Absolute Value Inequalities

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Instructor: Dr. Jo Steig

If you have not yet read through the Supplementary Reading on Absolute Value, it would be helpful to do so now before continuing.

The most common algebraic defintion of absolute value is as follows: 

ALGEBRAIC DEFINITION OF ABSOLUTE VALUE |a| = a , if a > 0 - a, if a < 0 That is, the absolute value of a number, a, is equal to the distance that 'a' is from zero.

There are two common types of inequalities involving absolute value that first crop up in algebra classes. They are:

What follows is a description of a way to look at these inequalities that than be used to solve both types. It moves away from simple memorization of a process (that is often forgotten or misapplied) to incorporate both the algebraic definition and the geometric properties of abolute value. Both are then applied to solve the inequalities.

This is NOT the only way to solve the inequalities. The power of this approach, however, is that it is allows the information contained within the absolute value defintion and a graph of that information to direct our steps. Memorization is not necessary. Rather, we rely on understanding.

TYPE 1: | a | < positive constant

| a | < 5 is read, "the absolute value of a is less than 5"

The solution of this inequality is the set of all real numbers whose distance from 0 on the number line is less than 5 units.

Graphically, this solution would look like the following:

That is, a represents all real numbers between -5 and 5. Algebraically, we can write this as:

- 5 < a < 5

Since this process is not dependent upon the expression inside the absolute value symbols, we can apply it to the following example.

Example 1: Solve the inequality: | 4x - 2 | < 5

Algebraically: Find all values of x so that 4x - 2 corresponds to a number that is less than 5 units from 0 on the number line.

Graphically:

That is,

- 5 < 4x - 5 < 5

- 3 < 4x < 7

-3/4 < x < 7/4

In interval notation, the solution would be: ( -3/4, 7/4).

Note: Since the graph drawn in step 2 contains only 1 segment, only one algebra statement (step 3) is needed to describe it.  

Summary of our steps: (1.) Read the inequality as a "distance from 0" statement (2.) Draw a line graph representing the distance statement formed in (1.) (3.) Translate the information contained on the graph to an algebra inequality statement that does not use absolute values. (4.) Solve the inequalities created in (3.)

And now let's apply this same graphical/definition approach to solving an inequality of the second type.

TYPE 2: | a | > positive constant

| a | > 5 is read, "the absolute value of a is more than 5"

The solution of this inequality is the set of all real numbers whose distance from 0 on the number line is more than 5 units.

Graphically, this solution would look like the following:

That is, a represents all real numbers that are either less than -5 or are greater than 5. Algebraically, we can write this as:

a < - 5 OR a > 5

Note: TWO statements are needed to describe the graph since it contains TWO segments. The graph CANNOT be described using one statement.

WARNING>>>>>>>>>>>> - 5 > a > 5 is NOT CORRECT!!!!!

Since this process is not dependent upon the expression inside the absolute value symbols, we can apply it to the following example.

 Example 2: Solve the inequality: | 3x + 1 | > 5

Algebraically: Find all values of x so that 3x + 1 corresponds to a number that is more than 5 units from 0 on the number line.

Graphically:

That is,

3x + 1 < - 5 OR 3x + 1 > 5

3x < - 6 OR 3x > 4

-x < - 2 OR x > 4/3

In interval notation, the solution would be: (- infinity, - 2) (4/3 , infinity).

 

Two trivial cases: I really should mention two trivial cases of inequalities that occur when

| a | < negative constant, and when

| a | > negative constant  

Consider: | a | < negative constant

This says that 'a' has a distance from zero that is is less than a negative number. Since distance cannot be less than a negative there is no solution.

Consider | a | > negative constant

This says that 'a' has a distance from zero that is greater than a negative number. But all distances are either 0 or positive therefore all numbers have a distance from 0 that is greater than a negative. The solution includes all real numbers.

Remember that your active participation is required before any of this has a chance to make an sense to you - it is within your mental context that learning must take place so you must ask why and why not. As you go back to your text and see if you can put into practice the steps that are outlined here when working any of these types of problems, ask questions and demand answers.

 

© 1999 Jo Steig

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