# Solving Absolute Value Inequalities Involving ‘Greater Than’

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

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:

$$\begin{gather} \cssId{s6}{|x|\gt 5} \\ \\ \cssId{s7}{|x + 1|\ge 3} \\ \\ \cssId{s8}{|2 - 3x| \gt 7} \end{gather}$$

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:

$$\begin{gather} \cssId{s22}{|x| \gt 2} \cr\cr \cssId{s23}{|x| \gt 5.7} \cr\cr \cssId{s24}{|x| \gt \frac{1}{3}} \end{gather}$$

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\,$ 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:

• Check that $\,k\,$ is a nonnegative number (greater than or equal to zero).
• The symbol $\,|x|\,$ represents the distance between $\,x\,$ and $\,0\,.$
• Thus, you want the numbers $\,x\,,$ whose distance from $\,0\,$ is greater than $\,k\,$:
• You can walk from $\,0\,$ in two directions: to the left, or to the right. Walk more than $\,k\,$ units to the left, and you get all the numbers to the left of $\,-k\,.$ Walk more than $\,k\,$ units to the right, and you get all the numbers to the right of $\,k\,.$ Throw these two sets together!
• Thus,  $\,|x| \gt k\$  is equivalent to  $\,x\lt -k\,$ or $\,x\gt k\,.$
• 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

$$\cssId{s46}{|x| \gt k \ \ \iff\ \ x\lt -k\ \ \text{ or }\ \ x\gt k}$$

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: $$\cssId{s50}{|\text{stuff}| \gt k \ \ \iff\ \ \text{stuff}\lt -k\ \ \text{ or }\ \ \text{stuff}\gt k}$$

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 $2-3x\lt -7$       or $\,2-3x \gt 7$ check that $\,k\ge 0\,$; use the theorem $-3x\lt -9$       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 $\,3$ $-6x + 7 \le -3$       or $\,-6x + 7\ge 3$ check that $\,k \ge 0\,$; use the theorem $-6x\le -10$       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.

## 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 $\,|2 - 3x| \gt 7\,$ 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 to toggle the graph.

Solve: