Basics of Inequalities

  • Thread starter Kartik.
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Can we explain the meaning of the modulus(absolute value) with these equations?
|x| > a
=>x > a or x < -a(if a [itex]\in[/itex] R+ and x [itex]\in[/itex] R if a [itex]\in[/itex] R-
|x|<a
=> -a < x < a if a [itex]\in[/itex] R+ and no solution if a [itex]\in[/itex] R-[itex]\cup[/itex]{0}
If yes, then examples please?(for instances in x and a)
Blindly apply these equations we can solve |x-1| >= 3 as x-1<= -3 or x-1 >=3
If yes then how can we solve a inequality like |x-1| - |x| + |2x+3| > 2x +4 using the same logical statements above?
 

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haruspex
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Can we explain the meaning of the modulus(absolute value) with these equations?
|x| > a
=>x > a or x < -a (if a [itex]\in[/itex] R+ and x [itex]\in[/itex] R if a [itex]\in[/itex] R-
|x|<a
=> -a < x < a if a [itex]\in[/itex] R+ and no solution if a [itex]\in[/itex] R-[itex]\cup[/itex]{0}
Yes, that all looks right except:
|x| > a => x > a or x < -a (if a [itex]\in[/itex] R+[itex]\cup[/itex]{0}) and etc.
(or, more simply, if a >= 0).
how can we solve a inequality like |x-1| - |x| + |2x+3| > 2x +4 using the same logical statements above?
Easiest way is to break it into the different cases: x <= -2, -2 <= x <= -3/2, -3/2 <= x <= 0, 0 <= x <= 1, x >= 1.
E.g.: x <= -2:
-(x-1) - (-x) + (-2x-3) > 2x + 4
Some of these will produce contradictions.
 

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