Equations with multiple absolute value operators

AI Thread Summary
The discussion centers on solving inequalities involving multiple absolute value operators, specifically the inequality |x + 2| - |x| > 1. The user initially struggles with the second scenario after confirming the positive case yields true for all x. They express confusion about rearranging the equation correctly, leading to an incorrect result. Another participant suggests rewriting the inequality as |x + 2| > |x| + 1 and recommends analyzing it across three intervals to find the solution. The method is confirmed as valid, and the issue is resolved.
pinsky
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Hello there. I'm having some problems with absolute values when they contain multiple "abs" operators and some other numbers outside the "abs"-es.

For example:

\left | x+2 \right | - \left | x \right | > 1

If i check it for the positive scenario, the result is true for all x-es.

x+2 - x > 1

I can't conceptually see how to arrange the equations for the second scenario.

I've tried the following

-x-2 < 1 + x

but that didn't give the right result. The correct result is -1/2<x
 
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pinsky said:
Hello there. I'm having some problems with absolute values when they contain multiple "abs" operators and some other numbers outside the "abs"-es.

For example:

\left | x+2 \right | - \left | x \right | &gt; 1

If i check it for the positive scenario, the result is true for all x-es.

x+2 - x &gt; 1

I can't conceptually see how to arrange the equations for the second scenario.

I've tried the following

-x-2 &lt; 1 + x

but that didn't give the right result. The correct result is -1/2<x

I would write the inequality as |x + 2| > |x| + 1.

You can get rid of the absolute values by looking at three separate intervals: -2 < x, -2 < x < 0, and x > 0. Each of these intervals defines one case. The final answer will be the union of the solution sets for the three cases.
 
So is that the general way of doing it?

It seams a bit like a non analytical method :P
 
This technique uses the definition of the absolute value, so there's nothing non-analytical about it.
 
Very well.

Thank you for helping.

SOLVED
 
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