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:
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\ $:
When you see a sentence of the form
$\,x \gt k\ $, here's what you should do:
Recall that ‘$\,\iff\,$’ is a symbol for ‘is equivalent to’.
The power of the sentencetransforming tool
‘$\,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:
$2  3x \gt 7$  (original sentence) 
$23x\lt 7\ \ \text{or}\ \ 23x \gt 7$  (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) 
$36x + 7 \ge 9$  (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) 
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.
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:
