2.2 Absolute value inequalities

Here’s one of my favorite problems:

Having seen that the solution to $\mid x\mid =\text{10}$ is $\mid x\mid =\pm \text{10}$ , many students answer this question . However, this is not only wrong: it is, as discussed above, relatively meaningless. In order to approach this question you have to—you guessed it!—step back and think.

Here are two different, perfectly correct ways to look at this problem.

1. What numbers work? 4 works. –4 does too. 0 works. 13 doesn’t work. How about –13? No: if $x=-\text{13}$ then $\mid x\mid =\text{13}$ , which is not less than 10. By playing with numbers in this way, you should be able to convince yourself that the numbers that work must be somewhere between –10 and 10. This is one way to approach the answer.
2. The other way is to think of absolute value as representing distance from 0 . $\mid 5\mid$ and $\mid -5\mid$ are both 5 because both numbers are 5 away from 0. In this case, means “the distance between x and 0 is less than 10”—in other words, you are within 10 units of zero in either direction. Once again, we conclude that the answer must be between –10 and 10.

It is not necessary to use both of these methods; use whichever method is easier for you to understand.

More complicated absolute value problems should be approached in the same three steps as the equations discussed above: algebraically isolate the absolute value, then think , then algebraically solve for $x$ . However, as illustrated above, the think step is a bit more complicated with inequalities than with equations.

Many students will still resist the think step, attempting to figure out “the rules” that will always lead from the question to the answer. At first, it seems that memorizing a few rules won’t be too hard: “greater-than problems always lead to OR answers” and that kind of thing. But those rules will fail you when you hit a problem like the next one.