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Thread: When? does the inequality solve

  1. #1
    Junior Member Freaky-Person's Avatar
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    When? does the inequality solve

    In what cases would you have to change the inequality sign?

    I mean, if you have an equation like,

    |x| + |2x-5| < 5

    You would have 3 cases right?

    So what would the sign (< and >) look like for those 3 and why does it change some times?
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  2. #2
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    Quote Originally Posted by Freaky-Person View Post
    In what cases would you have to change the inequality sign?

    I mean, if you have an equation like,

    |x| + |2x-5| < 5

    You would have 3 cases right?

    So what would the sign (< and >) look like for those 3 and why does it change some times?
    The traditional way you are taught in high-schools, that means look at the plus minus signs DOES NOT WORK in general. It only works in the problems they give you.
    This happens to be a more advanced problem.
    We need to use the following definition,
    $\displaystyle |n|=\left\{ \begin{array}{c}n \mbox{ for }n\geq 0 \\ -n\mbox{ for }n<0 \end{array} \right\} $
    Thus, we the the following cases,
    $\displaystyle |x|=x , -x $
    And it depends on, $\displaystyle x\geq 0, x<0$
    Also we have,
    $\displaystyle |2x-5|=2x-5,-2x+5$
    And it depends on, $\displaystyle 2x-5\geq 0,2x-5<0$
    Solve the inequality,
    $\displaystyle x\geq 2.5$ and $\displaystyle x<2.5$.
    In combination with,
    $\displaystyle x\geq 0$ and $\displaystyle x<0$.
    We have the following cases,

    1)$\displaystyle x<0\to x<2.5$
    $\displaystyle |x|=-x$ and $\displaystyle |2x-5|=-2x+5$

    2)$\displaystyle x\geq 0 \mbox{ and }x<2.5$
    $\displaystyle |x|=x$ and $\displaystyle |2x-5|=-2x+5$

    3)$\displaystyle x\geq 2.5 \to x\geq 0$
    $\displaystyle |x|=x$ and $\displaystyle |2x-5|=2x-5$.

    Thus, consider the inequality
    $\displaystyle |x|+|2x-5|<5$
    If, the first case,
    $\displaystyle -x-2x+5<5$
    $\displaystyle -3x+5<5$
    $\displaystyle -3x<0$
    $\displaystyle x>0$
    But that is a contradiction because $\displaystyle x<0$.
    If, the second case,
    $\displaystyle x-2x+5<5$
    $\displaystyle -x+5<5$
    $\displaystyle -x<0$
    $\displaystyle x>0$
    This will be a contradiction unless $\displaystyle 0<x<2.5$
    If, the third case,
    $\displaystyle x+2x-5<5$
    $\displaystyle 3x-5<5$
    $\displaystyle 3x<10$
    $\displaystyle x<10/3$
    This will be a contradiction unless $\displaystyle 2.5\leq x<10/3$
    Thus,
    $\displaystyle x=(0,2.5)\cup [2.5,10/3)=(0,10/3)$
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