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Homework Help: An inequality with absolute values

  1. Sep 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Solve Ix+3I>2
    *I is used for absolute value notation

    3. The attempt at a solution
    Considering both
    a) Ix+3I > 0 then Ix+3I= x+3
    b) Ix+3I < 0 then Ix+3I= -(x+3)

    when solved this would yield to;

    a) x>-3 and x>-1
    b) x<-5 and x<-3

    from my general reasoning i think the answer should be x>-1 and x<-5. Why are the solutions x>-3 and x<-3 omitted? Is it because the other two include a broader range?

    thanks for your help.
  2. jcsd
  3. Sep 6, 2012 #2


    User Avatar
    Homework Helper

    Hi theself, welcome to PF.

    You can find an absolute value key (vertical line) on the keyboard.

    So the problem is: Solve |x+3|>2

    The absolute value is never negative. The two cases are x+3≥0 (x>-3) and x+3<0 (x<-3).

    You wrote correctly that |x+3|=x+3 in the first case and |x+3|=-(x+3) in the second case.

    In the first case, you got x>-1 if x>-3 is true. x>-1 is more strict requirement than x>-3. So x>-1 is solution.

    In the second case, you got x<-5 if x<-3. x<-5 is more strict condition than x<-3 so x<-5 is solution.

    "Why are the solutions x>-3 and x<-3 omitted?"

    They are not solutions. Substitute x=-2 for example: (it is greater then -3). |x+3|=1 less then 2. Choose x=-4: (it is less then -3) |-4+3|=|-1|=1.

  4. Sep 6, 2012 #3


    Staff: Mentor

    Why? Use the | character instead.
    How did you get what you have just above?
    |x + 3| > 2
    <==> x + 3 > 2 OR x + 3 < -2
    Can you take it from here?
  5. Sep 6, 2012 #4


    User Avatar
    Science Advisor

    The simplest way to solve a general inequality is to first solve the corresponding equation. If |x+ 3|= 2 then either x+ 3= 2 or x+ 3= -2. In the first case x= -1 and in the second x= -5.

    The key point here is that, since absolute value is continuous, we can go from "<" to ">", and vice-versa, only at "=". That is, the two points, -5 and -1, divide the real numbers into three intervals and on each interval only one of "<" or ">" can apply. And we need only check one point in each interval to see which.

    For example, -6< -5 and |-6+3|= |-3|= 3 which is larger than 2, not less. The original inequality is false at x= -6 and so false for all x less than -5. -4 lies between -5 and -1 and |-4+3|= |-1|= 1 which is less than 2. The original inequality is true at x= -4 and so true for all x between -5 and -1. Finally, 0 is larger than -1 and |0+ 3|= |3|= 3 which is larger than 2, not less. The original inequality is false at x= 0 and so false for all x larger than -1.

    By the way, every keyboard I have ever seen has a "|" just above the return key. Are you using a keyboard that doesn't?
  6. Sep 6, 2012 #5
    Is the if condition there because we define it like that at the beginning ?
  7. Sep 6, 2012 #6
    Well, I kind of did it in a long way;
    If |x+3|> 0 then |x+3|= x+3, and substituted the term (x+3) to where |x+3| is found
    this gives
    if x> -3, x > -1
    and did the same for the other condition
  8. Sep 6, 2012 #7
    Wouldn't using the equation create a problem as you can not do everything you do to an equation to an inequality?
    For eg: dividing both sides by - 1

    * Also I apologize for my clumsiness for not finding the key for "|"
  9. Sep 6, 2012 #8


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    Homework Helper

    I think the answer is "yes"
    The result x>-1 came from the condition that x>-3. x>-3 and x>-1 are not two solutions. Those x values for which -3<x<-1 do not satisfy the original inequality. In similar way, when we supposed that x<-3 we got the result x<-5. Again, the original inequality is not true for -5<x<-3.

    Last edited: Sep 6, 2012
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