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Stuck on an inequality with absolute values!

  1. Sep 5, 2006 #1
    Hi, I'm stuck on the following problem:
    |(4/x)| > 5
    I split it up into two cases, case 1 is x > 0, case 2 is x < 0
    Code (Text):

    Case 1:                         Case 2:
    4/x > 5                       4/x < -5
    4 > 5x                          4 < -5x
    5x < 4                         -5x > 4
    x < 4/5                       *x < -4/5
    *This is where things go wrong, I tested on a number line and I know that the correct solution should be -4/5 < x < 4/5.. but doesn't the sign have to get switched back again because I'm dividing by a negative?

    I tried doing it another way:
    Code (Text):
    -5 < 4/x < 5            
    -5x < 4 < 5x          
    -5x < 4                  4 < 5x
    x > -4/5                 5x > 4
                              *x > 4/5
     
    *Now for some reason I'm getting x > 4/5 which is false.
    Any help would be very appreciated!
     
  2. jcsd
  3. Sep 5, 2006 #2

    0rthodontist

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    You have to be careful about the sign of x. If x is negative then multiplying by it switches the inequality--so in the second one above, the inequality is switched twice and comes out right.

    In your second code box, you are using the wrong approach. Does -5 < 4/x < 5 really mean |(4/x)| > 5?
     
  4. Sep 5, 2006 #3
    Oh, so in the second attempt, it should be -5 < 4/x > 5
    So in Case 2 of the first method it should be 4/-x > 5, right?
     
  5. Sep 5, 2006 #4

    0rthodontist

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    No on both counts--the second attempt just doesn't work. You can't express it in one inequality without using an absolute value. In the first attempt, it's not 4/-x > 5 -- it's just 4/x > 5 where x is negative.

    Actually, there's a slight quirk here. Let me go through it:

    Either (1) 4/x > 5 or (2) 4/x < -5 must be true
    It's important that you recognize that this is an "either-or" statement. You're not saying that both are true, just that at least one is true.

    Given (1), either (1a) x > 0 or (1b) x < 0. (x can't be 0 or it would be division by 0)
    (1b) implies 4 < 5x, or x > 4/5, which can't be true if x < 0. So if (1) is true, then (1a) is true and then you continue with your reasoning as you had it in your first post.

    Now given (2), either (2a) x > 0 or (2b) x < 0. (again x can't be 0).
    Can you eliminate one of these? Then once you do, continue with your reasoning as you had it in your first post--only perhaps with a switched inequality.

    Once you've done that can you sum all of that up into a statement of what x can be? Not by looking at the number line--by logically piecing together what you've shown. (First state what it is you've shown)
     
    Last edited: Sep 5, 2006
  6. Sep 5, 2006 #5
    Okay, I understand what you're saying.
    So:
    Code (Text):

    |(4/x)| > 5
    4/x > 5                                                  4/x < -5      
    Case 1: x > 0     Case 2: x < 0        Case 1: x > 0       Case 2: x < 0
    4/x > 5                    4/-x > 5           4/x < -5            4/-x < -5
    4 > 5x                     4 > -5x            4 < -5x               4 < 5x
    5x < 4                   -5x < 4               -5x > 4                5x >4
    x < 4/5                 *x > -4/5             x < -4/5              x > 4/5
    *How did you get  x > 4/5 here?
    Also, is the rest of the work correct so far?
     
  7. Sep 5, 2006 #6

    0rthodontist

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    You have to stop saying -x for no reason. If you know x is negative, you still just write it as x--you just also happen to know that it is negative. For example let's say x happens to be -5:
    x = -5
    Then I can write x * 7 / 2, and I know that will be equal to -35/2. I have no reason to write -x. In fact, if I DID write -x in this case, what it would mean is positive 5.

    Now try 4/x > 5, case 2, again. The first line is
    4 / x > 5
    Remember that if you multiply by a negative number (even if that negative number is a variable!) you have to change the direction of the inequality.
     
  8. Sep 5, 2006 #7
    Code (Text):

    |(4/x)| > 5
    4/x > 5                                                  4/x < -5      
    Case 1: x > 0     Case 2: x < 0        Case 1: x > 0       Case 2: x < 0
    4/x > 5                    4/x > 5           4/x < -5            4/x < -5
    4 > 5x                     4 < 5x            4 < -5x               4 > -5x
    5x < 4                    5x > 4              -5x > 4                -5x < 4
    x < 4/5                   x > 4/5             x < -4/5              x > -4/5
    x > 4/5 is false, because x < 0         x < -4/5 is false, because x > 0
    so x > 0 and x < 4/5                      so x < 0 and x > -4/5
    0 < x < 4/5                                          -4/5 < x < 0
    (-4/5, 0) U (0, 4/5)
    That's it, right!?!?
     
     
  9. Sep 5, 2006 #8

    0rthodontist

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    Yep, you got it.
     
  10. Sep 5, 2006 #9
    Thank you SO much for your help!!!! The hardest part was realizing that even though x was a variable it was still negative...
    Again, I REALLY appreciate your help! You were great!
     
  11. Sep 5, 2006 #10
    You know what's messed up? My teacher assigned multiple problem sets that she made up herself, and all the problem sets are supposed to have problems of similar difficulty. The first problem set contained these 3 problems:
    a) |(4/x)| > 5
    b) |2x + 1|<= 7
    c) (x + 1)(x - 5) < 0
    b) and c) are INCREDIBLY easy compared to a), no?
    Seeing as how we never even did an inequality with both absolute values and fractions I assume she accidently gave us an example that was "too mean". In class, she's done this at least 10 times within the first week. She'll teach us something, give us one of her own examples (that she already had written down in her notes), then realize that the example was too hard and tell us "Whoops.. this problem's too mean for you guys". The proccess repeats itself over and over. You know what's scary? This is a Calculus class, we're just reviewing now. I've no idea how I'm going to pass a semester with a teacher who messes up all the time. Oh, and she's a grad student.
     
  12. Sep 5, 2006 #11

    0rthodontist

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    a. isn't really that complicated either--on a test you might just write "... and clearly 4/x > 5 means that x > 0, so...." You also should know about how multiplying by a variable might change the inequality direction, and be comfortable about using a variable when it is negative. Anyway, now you do know about that, so it's a good thing she gave you this problem to solve or you might not know it!
     
  13. Sep 6, 2006 #12
    Good point.
     
  14. Sep 6, 2006 #13
    Here is how I would do this sort of problem.

    Clearly x can't be 0 so consider the following two cases: (a) x = a and (b) x = -a for some a > 0.

    If (a), then |4 / x| = |4 / a| = 4 / a < 5 so a < 4/5 which means that x < 4 / 5 (since x = a).
    If (b), then |4 / x| = |4 / -a| = 4 / a < 5 so a < 4/5 or -a > -4/5 which means that x > -4 / 5 (since x = -a).
     
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