Determine the set of solutions of $ |x| <2x+3 $ 
Determine the set of solutions of $ |x| <2x+3 $ where $|x| $ denotes
  the absolute value of $x$.

Forgive the banality of the question but when I try to solve this problem given standard techniques I just can't get an answer :
Assuming $x >0$ we have that $x<2x+3$ which gives $x>-3$
Then,assuming $x<0$, I have $-x<2x+3$ which gives $x>-1$
How do I extrapolate the solution from that ?
Now, I know that the solution is $x>-1$ just by using the fact that $y=2x+3$ must intersect $y=|x|$ in the second quadrant at $x=-1$ and for every $x>-1$ the inequality is satisfied ,but how do I get this solution just using the definition of absolute value ?

If $x \ge 0$ (you omitted $x = 0$) then $x < 2x + 3 \implies -3 < x$.  As $x \ge 0$ the combined result is $x \ge 0$ and this is true for all non-negative x.  (Which should be obvious as $2x > x$ for all positive x.)
If $x < 0$ the $-x < 2x + 3 \implies -1 < x$. Combining this with $x < 0$ the combined result is $-1 < x < 0$.  
Combining with the non-negative possibilities we get $-1 < x$ so any $x > -1$ will be a solution.
