SOLVING ABSOLUTE VALUE INEQUALITIES INVOLVING ‘LESS THAN’

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: $$\begin{gather} |x|\lt 5 \\ |x + 1|\le 3 \\ |2 - 3x| \lt 7 \end{gather} $$

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, $$ \begin{gather} |x| \lt k\ \ \ \text{ is equivalent to }\ \ -k \lt x \lt k \\ |x| \le k\ \ \ \text{ is equivalent to }\ \ -k \le x \le k \\ \end{gather} $$
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\ $’,   and   ‘$\,|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\ \text{ and }\ x < k\ $’.
Here's where those pieces are coming from: $$ \overset{\text{ first piece }}{\overbrace{-k \lt x}}\ \ \lt k $$ $$ -k \lt\overset{\text{second piece}}{\overbrace{x\lt k}} $$ 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 $\,k\gt 0\,$)

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   ‘$\,|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:   ‘$\,|\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 equation.
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.
Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Solving Absolute Value Inequalities
Involving ‘Greater Than’

 
 

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

For more advanced students, a graph is displayed.
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 if you prefer not to see the graph.

CONCEPT QUESTIONS EXERCISE:
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
PROBLEM TYPES:
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AVAILABLE MASTERED IN PROGRESS

Solve: