Solving Inequality: |x-1|+|x-2|>1

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Homework Help Overview

The problem involves solving the inequality |x-1| + |x-2| > 1, which falls under the subject area of inequalities and absolute values in algebra.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss different approaches to solving the inequality by considering the behavior of the absolute values at critical points (x=1 and x=2). Some suggest checking various intervals defined by these points, while others reflect on the implications of changing the inequality's constant.

Discussion Status

The discussion is active, with participants sharing their reasoning and methods for approaching the problem. Some guidance has been offered regarding checking intervals and test values, but there is no explicit consensus on a single method or solution.

Contextual Notes

Participants note the importance of the critical points where the absolute values equal zero and consider variations of the original inequality, which introduces additional complexity to the discussion.

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Homework Statement



So this is the question

|x-1|+|x-2|>1



Homework Equations



N/A

The Attempt at a Solution



I tried it, the solution seems right, but i don't know if my approach is correct.

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Think about these absolute values like this:

At x=1 or x=2, one of the absolute values is 0, so you can call these the "roots" (thought not technically roots, I can't think of a more appropriate name for them right now), so what you want to do is check all possibilities around those roots, and the roots themselves.

Check:

x<1
For this range, both [itex]x-1[/itex] and [itex]x-2[/itex] will be negative, so the inequality you need to solve would be [tex]-(x-1)-(x-2)>1[/tex]

1<x<2
Here you will have [itex]x-1>0[/itex] and [itex]x-2<0[/itex] so what you need to solve is [tex](x-1)-(x-2)>1[/tex]

x>2
For this value, both are positive so it should be clear what you need to solve here.

And then always check the "roots" themselves. Plug in the values of x=1 and x=2. By this point, you've checked all possible cases and should have your solution set.
 
The way I learned it:

To solve |ax+b|>k, solve ax+b>k and ax+b<-k. (This is basically what you did.)

Then I would plug test values into the original equation to see if it makes a true or false statement. You would use the intervals (-[itex]\infty[/itex],1);(1,2);(2,[itex]\infty[/itex]).

The values from those intervals that make true statements give you your solution set.
 
edit: I just realized that your question is a special case. What if it was instead [tex]|x-1|+|x-2|>2[/tex] or [tex]|x-1|+|x-2|>0[/tex] ? For the first what you need to do is check when [tex]|x-1|+|x-2|=2[/tex] and then check each interval around that.
 
Last edited:
Thanks Mentallic and Adaptation!
 

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