Absolute Value

Absolute value is a wolf in sheep’s clothing. What could be simpler than this: If a number is zero or more, its absolute value is just itself, and if the number is less than zero, to find its absolute value you change its sign.

And yet, when you put absolute value into an equation it gets confusing.

To find the confusion and find a way around it, let’s start from simple examples and build our way up.

Simple Problems

Consider this problem:

${\color{Blue} \left | a \right |=2}$

$a$ could be any real number, positive or negative. If it’s positive its absolute value is just itself:

${\color{Blue} a=2}$

If it’s negative you change its sign to find its absolute value.

${\color{Blue} a=-2}$

So there are two solutions: $a =\pm 2$ . No big deal, right?

So let’s step it up.

Try this:

${\color{Blue} |x+1|=3}$

Try the same approach as before. If it’s positive its absolute value is just itself:

${\color{Blue} x+1=3}$

${\color{Blue} x=2}$

And if it’s negative you change its sign to find its absolute value:

${\color{Blue} |x+1|=3}$

${\color{Blue} x+1=-3}$

${\color{Blue} x=-4}$

So it looks like the two solutions are $x=-4,$ $x=2.$ Check them:

For $x=2, |x+1|=3.$

For $x=-4, |x+1|=3.$

So that works.

Try something more challenging

${\color{Blue} |4x-3|-2=1}$

This one has another term on the same side of the equation as the absolute value term. The wolf is at the door.

Don’t mess with the contents of the absolute value bars until you have isolated the absolute value term. Failure to isolate the absolute value term is where most absolute value confusion comes from.

In an absolute value equation, if there is anything else on the same side of the equation as the absolute value term, the first step is to isolate the absolute value term.

In this case, add 2 to both sides of the equation:

${\color{Blue} |4x-3|=3}$

Now you can apply the absolute value definition: If a number is zero or more, its absolute value is just itself, and if the number is less than zero, to find its absolute value you change its sign. Assume both cases can happen.

And now you can solve:

4x-3=3	4x-3=-3
4x=6	4x=0
x=3/2	x=0

And check:

x=3/2	x=0
4x-3=4*3/2-3	4x-3=4*0/2-3
4x-3=3	4x-3=-3
\|4x-3\|=3	\|4x-3\|=3

So that works. And that’s the gist of it. If you’re doing absolute value for Algebra I, that should take care of absolute value equations for you. (You may also learn about absolute value inequalities. That’s a whole nother topic.) If you’re doing more advanced math, or if you’re curious, read on.

A More Complex Problem

To step it up yet one level more, try two absolute value expressions in one equation:

${\color{Blue} |2x+1|=|x-3|+1}$

I told you to isolate the absolute value expression on one side of the equation. But you can’t isolate both absolute value expressions.

A compromise is called for. Why do you have to isolate each absolute value expression? So you can separate the part of the equation that changes sign. Let’s look at that more closely.

Consider the equation $|a|=b.$ If $a$ is positive or equal to zero, that means $a=b.$ If $a$ is negative, we usually say $a=-b.$ But that amounts to the same thing as $-a=b.$ In other words, we are considering the possibility that $a$ is negative, so absolute value changes its sign, so we replace $|a|$ with $-a.$

What if you’re working with something like $|a|=|b|+c?$ Now $a$ and $b$ each have two possible values, one positive and one negative. Instead of the two possibilities that you have with $|a|=b,$ there are now four possibilities: $a$ and $b$ both positive, $a$ positive and $b$ negative, $a$ negative and $b$ positive, and $a$ and $b$ both negative.

You have to find a solution for each of those cases.

Back to $|2x+1|=|x-3|+1.$ Let’s get both absolute value expressions on the left. (This step is not mathematically necessary, but I think it’s a good way to organize the problem.)

${\color{Blue} |2x+1|=|x-3|+1}$

${\color{Blue} |2x+1|-|x-3|=1}$

Case 1: Both 2x+1 and x-3 are positive.

That means $|2x+1|=2x+1$ and $|x-3|=x-3.$ So the equation becomes:

${\color{Blue} (2x+1)-(x-3)=1}$

${\color{Blue} x=-3}$

Case 2: 2x+1 is positive and x-3 is negative.

Substitute $-(x-3)$ for $|x-3|$ and the equation becomes:

${\color{Blue} (2x+1)-(-(x-3))=1}$

${\color{Blue} (2x+1)+(x-3)=1}$

${\color{Blue} x=1}$

Case 3: 2x+1 is negative and x-3 is positive.

Substitute $-(2x+1)$ for $|2x+1|$ and $-(x-3)$ for $|x-3|,$ and the equation becomes:

${\color{Blue} -(2x+1)-(x-3)=1}$

${\color{Blue} x=1/3}$

Case 4: 2x+1 and x-3 are both negative.

Substitute $-(2x+1)$ for $|2x+1|$ and $-(x-3)$ for $|x-3|,$ and the equation becomes:

${\color{Blue} -(2x+1)-(-(x-3))=1}$

${\color{Blue} x=-5}$

Now check them:

For $x=-3, |2x+1|-|x-3|=-1.$

For $x=1, |2x+1|-|x-3|=1.$

For $x=1/3, |2x+1|-|x-3|=-1.$

For $x=-5, |2x+1|-|x-3|=1.$

Looks like only two of the solutions we found $x=1$ and $x=-5,$ work. The other two apparent solutions, $x=-3$ and $x=1/3,$ must be extraneous.

What gives?

Remember our assumptions: For example, in case 3 we assumed $x-3$ was positive, so $|x-3|=x-3.$ But the solution we got, $x=-3,$ makes $x-3=-6,$ a negative number, and then the assumption that $x-3$ is positive is no longer valid. The assumption is invalid, so the solution is extraneous. Let’s check all our solutions that way:

Case	1		2		3		4
	x=-3		x=1		x=1/3		x=-5
	2x+1	x-3	2x+1	x-3	2x+1	x-3	2x+1	x-3
Value	-5	-6	3	0	5/3	-8/3	-9	-8
Was it assumed to be positive or negative?	pos	pos	pos	neg	neg	pos	neg	neg
Does the value match the assumption?	no	no	yes	yes	no	no	yes	yes

Where the assumptions are not met, the solutions do not work. Where they are, they do.

Does this mean you always need to go back and check your assumptions? No, that assumption check was just me explaining why things may go wrong. But you do need to check your answers.

This sort of thing happens only in complex problems like this one. Compare a case like $|4x-3|-2=1$ above. Then $|4x-3|=3.$ As discussed above, the two possibilities are that $4x-3=3$ and that $4x-3=-3.$ There is no way to violate the assumptions that $4x-3$ is positive in the first case and negative in the second. It’s only when more variable terms come in that things can go wrong in this way.