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Thread: Absolute Value Inequality

  1. #1
    Super Member harpazo's Avatar
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    Absolute Value Inequality

    The statement "x is at least 6 units from 4" is expressed
    |x - 4| ≥ 6.

    Can you break the statement given in words and constants into the given absolute value inequality? I do not understand why |x - 4| ≥ 6 is the answer.

    My original answer was |x - 6| ≥ 4. Why is my original answer wrong?

    What about the question below?

    Statement: "x is more than 5 units from 2 "

    My answer is |x + 5| > 2. No answer given in the textbook for even number problems.
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  2. #2
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    Re: Absolute Value Inequality

    To say that "x is at least 6 units from 4" means that x must be larger than x= 4+ 6= 10 or smaller than x= 4- 6= -2. That "x= 4+ 6" is the same as "x- 4= 6" and "x= 4- 6" is the same as "x- 4= -6" so we are saying that x- 4> 6 or that x- 4< -6. Both cases can be expressed as |x- 4|> 6.

    Your "|x- 6|> 4" is wrong because x= 1 satisfies |1- 6|= 5> 4, but the distance from x= 1 to 4 is 3 which is NOT greater than 6. "|x- 6|> 4" is satisfied by all x whose distance from 6 is greater than 4, not the other way around.

    Remember that, in general, the distance between points "a" and "b", on the number line, is |a- b|. So, here, the distance of x "from 4" is |x- 4|.
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  3. #3
    Super Member harpazo's Avatar
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    Re: Absolute Value Inequality

    Quote Originally Posted by HallsofIvy View Post
    To say that "x is at least 6 units from 4" means that x must be larger than x= 4+ 6= 10 or smaller than x= 4- 6= -2. That "x= 4+ 6" is the same as "x- 4= 6" and "x= 4- 6" is the same as "x- 4= -6" so we are saying that x- 4> 6 or that x- 4< -6. Both cases can be expressed as |x- 4|> 6.

    Your "|x- 6|> 4" is wrong because x= 1 satisfies |1- 6|= 5> 4, but the distance from x= 1 to 4 is 3 which is NOT greater than 6. "|x- 6|> 4" is satisfied by all x whose distance from 6 is greater than 4, not the other way around.

    Remember that, in general, the distance between points "a" and "b", on the number line, is |a- b|. So, here, the distance of x "from 4" is |x- 4|.

    Your reply is a bit confusing.
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