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Math Help - [SOLVED] Rational inequality with an absolute value in the numerator

  1. #1
    nihil.ad.rem
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    [SOLVED] Rational inequality with an absolute value in the numerator

    Thanks!
    Last edited by nihil.ad.rem; July 21st 2007 at 07:54 AM.
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by nihil.ad.rem View Post
    Hi. How would I go about solving this problem?

    | x - 3 | / (x+4) is less than or equal to 5.

    I'm not really sure how to isolate the absolute value expression.

    Thanks!
    hello,

    \frac{|x-3|}{x+4}\leq 5 transform into:

    \frac{x-3}{x+4}\leq 5 ~ \wedge ~ x\geq 3~~\vee~~ \frac{-x+3}{x+4}\leq 5 ~ \wedge ~ x < 3

    Solve both inequalities separately. You should get as a final result:

    x < -4 ~~\vee~~x \geq -\frac{17}{6}
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by nihil.ad.rem View Post
    Hi. How would I go about solving this problem?

    | x - 3 | / (x+4) is less than or equal to 5.

    I'm not really sure how to isolate the absolute value expression.

    Thanks!
    I won't give the details, but another way would be to simply square the expression. (|x|)^2 = x^2. The downside is that you may have extra solutions, and instead of one inequality you have two:
    -25 \leq \frac{(x - 3)^2}{(x + 4)^2} \leq 25.

    However, despite the downsides I like the logic inherent in this method slightly better, even if it makes the problem more difficult to solve. And, of course, when you are done you get earboth's answer.

    -Dan
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