# Techniques as to solving absolute value equation

Solve absolute value equation with absolute value variable one one side or even both side, without a number outsides of absolute value signs are typically easy.

In my high school, I was taught to first separate the absolute variable and make it it to two different cases with different signs for one side.

But the real problem is: what happen if there is a constant outside of the absolute signs, like: $$|x-5| = |x+5|-1$$

There are three broad ways to approach a problem like this.

1. Draw the graphs of $y = \left | x-5 \right |$ and $y = \left | x + 5 \right | -1$ and see where they intersect so you know which regions you have to check for solutions, then just solve in those regions.

2. Check all possible combinations, as mookid suggests. Note that his step #2 is very important - there are four sets of solutions it will produce, and it's likely that several of them do not solve the original equation (actually this is true for many equations, and it's a good idea to do it at the end of just about any exercise).

3. It's a much uglier option, but you can use the fact that squaring an absolute value lets you remove the absolute value signs to do this:

$\left|x-5\right|=\left|x+5\right|-1 \\ \left(\left|x-5\right|\right)^2=\left(\left|x+5\right|-1\right)^2 \mbox{ (squaring both sides)}\\ x^2-10x+25=\left(\left|x+5\right|\right)^2-2\left|x+5\right|+1 \\ x^2-10x+25=x^2+10x+25-2\left|x+5\right|+1 \\ 20x+1=2\left|x+5\right| \mbox{ (collecting like terms)}\\ \left(\left|20x+1\right|\right)^2=4\left(\left|x+5\right|\right)^2 \mbox{ (squaring again)}\\ 400x^2+40x+1=4x^2+40x+100 \\ 396x^2=99 \\ x^2=\frac{1}{4} \\ x=\pm\frac{1}{2}$

And, once again you need to check both solutions, since one of them does not solve the original equation. Be careful with this approach, since for an arbitrary combination of terms it can result in having to solve a ridiculously large polynomial equation rather than a large number of separate linear equations as would occur if you just break it up into regions.

In this case there are three cases to consider: $x < -5$, $-5 \le x < 5$, and $5 \le x$. For example, if $x < -5$, $|x - 5| = -x + 5$ and $|x+5| = -x-5$ so the equation says $-x + 5 = -x - 5 - 1$. No solution there. Now try the other two cases.

This is exactly the same. The most simple general way to handle this kind of equations is to make this a two step journey:

1. If $|x−5|=|x+5|−1$: There are four cases: \begin{align} x-5 &= x+5 &- 1 &\iff x\in S_1\\ -x+5 &= x+5 &- 1 &\iff x\in S_2\\ x-5 &= -x-5 &- 1 &\iff x\in S_3\\ -x+5 &= -x-5 &- 1 &\iff x\in S_4\\ \end{align} Then you solve for each equation. What you proved in this step is that the set of solutions $S$ is such as $$S \subset \bigcup_i S_i$$

2. For each element of each set $S_i$, check whether this is a solution or not.

To this simple case of addition of an extra number, let me humbly add a fourth and purely graphic/intuitive method to @ConMan's answer.

Draw a line representing the $\mathbb{R}$ numbers. See my drawing:

You have two absolute value parts:

• The $|x-5|$ is the distance from $x$ to the number $5$.
• The $|x+5|=|x-(-5)|$ is the distance from $x$ to $-5$.

So note the numbers $5$ and $-5$ on your line. Your equation $|x-5|=|x+5|-1$ tells us that the distance from $x$ to $5$ must be one less than the distance from $x$ to $-5$.

Point $a$ would be the solution to $|x-5|=|x+5|$. Point $b$ would be the solution to $|x-5|=|x+5|-2$. And the solution to your original equation $|x-5|=|x+5|-1$ is found in the point $x=0.5$, where the difference between the two distances is exactly $1$.