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Abs(x-2) > -1 (That is, the absolute value of x-2 is greater than -1.) Now I have always been accustomed to setting up two inequalities, one x-2 > -1 and the other x-2 < 1. When I do this, I don't get the write answer (2.412). Is the method I use with abs value wrong? I use this whenever there is an inequality sign or an equals with an abs value on one end like in the problem above. Thank you.
$\displaystyle |x-2| > -1$Quote:
Originally Posted by benny92000
x can be any value ... $\displaystyle |anything| \ge 0 > -1$
sure you copied the problem correctly?
I copied the answer wrongly. My book has the symbol "Z" in the back. Is my method wrong? How did you go about solving it?.
no "method" required ... only common sense. the absolute value of anything is greater than any negative value. kapish?Quote:
Originally Posted by benny92000
I understand that. But does that mean the method I am accustomed to using is incorrect?
your method is correct, you just interpreted the results incorrectly.Quote:
Originally Posted by benny92000
$\displaystyle |x-2| > -1$
this means that ...
$\displaystyle x-2> -1$ or $\displaystyle -(x-2) > -1$
$\displaystyle x > 1$ or $\displaystyle x < 3$
the union of these two sets is the set of all real numbers.
