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Math Help - Double Absolute Value Inequalities ???

  1. #1
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    Question Double Absolute Value Inequalities ???

    Hey everyone- can anyone pls assist with how to solve these:


    -5< |x|<2


    This is nowhere in my book and I am not technically required to know this for my class but I am interested in how this would be solved, I do know how to do it if it were either/or:

    i.e. |x|<2 you would just do this: -2< x <2

    and obviously |x|> -5 would be all real #s

    but to I have no idea how to solve when its already compounded with two different values at once!

    Any help greatly appreciated!
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  2. #2
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    Re: Double Absolute Value Inequalities ???

    Quote Originally Posted by cds333 View Post
    -5< |x|<2
    This is nowhere in my book and I am not technically required to know this for my class but I am interested in how this would be solved, I do know how to do it if it were either/or:
    i.e. |x|<2 you would just do this: -2< x <2
    and obviously |x|> -5 would be all real #s
    Because that is true then the problem is just
    |x|<2.
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  3. #3
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    Double / Compound Absolute Value Inequalities ???

    Hey thanks plato!

    So what you're saying is the left side gives (-∞,∞), and the right side gives (-2,2), but since the right side is more specific it overrules the left?

    What if the problem were something like: 2 ≤ |x| ≤ 5

    Thanks again!
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  4. #4
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    Re: Double / Compound Absolute Value Inequalities ???

    a\le x\le b should be treated as two inequalities: a\le x and x\le b. Notice the "and"- they must both be true which means that the solution set of the "double" inequality is the intersection of the two separate inequalities.

    Further, a\le |x| itself means a \le x if x is non-negative or a\le -x if x is negative. Here, we have "or" so the solution set is the union of the two separate solution sets.

    2\le |x|\le 5 gives the two inequalities 2\le |x| and |x|\le 5.

    (1) 2\le |x| means
    (a) 2\le x as long as x is non-negative which it certainly is if it is larger than or equal to 2: [tex]2\le x.
    (b) 2\le -x which is the same as -2\ge x as long as x is negative, which it is as long as x is less than -2: x\le -2.
    The union of those two sets is the two separate intervals, [-\infty, -2)\cup[2, \infty)

    (2) |x|\le 5 means
    (a) x\le 5 as long as x is non-negative: 0\le x\le 5
    (b) -x\le 5 which is the same as x\ge -5 as long as x is negative: -5\le x< 0
    The union of those two sets is the single interval, [-5, 5].

    Now take the intersection of those two sets. The interval (-\infty, -2] intersect [-5, 5] is [-5, -2]. The interval [2, \infty) intersect [-5, 5] is [2, 5]. The solution set for [tex]2\le x\le 5[/itex] is [-5, -2]\cup [2, 5].

    You probably could have done that more simply by reducing the absolute value first: 2\le x\le 5 and 2\le -x\le 5
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