What is the best way to solve an equation involving multiple absolute values? An absolute value expression such as $|ax-b|$ can be rewritten in two cases as $|ax-b|=\begin{cases}
ax-b & \text{ if } x\ge \frac{b}{a} \\ 
b-ax & \text{ if } x< \frac{b}{a} 
\end{cases}$, so an equation with $n$ separate absolute value expressions can be split up into $2^n$ cases, but is there a better way?
For example, with $|2x-5|+|x-1|+|4x+3|=13$, is there a better way to handle all the possible combinations of $x\ge\frac{5}{2}$ versus $x<\frac{5}{2}$, $x\ge 1$ versus $x< 1$, and $x\ge-\frac{3}{4}$ versus $x<-\frac{3}{4}$?

For an equation with $n$ absolute values, the $n$ places where each absolute value splits into 2 cases divide the number line into $n+1$ regions where, within each region, each absolute value can be replaced by either the expression inside the absolute value or its opposite. Each resulting equation can then be solved, restricting solutions to the corresponding region on the number line.
In the given example,
$\small{\begin{matrix}
\leftarrow  & -\frac{3}{4} & \text{---}   & 1 & \text{---} & \frac{5}{2} & \rightarrow
\\
\begin{matrix}-(2x-5)-(x-1)\\-(4x+3)=13\end{matrix}
& \begin{matrix}|\\|\end{matrix} &
\begin{matrix}-(2x-5)-(x-1)\\+(4x+3)=13\end{matrix} 
& \begin{matrix}|\\|\end{matrix} &
\begin{matrix}-(2x-5)+(x-1)\\+(4x+3)=13\end{matrix} 
& \begin{matrix}|\\|\end{matrix} &
\begin{matrix}(2x-5)+(x-1)\\+(4x+3)=13\end{matrix} 
\\
-7x+3=13
& | &
x+9=13
& | &
3x+7=13
& | &
7x-3=13
\\
x=-\frac{10}{7}
& | &
x=4\notin[-\frac{3}{4},1]
& | &
x=2
& | &
x=\frac{16}{7}\notin[\frac{5}{2},\infty)
\end{matrix}}$
So, the solutions are $x=-\frac{10}{7}$ and $x=2$ (the values that were solutions to an equation for a particular region and were within that region).
