. Different powers.

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.1On the dots you have to put numbers for which the sum is 1 and the product is 20. Does that sound familiar?

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.1OK, then find these numbers!

andreadesirepen
 one year ago
Best ResponseYou've already chosen the best response.0is it 2x/y^2 ?

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.1No, first you get: y²+y20=(y4)(y+5). That is just the numerator.

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.1Now you have to do the same with the denominator: y²y12=(y ...)(y ...) On the dots are now numbers with sum 1 and product 12. What are they?

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.1I get the feeling you don't understand what is going on. If I want to factor y²y12, 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 +3y4y12 = y²y12. 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
 one year ago
Best ResponseYou've already chosen the best response.1So after all this factoring, we now have:\[\frac{ y^2+y20 }{ y^2y12 }=\frac{ (y+5)(y4) }{ (y+3)(y4) }\] 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 x4 as a common factor. That can therefore be divided out, leaving only:\[\frac{ y+5 }{ y+3 }\]

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.1So, 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!

andreadesirepen
 one year ago
Best ResponseYou've already chosen the best response.0it´ll take me a while to figure this. thanks though!

andreadesirepen
 one year ago
Best ResponseYou've already chosen the best response.0do you know this one ? http://learn.flvs.net/webdav/assessment_images/educator_algebra2_v10/09_14_31.gif

andreadesirepen
 one year ago
Best ResponseYou've already chosen the best response.0THIS one is tuff

tyteen4a03
 one year ago
Best ResponseYou've already chosen the best response.0@andreadesirepen You need to download the file and reupload it on OpenStudy.

tyteen4a03
 one year ago
Best ResponseYou've already chosen the best response.0@andreadesirepen Same drill. Remember that this expression can also be written as \(\frac{12x}{x^2+6x+9} \times \frac{x^29}{4x^2}\).

andreadesirepen
 one year ago
Best ResponseYou've already chosen the best response.0stay, change, flip?

andreadesirepen
 one year ago
Best ResponseYou've already chosen the best response.0or do we cross multiply?
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