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babyjay

  • 2 years ago

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

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  1. precal
    • 2 years ago
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    restriction is the value that would make the denominator equal 0

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

  3. precal
    • 2 years ago
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    |dw:1359076743638:dw|

  4. precal
    • 2 years ago
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    do you know see what number x can not be

  5. precal
    • 2 years ago
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    |dw:1359076865546:dw|

  6. precal
    • 2 years ago
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    do you see what x can not be?

  7. babyjay
    • 2 years ago
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    are you talking about 3

  8. precal
    • 2 years ago
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    yes but also, another number

  9. babyjay
    • 2 years ago
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    3 and -3

  10. precal
    • 2 years ago
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    |dw:1359077321883:dw|

  11. precal
    • 2 years ago
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    bingo, you got it

  12. precal
    • 2 years ago
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    |dw:1359077374821:dw|

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

  14. precal
    • 2 years ago
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    yes, keep going

  15. babyjay
    • 2 years ago
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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

  16. precal
    • 2 years ago
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    yes perfect, gotta go........

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