Possible Values of the Inequality.

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The discussion centers on determining the possible values of |2x−3| given the constraint 0<|x−1|<2. Two cases are analyzed: one where 1<x<3 leading to -1<2x-3<3, and another where -1<x<1 resulting in -5<2x-3<-1. It is concluded that the valid range for |2x-3| is |2x-3|<3, as the inequality -5<|2x-3|<-1 does not hold true. Participants emphasize the importance of correctly interpreting the absolute value constraints. The final assertion confirms that the only relevant solution is |2x-3|<3.
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Homework Statement



What are the possible values of |2x−3| when 0<|x−1|<2?

Homework Equations





The Attempt at a Solution



We know \left|x-1\right| becomes x-1 if x-1≥0 and -(x-1) if x-1<0.
Now consider two cases.
Case 1:
0<x-1<2 \Rightarrow 1<x<3 \Rightarrow -1<2x-3<3.

Case 2:
0<-x+1<2 \Rightarrow -1<x<1 \Rightarrow -2<2x<2 \Rightarrow -5<2x-3<-1.

Then the possible value include -1<|2x-3|<3 and -5<|2x-3|<-1.

Am I going about this problem correctly?
 
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NATURE.M said:
Then the possible value include -1<|2x-3|<3 and -5<|2x-3|<-1.

Saying -5< |2x-3| is not a particularly bold statement to make... take another look at the inequalities you got before you took the absolute value and think a bit harder about what they tell you about the absolute value.
 
Office_Shredder said:
Saying -5< |2x-3| is not a particularly bold statement to make... take another look at the inequalities you got before you took the absolute value and think a bit harder about what they tell you about the absolute value.

Then the possible solutions should probably just be |2x-3|<3, since no solution exists to the inequality -5<|2x-3|<-1 since |2x-3| is bounded by negative values.
Am I right?
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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