audio read-through Solving Absolute Value Inequalities Involving ‘Less Than’

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

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:

$$\begin{gather} \cssId{s12}{|x|\lt 5} \\ \\ \cssId{s13}{|x + 1|\le 3} \\ \\ \cssId{s14}{|2 - 3x| \lt 7} \end{gather} $$

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:

THEOREM solving absolute value inequalities involving ‘less than’
Let $\,x\in\mathbb{R}\,$, and let $\,k\ge 0\,.$   Then: $$ \begin{gather} \cssId{s27}{|x| \lt k\ \ \ \text{ is equivalent to }\ \ -k \lt x \lt k} \\ \\ \cssId{s28}{|x| \le k\ \ \ \text{ is equivalent to }\ \ -k \le x \le k} \\ \end{gather} $$

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:

$$ \cssId{s43}{\overset{\text{ first piece }}{\overbrace{-k \lt x}}\ \ \lt k} $$ $$ \cssId{s44}{-k \lt\overset{\text{second piece}}{\overbrace{x\lt k}}} $$

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 $\,k\gt 0\,$)

values of x between -k and k, for k greater than zero

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

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

The power of the sentence-transforming tool

$$\cssId{s70}{|x| \lt k \ \iff\ -k \lt x \lt k}$$

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:

$$\cssId{s75}{|\text{stuff}| \lt k \ \iff\ -k \lt \text{stuff} \lt k}$$

See how this idea is used in the following examples:

Example

Solve: $|2 - 3x| \lt 7$
Solution: Write a nice, clean list of equivalent sentences:
$|2 - 3x| \lt 7$ original sentence
$-7 \lt 2-3x \lt 7$ check that $\,k\gt 0\,$; use the theorem
$-9 \lt -3x \lt 5$ subtract $\,2\,$ from all three parts of the compound inequality
$\displaystyle 3 \gt x \gt -\frac{5}{3}$ divide all three parts by $\,-3\,$; change direction of inequality symbols
$\displaystyle -\frac{5}{3} \lt x \lt 3$ write in the conventional way

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

Solve: $3|-6x + 7| \le 9$
Solution: To use the theorem, you must have the absolute value all by itself on one side of the sentence. Thus, your first job is to isolate the absolute value:
$3|-6x + 7| \le 9$ original sentence
$|-6x + 7| \le 3$ divide both sides by $\,3$
$-3 \le -6x + 7 \le 3$ check that $\,k \ge 0\,$; use the theorem
$-10 \le -6x \le -4$ subtract $\,7\,$ from all three parts of the compound inequality
$\displaystyle \frac{10}{6} \ge x \ge \frac{4}{6}$ divide all three parts by $\,-6\,$; change direction of inequality symbols
$\displaystyle \frac{2}{3} \le x \le \frac{5}{3}$ simplify fractions; write in the conventional way

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

Solve: $|5 - 2x| \lt -3$
Solution: The theorem can't be used here, since $\,k\,$ is negative. In this case, you need to stop and think.

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.

Concept Practice

Solve the given absolute value inequality. Write the result in the most conventional way.

For more advanced students, a graph is available. For example, the inequality $\,|2 - 3x| \lt 7\,$ is optionally accompanied by the graph of $\,y = |2 - 3x|\,$ (the left side of the inequality, dashed green) and the graph of $\,y = 7\,$ (the right side of the inequality, solid purple). In this example, you are finding the values of $\,x\,$ where the green graph lies below the purple graph.

Click the ‘Show/Hide Graph’ button to toggle the graph.


Solve: