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 ?

Your computations are wrong. 
I) $x < 2x + 3 \Rightarrow 0 < x + 3 \Rightarrow x > -3$
II) $-x < 2x + 3 \Rightarrow 0 < 3x + 3 \Rightarrow x > -1$
Since we assumed $x > 0$ for I), the solution is $x > -1$
