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Math Help - Inequalities with absolute values and fractions

  1. #1
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    Inequalities with absolute values and fractions

    hi,

    I am facing a couple of issues with two inequalities, would appreciate help here.

    1) |2x + 3| ≤ 3x − 5

    the first case would be 2x + 3 ≤ 3x - 5, solves easily, but no approach to the second case makes sense to me:
    -2x - 3 ≤ 3x - 5 ??
    2x + 3 >= -3x + 5 ??
    => x >= 2/5 but that doesn't even come close to fitting the original inequality?

    If I square them, I get
    4x^2 + 12x + 9 <= 9x^2 - 30x + 25, solving that with the usual formula I end up with x1 = 2 and x2 = 8/5, which is also somewhat strange.


    2) 3 / | x - 9 | > 2 / (x + 2)
    The essential questions here are how many cases I have to distinguish between and if anyone could be so kind to formulate a general rule when the inequality sign changes direction and the signs on one (or both?) sides of the inequality change from + > - and vice versa.

    Thank you.
    Oz.
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  2. #2
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    Quote Originally Posted by ozfingwe View Post
    hi,

    I am facing a couple of issues with two inequalities, would appreciate help here.

    1) |2x + 3| ≤ 3x − 5

    ...
    Use the definition of the absolute value:

    |2x+3|=\left\{\begin{array}{l}2x+3\ if\ 2x+3 \geq 0~\implies x\geq -\dfrac32 \\ -(2x+3)\ if\ 2x+3 < 0~\implies x < -\dfrac32 \end{array}\right.

    So you have to solve 2 different equations:

    2x+3 \leq 3x-5~\wedge~x\geq -\dfrac32

    \boxed{8 \leq x}~\wedge~x\geq -\dfrac32
    -------------------------------------------------------------------------
    -(2x+3) \leq 3x-5~\wedge~x < -\dfrac32

    2 \leq 5x~\wedge~x < -\dfrac32

    \dfrac25 \leq x~\wedge~x < -\dfrac32~\implies~x \notin \mathbb{R}
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