Question 1108359:  Solve :  
1/|2x-1| greater than or equal to 1 
 Found 2 solutions by  greenestamps, TeachMath: Answer by greenestamps(13214)      (Show Source): 
You can  put this solution on YOUR website! 
  
Solve:   
  
I'll show a couple of different ways you could solve this.  You should know and understand both methods; in any given problem one or the other might be easier to use.
  
And there are undoubtedly other methods; perhaps you will get answers from other tutors that show you method(s) that are different than these two.
  
(1) One thing you can do is determine the values of x for which the expression is undefined and the values for which equality holds.  That will divide the number line into segments; you can then check in which of those segments of the number line the inequality is satisfied.
  
The left side of the inequality is undefined when the denominator is 0; that is at x = 1/2.
  
The equation   is satisfied when  , which is when x = 0 and when x = 1.
  
So the number line is divided into the segments 
(-infinity, 0], [0,1/2), (1/2,1], and [1, infinity).
  
Picking values in each of these segments shows the inequality is satisfied on [0,1/2) and (1/2,1].
 
  
(2) A more traditional algebraic approach is to separate the solution into two cases, depending on whether 2x-1 is positive or negative.  (We already know we don't need to check the case where 2x-1 is 0, because that makes the inequality invalid.)
  
(a) If   then  ; then   and the inequality is 
  
  
  
 
  
So in this case, where we are only considering values of x greater than 1/2, the solution set is all numbers less than or equal to 1; that gives us the (1/2,1] part of the solution.
  
(b) If   then  ; then   and the inequality is 
  
  
  
 
  
So in this case, where we are only considering values of x less than 1/2, the solution set is all numbers greater than or equal to 0; that gives us the [0,1/2) part of the solution.
 
  
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I don't know why tutor teachmath bothered to give you an incorrect answer without showing any work....  That's a strange way of interpreting "teach math".
  
The answer that tutor shows is [0,.5)U[.5,1].  There are two things wrong with that answer: 
(1) x=.5 is not included in the solution set, as shown in the second interval of the answer. 
(2) the answer is equivalent to [0,1] 
 Answer by TeachMath(96)      (Show Source): 
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