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 ?
 A: You made a mistake in the computation:
$$-x<2x+3$$
implies $x>-1$. In the first case you have $x\ge0$ and $x>-3$ so: $x\ge0$.In the second case you have $x<0$ and $x>-1$ so : $-1<x<0$. Taking the union of the two cases you obtain $x\ge0$ or $-1<x<0$ so : $-1<x.$
A: 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 $x\in (-1,0]\cup (0,\infty) = (-1,\infty)$.
A: 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 $x > -1$
A: 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.
A: The solutions of your inequality can be seen as the union of the solution sets of
\begin{cases}
x\ge0\\
|x|<2x+3
\end{cases}
and
\begin{cases}
x<0\\
|x|<2x+3
\end{cases}
The first system can be written, as $|x|=x$ because of $x\ge0$,
\begin{cases}
x\ge0\\
x<2x+3
\end{cases}
that easily gives
\begin{cases}
x\ge0\\
x>-3
\end{cases}
So the solution set is $[0,\infty)$.
The second system becomes, as $|x|=-x$ because $x<0$,
\begin{cases}
x<0\\
-x<2x+3
\end{cases}
that has as solution set $(-1,0)$.
So the solution set of the initial inequality is
$$
(-1,0)\cup[0,\infty)=(-1,\infty)
$$

A different, but more complicated, strategy is to notice that the inequality is only satisfied if $2x+3\ge0$, so we can square and get
\begin{cases}
x^2<4x^2+12x+9\\
2x+3\ge0
\end{cases}
Rewriting the quadratic as $x^2+4x+3>0$, we get
\begin{cases}
x<-3\text{ or } x>-1\\
x>-3/2
\end{cases}
that reduces to $x>-1$.
