Consider the following inequality that includes an absolute value:

$|x| < 10$

Knowing that the solution to $\abs{x}=10$ is $x = ± 10$, many students answer this question $x < ± 10$. However, this is wrong.

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

Trial and Error

What numbers work? That is to say, for what numbers is $\abs{x} < 10$ a true statement? Let's test some out. 

4 works. -4 does too. 13 doesn't work. How about -13? No: If $x = -13$, then $\abs{x} = 13$, 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 finding the answer.

Absolute Value as Distance

The other way is to think of absolute value as representing distance from 0. $\abs{5}$and $\abs{-5}$ are both 5 because both numbers are 5 away from 0. 

In this case, $\abs{x} < 10$ 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.

This answer can be visualized on the number line as shown below, in which all numbers whose absolute value is less than 10 are highlighted.

Solution to $\abs{x} < 10$

All numbers whose absolute value is less than 10.

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

Solving Inequalities with Absolute Value

More complicated absolute value problems should be approached in the same way as equations with absolute values: algebraically isolate the absolute value, and then algebraically solve for $x$. 

For example, consider the following inequality:

 $\abs{2x} + 3>8$

It is difficult to immediately visualize the meaning of this absolute value, let alone the value of $x$ itself. It is necessary to first isolate the inequality:

$\begin{aligned} \abs{2x} + 3 - 3 &> 8 - 3 \\ \abs{2x} &> 8 \end{aligned}$

Now think about the number line. In those terms, this statement means that the expression $2x$ must be more than 8 places away from 0. Therefore, it must be either greater than 8 or less than -8. Expressing this with inequalities, we have:

$2x>8$ or $2x < -8$

We now have 2 separate inequalities. If each one is separately solved for $x$, we will see the full range of possible values of $x$. Consider them independently. First:

$\begin{aligned} 2x &>8 \\ \dfrac{2x}{2} &> \dfrac{8}{2} \\ x &> 4 \end{aligned} $

Second:

$\begin{aligned} 2x &< -8 \\ \dfrac{2x}{2} &< \dfrac{-8}{2} \\ x &< -4 \end{aligned}$

We now have two ranges of solutions to the original absolute value inequality: 

$x > 4$ and $x < -4$

This can also be visually displayed on a number line: 

Solution to $\abs{2x} + 3>8$

The solution is any value of $x$ less than -4 or greater than 4.

Example 

Solve the following inequality:

$\abs{x-2} + 10 > 7$

First, algebraically isolate the absolute value:

$\begin{aligned} \abs{x-2} + 10 - 10 &> 7 - 10 \\ \abs{x-2} &> - 3 \end{aligned}$

Now think: the absolute value of the expression is greater than –3. What could the expression be equal to? 2 works. –2 also works. And 0. And 7. And –10. Absolute values are always positive, so the absolute value of anything is greater than –3! All numbers therefore work.