# 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 $x<-\frac{3}{4}$?

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For someone who has to solve a lot of equations of this sort, I would always recommend that the first thing to do is to make a plot of the functions of interest, if only to reckon which intervals one should be looking at. –  Guess who it is. Aug 23 '10 at 22:41
@J. Mangaldan: The plot is also a good way to verify the answers. There are something like 24 signed-number operations involved in my algebraic solution below, so even with 99% accuracy per operation, that's only a 78.6% chance of correct calculations throughout. –  Isaac Aug 23 '10 at 23:32

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).

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This table is too wide to display correctly for me :/ –  Larry Wang Aug 24 '10 at 0:22
@Kaestur Hakarl: Sorry--with the \small{} wrapper, it just barely fits for me, but I couldn't think of any way to make it smaller and still legible. I'm open to suggestions. –  Isaac Aug 24 '10 at 0:33
See my answer: what you propose is a special case of the CAD algorithm. –  Bill Dubuque Aug 24 '10 at 0:49

This is merely a very special case of the powerful CAD (cylindrical algebraic decomposition) algorithm for quantifier elimination in real-closed fields, e.g. see Jirstrand's paper [1] for a nice introduction.

[1] M. Jirstrand. Cylindrical algebraic decomposition - an introduction. 1995
Technical report S-58183, Automatic Control group, Department of Electrical Engineering