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babyjay Group Title

Can someone help me I have the problem solved but do not know how to determine the restriction variables. x-2/x+3 plus 10x/x^2-9

  • one year ago
  • one year ago

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  1. precal Group Title
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    restriction is the value that would make the denominator equal 0

    • one year ago
  2. precal Group Title
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    one of the math commandments is "thou shall not divide by zero"

    • one year ago
  3. precal Group Title
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    |dw:1359076743638:dw|

    • one year ago
  4. precal Group Title
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    do you know see what number x can not be

    • one year ago
  5. precal Group Title
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    |dw:1359076865546:dw|

    • one year ago
  6. precal Group Title
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    do you see what x can not be?

    • one year ago
  7. babyjay Group Title
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    are you talking about 3

    • one year ago
  8. precal Group Title
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    yes but also, another number

    • one year ago
  9. babyjay Group Title
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    3 and -3

    • one year ago
  10. precal Group Title
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    |dw:1359077321883:dw|

    • one year ago
  11. precal Group Title
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    bingo, you got it

    • one year ago
  12. precal Group Title
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    |dw:1359077374821:dw|

    • one year ago
  13. babyjay Group Title
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    do you work the problem out further like (x^2-5x+6)+10x/(x+3)(x-3)

    • one year ago
  14. precal Group Title
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    yes, keep going

    • one year ago
  15. babyjay Group Title
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    then x^2+5x+6/(x+3)(x-3) (x+3)(x+2)/(x+3)(x-3) answer x=2/x-3;x=/-3,3

    • one year ago
  16. precal Group Title
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    yes perfect, gotta go........

    • one year ago
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