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Math Help - What am I doing wrong,inequalitie again

  1. #1
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    What am I doing wrong,inequalitie again

    Solve following:
    Ix-4I+2x<7
    Those I:s mean absolute value...
    Well I did this..

    (-7)<x-4+2x<7 /+4
    (-3)<3x<11 / /3
    (-1)<x<11/3 but that is apparently very wrong...correct answer is x<3
    Any idea where I wen´t wrong?
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by Henryt999 View Post
    Solve following:
    Ix-4I+2x<7
    Those I:s mean absolute value...
    Well I did this..

    (-7)<x-4+2x<7 /+4
    This is wrong, you should've done this: |x-4|<7-2x\implies 2x-7<x-4<7-2x, and solve each one separately.

    By the way, can't you generate the | ?
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  3. #3
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Henryt999 View Post
    Solve following:
    Ix-4I+2x<7
    Those I:s mean absolute value...
    Well I did this..

    (-7)<x-4+2x<7 /+4
    (-3)<3x<11 / /3
    (-1)<x<11/3 but that is apparently very wrong...correct answer is x<3
    Any idea where I wen´t wrong?
    You can use the | key to mark the beginning and end of the modulus. On an English (UK) keyboard it's assigned to Shift+\ which is in the bottom left, next to shift.

    |x-4| + 2x < 7

    |x-4| < 7 - 2x

    \pm (x-4) < 7-2x



    A)
    (x-4) < 7-2x

    3x < 11

    x < \frac{11}{3}


    B)
    -(x-4) < 7-2x

    Multiply by -1: (x-4) > 2x-7

    Collect like terms: -x > -3

    Multiply by -1: x < 3


    Since  3 < \frac{11}{3} discard solution A.
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