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Homework Help: Precalculus Algebraic Inequality

  1. Sep 8, 2010 #1
    1. The problem statement, all variables and given/known data

    -3<(1/X)≤ 1

    Solve.


    2. Relevant equations



    3. The attempt at a solution
    Here's my attempt at it:
    (1/X)≤1 and (1/X)> -3

    X≥ 1 and X> (-1/3)

    Am I doing something wrong here? Is this the complete solution? Looking at my answer, is there more that I should I add?
     
  2. jcsd
  3. Sep 8, 2010 #2

    cepheid

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    Welcome to PF

    For each inequality, you have to consider two different cases: X > 0 or X < 0. You also have to be careful. When you multiply both sides by a negative number, the sign of the inequality flips. Therefore, if X < 0, one of your solutions is wrong. Let me show you. We start with:

    1/X > -3

    If you just multiply both sides by X and then divide both sides by -3 without ever flipping the inequality sign, you may get the wrong answer, depending on whether X is positive or negative. Consider, for example, X = -1/4. It satisfies your solution that X > -1/3. However, it means that 1/X = -4, which is NOT greater than -3. So in this case, your solution is wrong.
     
  4. Sep 8, 2010 #3
    You have a very responsive community here!

    So if I fix the error that you pointed out, will my solution be correct?
     
  5. Sep 8, 2010 #4

    cepheid

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    Did you read what I posted? For each of the two inequalities, you have to consider two different cases: X > 0 and X < 0. Therefore you're doing four computations. Each case will in general lead to a different solution (meaning a different range of allowable X values). So your solution is incomplete.

    EDIT: To clarify, your final solution will then be the intersection of these four sets (allowed x ranges)
     
  6. Sep 8, 2010 #5
    I considered your scenarios and this is what I have:

    X ≥ 1 , X< -1/3, and -1/3 <X<0
     
  7. Sep 8, 2010 #6

    cepheid

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    I think you're close, but that's not quite right. Since you've clearly given it a number of attempts let me run through it for you. Let me know if you have any questions:

    CASE 1: X > 0

    As you know, our original inequality is actually two separate ones:

    i) 1/X ≤ 1
    ii) -3 < 1/X

    To solve (i), I'm going to multiply both sides by X. Since X is positive, I don't have to flip any signs, and I end up with:

    X ≥ 1

    To solve (ii), I'm going to multiply both sides by X:

    -3X < 1

    then I'm going to divide both sides by -3, which requires a sign flip, since -3 is negative:

    X > -1/3

    Since the above inequality is satisfied by ANY X > 0, this result doesn't actually provide any useful constraint. So the result is:

    Solution for CASE 1: X ≥ 1

    CASE 2: X < 0

    To solve inequality (i) for X, we have to multiply both sides by X, which requires a sign flip since X is negative:

    X ≤ 1

    Since this result is satisfied by ANY X < 0, it does not provide a useful constraint. Solving inequality (ii), we have to first multiply both sides by X, and then divide both sides by -3. Since we multiply by TWO negative numbers, we have to flip the inequality sign twice, which is the same as not flipping it at all:

    X < -1/3

    Solution for CASE 2: X < -1/3

    The final solution is therefore the union of the solutions for the two cases:

    X ≥ 1 OR X < -1/3
     
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