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ZeHanz
 2 years 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!

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

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

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

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

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

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

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

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

tyteen4a03
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0stay, change, flip?

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