Solving Absolute Value Inequalities Involving ‘Greater 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
 Solving Absolute Value Inequalities Involving ‘Less 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:
$$\begin{gather} \cssId{s6}{x\gt 5} \\ \\ \cssId{s7}{x + 1\ge 3} \\ \\ \cssId{s8}{2  3x \gt 7} \end{gather} $$Each of these inequalities has only a single set of absolute value symbols which is by itself on the lefthand 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:
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 lesstroublesome compound inequality.
Recall that ‘$\iff$’ is a symbol for ‘is equivalent to’.
The power of the sentencetransforming 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
$2  3x \gt 7$  original sentence 
$23x\lt 7$ or $\,23x \gt 7$ 
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
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:
$36x + 7 \ge 9$  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:
$36(\frac{2}{3}) + 7 = 34 + 7 = 33 = 9$
Check!
$36(\frac{5}{3}) + 7 = 310 + 7 = 33 = 9$
Check!
Example
No matter what number you substitute for $\,x\,,$ the lefthand 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.