Absolute value definition Is it true that $\dots$
$$
\left| y \right| = 
\begin{cases}
   y \hspace{1cm} y \geq 0 \\ -y \hspace{0.7cm} y < 0
\end{cases}
$$
I'm a little bit confused with the second case, where $|y| = -y$ then $y<0$, 
for example :
$$
\left| 2x-4 \right|=-(2x-4)
$$
if we assume that $ y=2x-4 $ then 
$$
\begin{align*}
y&<0 \\
2x-4&<0 \\
2x&<4 \\
x&<2
\end{align*}
$$
in the other way, we can solve it like this
$$
\begin{align*}
|y| \geq 0 \\
|2x-4| \geq 0 \\
-(2x-4) \geq 0 \\
2x-4 \leq 0 \\
x \leq 2
\end{align*}
$$
why is it giving the different answers?
 A: You should train to read formulas without reference to specific variables. The definition
$$
|x|=\begin{cases}
x & x\ge0 \\[4px]
-x & x<0
\end{cases}
$$
it is meant that

the absolute value of a number is
  
  
*
  
*the number itself if it is greater than or equal to $0$,
  
*the negative of the number if it is less than $0$.
  

You should also avoid using a variable with two different meanings in the same statement.
It seems that you want to see when $|2x-4|=-(2x-4)$. According to the definition, this happens if and only if


*

*$2x-4<0$, or

*$2x-4=0$.


Why the second case? Because $0=-0$. On the other hand, if $2x-4>0$, then we cannot have $(2x-4)=-(2x-4)$, because one term is positive and the other one is negative.
One might make the initial definition more symmetric by declaring
$$
|x|=\begin{cases}
x & x>0 \\[4px]
0 & x=0 \\[4px]
-x & x<0
\end{cases}
$$
but you can also note that
$$
|x|=\begin{cases}
x & x>0 \\[4px]
-x & x\le0
\end{cases}
$$
would be a completely equivalent definition.
A: Note that in general the definition is
$$\left| f(x) \right| = 
\begin{cases}
   f(x) \hspace{1cm} f(x) \geq 0 \\ -f(x)  \hspace{0.7cm} f(x) < 0
\end{cases}$$
A: The absolute value make a function simmetrical with respect to the $x$-axis.
Then we have pay attention when the function assumes negative value.
In your example we have:
$$|2x−4|=\begin{cases}2x-4 \quad\text{if} \quad 2x-4\ge0\implies x\ge2\\
4-2x\quad \text{if} \quad 2x-4 <0\implies x<2\end{cases}$$
The graphic of your function in fact is:
$\hspace{6cm}$
A: It is a bit confusing to use the same letter $x$ for your substitution.
Let set $y=2x-4$.
Then $|y|$ is $\begin{cases}+y&=2x-4&\quad\text{when}\quad y\ge0\iff 2x-4\ge 0\iff x\ge 2\\-y&=4-2x&\quad\text{when}\quad y<0\iff 2x-4<0\iff x<2\end{cases}$

Indeed for $x=1$ then $x<2$ then $|y|=|2\times 1-4|=|-2|=2=4-2\times 1$
And for $x=3$ then $x\ge 2$ then $|y|=|2\times 3-4|=|2|=2=2\times 3-4$
