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:
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\,$’ is a symbol for ‘is equivalent to’.
The power of the sentence-transforming tool
‘$\,|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:
$|2 - 3x| \gt 7$ | (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) |
$3|-6x + 7| \ge 9$ | (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) |
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.