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

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The discussion focuses on solving the inequality |x-1| + |x-2| > 1 by analyzing different intervals based on the critical points x=1 and x=2. It suggests checking three cases: when x is less than 1, between 1 and 2, and greater than 2, to determine the behavior of the absolute values. The approach involves rewriting the inequality for each interval and testing the boundaries to find the solution set. Additionally, the poster reflects on how to handle variations of the inequality, such as |x-1| + |x-2| > 2 or |x-1| + |x-2| > 0, emphasizing the importance of checking the intervals around the critical points. This methodical approach ensures a comprehensive understanding of the inequality's solution.
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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 x-1 and x-2 will be negative, so the inequality you need to solve would be -(x-1)-(x-2)&gt;1

1<x<2
Here you will have x-1&gt;0 and x-2&lt;0 so what you need to solve is (x-1)-(x-2)&gt;1

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 (-\infty,1);(1,2);(2,\infty).

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 |x-1|+|x-2|&gt;2 or |x-1|+|x-2|&gt;0 ? For the first what you need to do is check when |x-1|+|x-2|=2 and then check each interval around that.
 
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Thanks Mentallic and Adaptation!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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