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Thread: Inequalites involving absoulte values [resolved]

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  1. Dec 10th, 2001, 07:00 PM #1
    Dillinger4
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    Inequalites involving absoulte values [resolved]

    Help~* I happened to see this on a test i have to take next week so i thought that i would try and learn how to do these but i am having trouble.

    For a postive real number a |x| < a breaks into -a < x and x < a which can be treated as a single compound inequality -a<x<a

    |x| >a breaks into x < -a or x > a.

    Now i just saw two problems which i donot understand.
    |4x+2| < 6 // are they saying |x| < a ?
    |2x - 1| > 3 // are they saying that |x| > a?

    I think that the defination of an absoulte value that i had i my mind is wrong. Isn't the absoulte value of a postive number itself and the absoulte value of a negative number it's positive complement? So in this case how can |4x+2| < 6 be correct?

    Thanks for any help guys.
    Last edited by Dilenger4; Dec 12th, 2001 at 01:51 PM.
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