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!

The absolute value has a discontinuous slope change at zero. Hence, f(x) has discontinuous slope changes at -2 from $|2x+4|$ and at 1/2 from $|2x-1|$. So there are three cases: $-\infty < x \le -2$, $-2 < x \le 1/2 $ and $1/2 < x < \infty$.
Now for $x$ sufficiently negative $|2x+4| = -2x-4$ and we know that $|2x+4|$ had discontinuous slope at $2x+4=0$. Similarly for sufficiently negative $x$ $|2x-1| = -2x+1$ and changes slope when $2x-1=0$. Hence
$$
f(x) = \left{ 
 \begin{array}{ll}
    +3(2x+4)-(2x-1)+7, & x  \le -2; \\
    -3(2x+4) -(2x-1)+7, & -2 <x \le 1/2; \\
   -3(2x+4)+(2x-1)+7, & 1/2 < x
\right.
$$
Sorry! Don't see why the formatting is messed up. Hope it makes sense!  I tried to use plain vanilla latex commands. I need to learn MathJaX
