SOLVING ABSOLUTE VALUE INEQUALITIES INVOLVING ‘GREATER THAN’

This section should feel remarkably similar to the previous ones.

This section presents the tools needed
to solve absolute value inequalities involving ‘greater than’, like these: $$\begin{gather} |x|\gt 5 \\ |x + 1|\ge 3 \\ |2 - 3x| \gt 7 \end{gather} $$

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   ‘$\,\gt\,$’ (greater than)   or   ‘$\,\ge\,$’ (greater than or equal to).

As in the previous sections, 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 ‘greater than’
Let $\,x\in\mathbb{R}\,$, and let $\,k\ge 0\,$.   Then, $$ \begin{gather} |x| \gt k\ \ \ \text{ is equivalent to }\ \ x\lt -k\ \ \text{ or }\ \ x\gt k \\ |x| \ge k\ \ \ \text{ is equivalent to }\ \ x\le -k\ \ \text{ or }\ \ x\ge k \\ \end{gather} $$
TRANSLATING THE THEOREM

Recall first that normal mathematical conventions dictate that   ‘$\,|x| \gt k\ $’   represents an entire class of sentences,
including the members   ‘$\,|x| \gt 2\ $’,   ‘$\,|x| \gt 5.7\ $’,   and   ‘$\,|x| \gt \frac{1}{3}\,$’.
The variable $\,k\ $ changes from sentence to sentence, but is constant within a given sentence.

Recall that   ‘$\,x\lt -k\ \ \text{ or }\ \ x\gt k\ $’   is an   ‘or’   sentence, where that's the mathematical word ‘or’.
An ‘or’ sentence is true when at least one of the subsentences is true.
Thus, the sentence ‘$\,x\lt -k\ \ \text{ or }\ \ x\gt k\ $’ is true
for all the numbers to the left of $\,-k\,$, put together with all the numbers to the right of $\,k\ $:

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

When you see a sentence of the form $\,|x| \gt 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| \gt k \iff x\lt -k\ \ \text{ or }\ \ x\gt k\ $’
goes far beyond solving simple sentences like $\,|x| \gt 5\,$!

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}| \gt k \iff \text{stuff}\lt -k\ \ \text{ or }\ \ \text{stuff}\gt k\ $’

See how this idea is used in the following examples:

EXAMPLE:
Solve: $|2 - 3x| \gt 7$
Solution:
Write a nice, clean list of equivalent sentences:
$|2 - 3x| \gt 7$ (original sentence)
$2-3x\lt -7\ \ \text{or}\ \ 2-3x \gt 7$ (check that $\,k\ge 0\,$; use the theorem)
$-3x\lt -9\ \ \text{or}\ \ -3x \gt 5$ (subtract $\,2\,$ from both sides of both subsentences)
$\displaystyle x\gt 3\ \ \text{or}\ \ x\lt -\frac{5}{3}$ (divide by $\,-3\,$; change direction of inequality symbols)
$\displaystyle x\lt -\frac{5}{3}\ \ \text{or}\ \ x\gt 3$ (in the web exercise, the ‘less than’ part is always reported first)
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: $3|-6x + 7| \ge 9$
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:
$3|-6x + 7| \ge 9$ (original sentence)
$|-6x + 7| \ge 3$ (divide both sides by $\,3$)
$-6x + 7 \le -3\ \ \text{or}\ \ -6x + 7\ge 3$ (check that $\,k \ge 0\,$; use the theorem)
$-6x\le -10\ \ \text{or}\ \ -6x\ge -4$ (subtract $\,7\,$ from both sides of both subsentences)
$\displaystyle x\ge\frac{10}{6}\ \ \text{or}\ \ x\le \frac{4}{6}$ (divide by $\,-6\,$; change direction of inequality symbols)
$\displaystyle x\ge\frac{5}{3}\ \ \text{or}\ \ x\le \frac{2}{3}$ (simplify fractions)
$\displaystyle x\le \frac{2}{3}\ \ \text{or}\ \ x\ge\frac{5}{3}$ (in the web exercise, the ‘less than’ part is always reported first)
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: $|5 - 2x| \gt -3$
Solution:
The theorem can't be used here, since $\,k\,$ is negative.
In this case, you need to stop and think.
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 will always be greater than $\,-3\,$.
Therefore, this sentence has all real numbers as solutions. It is always true.
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 Sentences, All Types

 
 
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.

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 $\,|2 - 3x| \gt 7\,$
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 above the purple graph.
Click the “show/hide graph” button if you prefer not to see the graph.

Solve: