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Math Help - [SOLVED] Inequalities

  1. #1
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    [SOLVED] Inequalities

    2/x<x-1
    Since Im not allowed to multiply by x I canīt work it out.
    Never done this kind of stuff before.

    2/x<x-1 -> 2/x -x +1< 0
    I can wright it as (2-x^2+x)/x <0

    Multiplying (-1) gives me (x^2-x-2)/x>0

    Whatever I do I canīt get it right....
    All help apprecieted.
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  2. #2
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    I got it

    Never mind solved
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  3. #3
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    Krizalid's Avatar
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    Just wondering, can you tell me what was your solution set?
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  4. #4
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    Sure

    Sure I did it this way ( if I understood it correctly.

    2/x > x-1 / add: -x+1
    2/x-x+1 > 0 /make denominator same
    (2-x^2+x)/x > 0 /*(-1)
    (x^2-x-2)/x < 0

    Now I solved for x with: X^2-x-2 -> x = 1/2 +- 1.5
    X1 = 2
    X2 = (-1)

    Now So I got the intervalls (-1);(0);(2)
    So i picked a number in all intervalls + a number below (-1) and one above (2).

    That gave me x>(-1)
    or x>2
    and not 0 cause that is undefined...
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  5. #5
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    Quote Originally Posted by Henryt999 View Post
    2/x<x-1
    Note that you solved this inequality flipped.
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  6. #6
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    Iīm sorry what

    Im sorry Iīm not too familiar with the english expression?
    What did you mean?
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  7. #7
    Math Engineering Student
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    Instead of solving the one for <, you solved it for >.
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  8. #8
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    Any other way?

    Any other way of solving it?
    Ill be more then happy to learn it...
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  9. #9
    Math Engineering Student
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    I'll do it when I get back.

    I have to leave now.
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  10. #10
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    Quote Originally Posted by Henryt999 View Post
    2/x<x-1
    Since Im not allowed to multiply by x I canīt work it out.
    Never done this kind of stuff before.

    2/x<x-1 -> 2/x -x +1< 0
    I can wright it as (2-x^2+x)/x <0

    Multiplying (-1) gives me (x^2-x-2)/x>0

    Whatever I do I canīt get it right....
    All help apprecieted.
    1. Multiply both sides by X
    (2/x)(x) < (X)(X-1) \implies 2 < X^2 -1x

    Domain: All real numbers greater than 2
    Last edited by Masterthief1324; January 22nd 2010 at 12:36 PM.
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