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andreadesirepen

  • 3 years ago

how would i simplify this?

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  1. andreadesirepen
    • 3 years ago
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  2. ZeHanz
    • 3 years ago
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    First, try to factor both numerator and denominator. If you're lucky, you can then divide out the common factors!

  3. ZeHanz
    • 3 years ago
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    Can you factor y²+y-20?

  4. andreadesirepen
    • 3 years ago
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    i woul eliminate the y^2 right?

  5. andreadesirepen
    • 3 years ago
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    y^3-20

  6. ZeHanz
    • 3 years ago
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    No, write it as: y²+y-20=(y ...)(y ...)

  7. tyteen4a03
    • 3 years ago
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    @andreadesirepen y^2 + y != y^3. Different powers.

  8. ZeHanz
    • 3 years ago
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    On the dots you have to put numbers for which the sum is 1 and the product is -20. Does that sound familiar?

  9. andreadesirepen
    • 3 years ago
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    yes

  10. ZeHanz
    • 3 years ago
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    OK, then find these numbers!

  11. ZeHanz
    • 3 years ago
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    (please)

  12. andreadesirepen
    • 3 years ago
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    is it 2x/y^2 ?

  13. ZeHanz
    • 3 years ago
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    No, first you get: y²+y-20=(y-4)(y+5). That is just the numerator.

  14. ZeHanz
    • 3 years ago
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    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?

  15. andreadesirepen
    • 3 years ago
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    what?

  16. andreadesirepen
    • 3 years ago
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    x^2/2y?

  17. ZeHanz
    • 3 years ago
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    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.

  18. ZeHanz
    • 3 years ago
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    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:\[\frac{ y+5 }{ y+3 }\]

  19. ZeHanz
    • 3 years ago
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    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!

  20. andreadesirepen
    • 3 years ago
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    it´ll take me a while to figure this. thanks though!

  21. ZeHanz
    • 3 years ago
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    yw!

  22. andreadesirepen
    • 3 years ago
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    do you know this one ? http://learn.flvs.net/webdav/assessment_images/educator_algebra2_v10/09_14_31.gif

  23. andreadesirepen
    • 3 years ago
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    THIS one is tuff

  24. tyteen4a03
    • 3 years ago
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    @andreadesirepen You need to download the file and reupload it on OpenStudy.

  25. andreadesirepen
    • 3 years ago
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  26. andreadesirepen
    • 3 years ago
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    @tyteen4a03

  27. tyteen4a03
    • 3 years ago
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    @andreadesirepen Same drill. Remember that this expression can also be written as \(\frac{12x}{x^2+6x+9} \times \frac{x^2-9}{4x^2}\).

  28. andreadesirepen
    • 3 years ago
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    stay, change, flip?

  29. andreadesirepen
    • 3 years ago
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    or do we cross multiply?

  30. andreadesirepen
    • 3 years ago
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    @tyteen4a03

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