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Inequalities
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Concept Version 9
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Inequalities with Absolute Value

Inequalities with absolute values can be solved by thinking about absolute value as a number's distance from 0 on the number line.

Learning Objective

  • Solve inequalities with absolute value


Key Points

    • Problems involving absolute values and inequalities can be approached in at least two ways: through trial and error, or by thinking of absolute value as representing distance from 0 and then finding the values that satisfy that condition.
    • When solving inequalities that involve an an absolute value within a larger expression (for example, $\abs{2x} + 3>8$), it is necessary to algebraically isolate the absolute value and then algebraically solve for the variable.

Terms

  • absolute value

    The magnitude of a real number without regard to its sign; formally, -1 times a number if the number is negative, and a number unmodified if it is zero or positive.

  • inequality

    A statement that of two quantities one is specifically less than or greater than another.

  • number line

    A line that graphically represents the real numbers as a series of points whose distance from an origin is proportional to their value.


Full Text

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.

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