Help with a system of inequalities with absolute values I'm trying to solve this system on inequalities
$$
\left\{ 
\begin{array}{c}
|x-3|<2x \\ 
|2x+5|>3 
\end{array}
\right. 
$$
The steps I'm taking are:
Finding the absolute values sings, so for 
$x-3 \geq 0$ we have  $x \geq 3$ therefore
$$|x-3| = 
\left\{ 
\begin{array}{c}
x-3 & \text{for $x \geq 3$} \\ 
-x+3 & \text{for $x < 3$}  
\end{array}
\right. $$ 
and 
$2x+5 \geq 0$ we have  $x \geq \frac{-2}{5}$ therefore
$$|2x+5| = 
\left\{ 
\begin{array}{c}
2x+5 & \text{for $x \geq \frac{-2}{5}$} \\ 
-2x-5 & \text{for $x<\frac{-2}{5}$} \\ 
\end{array}
\right. $$ 
So I build a few systems with the complete inequalities, for the first one we have:
$$
\left\{ 
\begin{array}{c}
x \geq 3  \\ 
x-3<2x = x>-3 
\end{array}
\right. $$ 
So the solution here would be $x>3$, then:
$$
\left\{ 
\begin{array}{c}
x<3  \\ 
-x+3<2x = x>1 
\end{array}
\right. $$ 
The solution would be $1<x<3$. Then
$$
\left\{ 
\begin{array}{c}
x \geq \frac{-2}{5}  \\ 
2x+5>3 = x>-1
\end{array}
\right. $$ 
So the solution of the system is $x>-1$, then
$$
\left\{ 
\begin{array}{c}
x< \frac{-2}{5}  \\ 
-2x-5>3 = x<-4
\end{array}
\right. $$ 
And the solution is $x<-4$
Now the solution my book gives is x>1 for the initial system. But I can't find that one. I can't get a solution at all. I tried finding a common point between the 4 solutions I found (as if it was a 4-inequalities system), but there isn't one really. What am I doing wrong?

We have $$|x-3|<2x\implies 3x^2+6x-9>0\implies (x+3)(x-1)>0$$ and $$|2x+5|>3\implies 4x^2+20x+16>0\implies (x+4)(x+1)>0$$. I think you could proceed now? For a quick method on how to solve further, see here
