Basics of Inequalities: Solving |x-1| - |x| + |2x+3| > 2x +4

[Kartik.]

Can we explain the meaning of the modulus(absolute value) with these equations?
|x| > a
=>x > a or x < -a(if a $\in$ R⁺ and x $\in$ R if a $\in$ R^-
|x|<a
=> -a < x < a if a $\in$ R⁺ and no solution if a $\in$ R^-$\cup${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?

 

[haruspex]

Can we explain the meaning of the modulus(absolute value) with these equations?
|x| > a
=>x > a or x < -a (if a $\in$ R⁺ and x $\in$ R if a $\in$ R^-
|x|<a
=> -a < x < a if a $\in$ R⁺ and no solution if a $\in$ R^-$\cup${0}

Yes, that all looks right except:
|x| > a => x > a or x < -a (if a $\in$ R⁺$\cup${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.