Question 1108359
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Solve:  {{{1/abs(2x-1)<=1}}}<br>
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.<br>
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.<br>
(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.<br>
The left side of the inequality is undefined when the denominator is 0; that is at x = 1/2.<br>
The equation {{{1/abs(2x-1)=1}}} is satisfied when {{{abs(2x-1)=1}}}, which is when x = 0 and when x = 1.<br>
So the number line is divided into the segments
(-infinity, 0], [0,1/2), (1/2,1], and [1, infinity).<br>
Picking values in each of these segments shows the inequality is satisfied on [0,1/2) and (1/2,1].<br><br>
(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.)<br>
(a) If {{{x > 1/2}}} then {{{2x-1>0}}}; then {{{1/abs(2x-1) = 1/(2x-1)}}} and the inequality is
{{{1/(2x-1) >= 1}}}
{{{1 >= 2x-1}}}
{{{2 >= 2x}}}
{{{x <= 1}}}<br>
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.<br>
(b) If {{{x < 1/2}}} then {{{2x-1<0}}}; then {{{1/abs(2x-1)= 1/(1-2x)}}} and the inequality is
{{{1/(1-2x) >= 1}}}
{{{1 >= 1-2x}}}
{{{2x >= 0}}}
{{{x >= 0}}}<br>
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.<br><br>
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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".<br>
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]