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Fanduekisses

  • one year ago

**CHECK MY WORK PWEASE?? **Am I doing this right? SOLVING INEQUALITY.

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  1. Fanduekisses
    • one year ago
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    \[\frac{ 3 }{ x-1 }+\frac{ 2x}{ x+1 }>-1\]

  2. Fanduekisses
    • one year ago
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    \[\frac{ 3(x+1)+2x(x-1)+(x-1)(x+1) }{ (x-1)(x+1) } >0\]

  3. Fanduekisses
    • one year ago
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    \[\frac{ 3x^2+x-2 }{ (x-1)(x+1) }\]

  4. Fanduekisses
    • one year ago
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    I did a sign chart but I get all positive, I know I must've done something wrong because the answer is \[(-\infty, -1) U (1,\infty)\]

  5. Fanduekisses
    • one year ago
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    I did a sign chart but I get all positive, I know I must've done something wrong because the answer is \[(-\infty, -1) U (1,\infty)\]

  6. Fanduekisses
    • one year ago
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    @countonme123 @dan815 @Data_LG2

  7. misty1212
    • one year ago
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    HI!!

  8. misty1212
    • one year ago
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    there is some mistake here, lets see if we can find it

  9. Fanduekisses
    • one year ago
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    :)

  10. misty1212
    • one year ago
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    first lets check the algebra

  11. Fanduekisses
    • one year ago
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    ok

  12. misty1212
    • one year ago
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    ooh i see it!!

  13. misty1212
    • one year ago
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    your numerator is wrong you wrote \[\frac{ 3x^2+x-2 }{ (x-1)(x+1) }\] but it should be \[\frac{ 3x^2+x+2 }{ (x-1)(x+1) }\]

  14. Fanduekisses
    • one year ago
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    Ohhhhhh I see it now too lol

  15. misty1212
    • one year ago
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    now the numerator in this case is always positive so you can ignore it

  16. misty1212
    • one year ago
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    and \[(x+1)(x-1)>0\] outside the zeros, on \(x<-1\) and \(x>1\)

  17. Fanduekisses
    • one year ago
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    So I only look at x=+1 then

  18. misty1212
    • one year ago
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    zeros are at \(1\) and \(-1\)

  19. Fanduekisses
    • one year ago
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    Thank you! :D

  20. Fanduekisses
    • one year ago
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    I have another question, how do you find the domain of a function that has a fraction inside a square root? As in:\[f(x)=\sqrt{\frac{ x }{ x^2-2x-35 }}\]

  21. Fanduekisses
    • one year ago
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    I factored the denominator: (x-7)(x+5)

  22. Fanduekisses
    • one year ago
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    so x cannot equal 7 or -5

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