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Thread: What am I doing wrong,inequalitie again

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

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

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

    $\displaystyle (-7)<x-4+2x<7$ $\displaystyle /+4$
    This is wrong, you should've done this: $\displaystyle |x-4|<7-2x\implies 2x-7<x-4<7-2x,$ 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:
    $\displaystyle Ix-4I+2x<7$
    Those I:s mean absolute value...
    Well I did this..

    $\displaystyle (-7)<x-4+2x<7$ $\displaystyle /+4$
    $\displaystyle (-3)<3x<11$ $\displaystyle / /3$
    $\displaystyle (-1)<x<11/3$ 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.

    $\displaystyle |x-4| + 2x < 7$

    $\displaystyle |x-4| < 7 - 2x$

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



    A)
    $\displaystyle (x-4) < 7-2x$

    $\displaystyle 3x < 11$

    $\displaystyle x < \frac{11}{3}$


    B)
    $\displaystyle -(x-4) < 7-2x$

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

    Collect like terms: $\displaystyle -x > -3$

    Multiply by -1: $\displaystyle x < 3$


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