Solving Absolute Value Inequalities Involving ‘Less Than’

Need some simpler practice with absolute value and related concepts first?

• Simplifying Basic Absolute Value Expressions
• Solving Simple Absolute Value Sentences
• Solving Sentences involving ‘Plus or Minus’
• Solving Absolute Value Equations

This section should feel remarkably similar to the previous one.

Instead of solving absolute value equations, this section presents the tools needed to solve absolute value inequalities involving ‘less than’, like these:

Each of these inequalities has only a single set of absolute value symbols which is by itself on the left-hand side of the sentence, and has a variable inside the absolute value. The verb is either  ‘$\,\lt\,$’ (less than)   or  ‘$\,\le\,$’ (less than or equal to).

As in the previous section, solving sentences like these is easy, if you remember the critical fact that absolute value gives distance from $\,0\,.$

Keep this in mind as you read the following theorem:

Translating the Theorem

Recall first that normal mathematical conventions dictate that   ‘$\,|x| \lt k\ $’   represents an entire class of sentences, including the members:

$|x| \lt 2$
$|x| \lt 5.7$
$|x| \lt \frac{1}{3}$

The variable $\,k\ $ changes from sentence to sentence, but is constant within a given sentence.

Also recall that  ‘$\,-k \lt x \lt k\ $’   is a shorthand for  ‘$\,-k \lt x\,$ and $\,x \lt k\,$’. Here's where those pieces are coming from:

Thus,   ‘$\,-k \lt x \lt k\ $’   is an   ‘and’   sentence in disguise, where that's the mathematical word ‘and’.

The sentence   ‘$\,-k \lt x \lt k\ $’   is a great shorthand, because when you look at it, you see $\,x\,$ ‘smushed between’ $\,-k\ $ and $\,k\ $; and these are precisely the values of $\,x\,$ that make this sentence true:

The values of $\,x\,$ that make the sentence ‘$\,-k \lt x \lt k\ $’ true (for $\,k\gt 0\,$)

When you see a sentence of the form $\,|x| \lt k\ ,$ here's what you should do:

• Check that $\,k\,$ is a positive number (greater than zero). (If $\,k\ $ is zero, there aren't any solutions, so it's kind of boring.)
• The symbol $\,|x|\,$ represents the distance between $\,x\,$ and $\,0\,.$
• Thus, you want the numbers $\,x\,,$ whose distance from $\,0\,$ is less than $\,k\ $:
  
• You can walk from $\,0\,$ in two directions: to the right, or to the left. Walk to the right a distance less than $\,k\,,$ and you get all the numbers from $\,0\,$ to $\,k\,.$ Walk to the left a distance less than $\,k\,,$ and you get all the numbers from $\,-k\,$ to $\,0\,.$ So, you want all the numbers between $\,-k\ $ and $\,k\ $:
  
• Thus,   $\,|x| \lt k\ $   is equivalent to   $\,-k \lt x \lt k\,.$
• Equivalent sentences are completely interchangeable, and you can use whichever is easiest to work with. In this case, you're getting rid of the troublesome absolute value in exchange for a less-troublesome compound inequality.

Recall that  ‘$\,\iff\,$’   is a symbol for ‘is equivalent to’.

The power of the sentence-transforming tool

goes far beyond solving simple sentences like $\,|x| \lt 5\,$!

Since $\,x\,$ can be any real number, you should think of $\,x\,$ as merely representing the stuff inside the absolute value symbols. Thus, you could think of rewriting the tool as:

See how this idea is used in the following examples:

Example

It's a good idea to check the ‘boundaries’ of the solution set:

$|2 - 3(-\frac{5}{3})| = |2 + 5| = 7$
Check!

$|2 - 3(3)| = |2 - 9| = |-7| = 7$
Check!

Example

Check the ‘boundaries’ of the solution set:

$3|-6(\frac{2}{3}) + 7| = 3|-4 + 7| = 3|3| = 9$
Check!

$3|-6(\frac{5}{3}) + 7| = 3|-10 + 7| = 3|-3| = 9$
Check!

Example

Can absolute value ever be negative?  No! No matter what number you substitute for $\,x\,,$ the left-hand side of the inequality will always be a number that is greater than or equal to zero, so it can't possibly be less than $\,-3\,.$ Therefore, this sentence has no solutions. It is always false.