Let $x$ be in the set of real numbers $\mathbb{R}$ and let $f(x)=|2x-1|-3|2x+4|+7$ be a function, write $f(x)$ without the absolute value. Let $x$ be in the set of real numbers $\mathbb{R}$ and let $f(x)=|2x-1|-3|2x+4|+7$ be a function, write $f(x)$ without the absolute value.
I thought of it this way: $$f(x)=\begin{cases}2x-1-3(2x+4)+7 \,(\text{then I simplify)} & \text{if $x>0$}\\
-(2x-1-3(2x+4)+7)\,(\text{then I simplify)} & \text{if $x\le0$}\end{cases}$$
But is there some way without having to use the cases?
Edit: NEW work on this problem! I found three cases;
If $x\in ]-\infty,-2]$ then f(x)=$4x+20$
If $x\in]-2,1/2]$ then f(x)=$-8x-4$ 
If $x\in]1/2,+\infty[$ then f(x)=$-4x-6$
IS THIS TRUE?
Thank you very much!

Let f(x)=|x|.We know $$f(x)=\begin{cases}
-x \text{  if x<0}\\
x\text{ if x>0}
\end{cases}$$
What if f(x)=|x-1|. We must have,
$$f(x)=\begin{cases}
-(x-1) \text{  if x<1}\\
x-1\text{ if x>1}
\end{cases}$$
Note that there are 3 terms in the given function.
$|2x-1|$, $3|2x+4|$ and 7.
The first term is postive if x$\gt \frac{1}{2}$. The second term is positive if $x\gt -2$. Hence we must have,
$$f(x)=2x-1-3(2x+4)+7 \text{   whenever   x$\gt \frac{1}{2}$}$$
The first term is negative if x$\lt \frac{1}{2}$ and the second term is negative if x$\lt -2$.Hence we have a second condition,
$$f(x)=-(2x-1)+3(2x+4)+7 \text{   whenever   x$\lt -2$}$$
Can you work out what will happen if x $\in (-2,\frac{1}{2})$?


! If x $\in (-2,\frac{1}{2})$ surely |2x-1|<0,hence |2x-1|=-(2x-1) and |2x+4|>0,hence |2x+4|=2x+4.
  Finally we have the following for f(x)
  $$f(x)=
\begin{cases}
2x-1-3(2x+4)+7~ \text{, x$\gt \frac{1}{2}$}\\
-(2x-1)+3(2x+4)+7~ \text{, x$\lt -2$}\\
-(2x-1)-3(2x+4)+7 ~\text{, x$\in(-2,\frac{1}{2})$}
\end{cases}
$$ 

