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pinsky
Oct23-10, 09:59 AM
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

Mark44
Oct23-10, 10:51 AM
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

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.

pinsky
Oct23-10, 11:06 AM
So is that the general way of doing it?

It seams a bit like a non analytical method :P

Mark44
Oct23-10, 08:28 PM
This technique uses the definition of the absolute value, so there's nothing non-analytical about it.

pinsky
Oct24-10, 04:59 AM
Very well.

Thank you for helping.

SOLVED