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I honestly cannot believe I am stuck on this.

  1. Aug 28, 2011 #1
    1. The problem statement, all variables and given/known data


    Solve

    [tex]\left | \frac{5}{x + 2} \right | < 1[/tex]

    3. The attempt at a solution

    All I know is that x= -2 is discontinuous. I thought about doing test points, but it seemed rather useless seeing I will probably end up with (-2, infinity) or (-inf, -2)

    Computer says the solution is x < -7, x >3

    I then thought about doing this

    [tex]\left |5\right | < \left|x + 2\right|[/tex]
    [tex]-|x + 2| < -5[/tex]
    [tex]5 < -(x + 2) < -5[/tex]
    [tex]-5 > x + 2 > 5[/tex]
    [tex]-7 > x > 3[/tex]

    I am sorry lol
     
  2. jcsd
  3. Aug 28, 2011 #2
    I don't understand how you get the third step? :confused:

    You should go by taking two cases.
    What do get when you solve |x+2|? Definitely (x+2) and -(x+2). :smile:

    If you solve these two equations:-
    5<(x+2)
    and
    5<-(x+2)
    You will get the desired result. :smile:
     
  4. Aug 28, 2011 #3
    [tex]-7 > x > 3[/tex]

    ?????????????????????????????????????????

    x is greater than 3 and less than -7?
     
  5. Aug 28, 2011 #4
    See I don't know what I am doing lol
     
  6. Aug 28, 2011 #5

    Dembadon

    User Avatar
    Gold Member

    You can set up two inequalities (see Pranav-Anora's post). Or you can set up a double inequality since this problem is of the "less-than" variety. You attempted to do the double inequality method, but you shouldn't have a negative on your absolute value expression. I think you mixed two different methods into one. :wink:

    Here's a general example of how you should've setup the double inequality:

    Given |expression| < k, set up the double inequality, -k < |expression| < k and solve.
     
  7. Aug 28, 2011 #6

    uart

    User Avatar
    Science Advisor

    Why don't you try making a sketch of y= |5/(x+2)| and y=1 on the same set of axes.
     
  8. Aug 28, 2011 #7

    Mark44

    Staff: Mentor

    No, the rational expression is undefined at x = -2.
    Your inequality is equivalent to
    |x + 2| > 5 and x [itex]\neq[/itex] -2

    In general, if you have an inequality of the form |x + a| > b, you can get rid of the absolute values by rewriting this as
    x + a > b or x + a < - b

    For your problem, since x cannot be -2, any interval you end up with cannot include -2.
     
  9. Aug 29, 2011 #8

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    For real A, the inequality |A| < 1 is the same as -1 < A < 1.

    RGV
     
  10. Aug 30, 2011 #9

    NascentOxygen

    User Avatar

    Staff: Mentor

    Last edited by a moderator: Apr 26, 2017
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