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?
 A: Since $2x>|x-3|$ we get $x>0$ so $2x+5>0$ and so $2x+5>3$ so $x>-2$ which is nothing new. So $x>0$ and $|x-3|<2x$ so after squaring we get $$ x^2-6x+9<4x^2\implies 3x^2+6x-9>0$$
or $$(x+3)(x-1)>0\implies x-1>0$$ so $\boxed{x>1}$.
A: 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
A: Since $0\le\left|x-3\right|$ and $\left|x-3\right|\le2x$ we conclude $2x\ge0$ and then $x\ge0$.
From $\left|x-3\right|<2x$ and since $x\ge0$, we get $-2x< x-3<2x$ and therefore we have $\left\{\begin{array}{c}x-3<2x \\x-3>-2x\end{array}\right.$ and therefore $\left\{\begin{array}{c}x>-3 \\3x>3\end{array}\right.$ and then $\left\{\begin{array}{c}x>-3 \\x>1\end{array}\right.$, so $x>1$ which satisfies the initial statement $x\ge0$.
So far we see $x$ has to be greater than $1$ to satisfy  $\left|x-3\right|<2x$.
 Let us consider the next inequality $|2x+5|>3$ which leads to $2x+5>3$ or $2x+5<-3$. Then we get $x>-1$ or $x<-4$. And finally if we combine this with the result of first inequality, $x>1$ we get $x>1$.
A: There is actually only the first inequality to be solved:
$$ |2x+5| \stackrel{\color{blue}{0}\leq|x-3| \color{blue}{< 2x}}{>} |x-3|+5 > 3$$
So, you only need to solve $|x-3| < 2x$ while $x>0$:
$$
-2x < x-3 < 2x \Leftrightarrow
\begin{cases}
 3x > 3 \Leftrightarrow  \color{blue}{x> 1} \\
 x > -3 \mbox{ does not extend the solution}
\end{cases}
$$
