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 ?

When we assume $x > 0$ we get $x < 2x + 3$ wich yields $x > -3$. But we assumed $x > 0$. So far our set of solutions is $x > 0$ since it both satisfies $x > 0$ and $x > 3$ (both our conditions). In terms of set theory, one condition is satisfied when $x \in (0,\infty)$ and the other one when $x\in (-3,\infty)$ so both are satisfied when $x\in (0,\infty)\cap (-3,\infty) = (0,\infty)$.
When we assume $x \leq 0$ we get $x >  -1$. To satisfy both conditions we now need $-1 < x \leq 0$. In terms of set theory, one condition is satisfied when $x \in (-\infty,0]$ and the other one when $x\in (-1,\infty)$ so both are satisfied when $x\in (-\infty,0]\cap (-1,\infty) = (-1,0]$.
We now combine both solutions to get $x > -1$. In terms of set theory, we will have a solution if $x\in (-1,0]\cup (0,\infty) = (-1,\infty)$.
