SOLVING ABSOLUTE VALUE SENTENCES, ALL TYPES
THEOREM solving absolute value sentences
Let [beautiful math coming... please be patient] $\,x\in\mathbb{R}\,$, and let [beautiful math coming... please be patient] $\,k\ge 0\,$.   Then, [beautiful math coming... please be patient] $$ \begin{gather} |x| = k\ \ \text{ is equivalent to }\ \ x = \pm k \\ \\ |x| \lt k\ \ \ \text{ is equivalent to }\ \ -k \lt x \lt k \\ |x| \le k\ \ \ \text{ is equivalent to }\ \ -k \le x \le k \\ \\ |x| \gt k\ \ \ \text{ is equivalent to }\ \ x\lt -k\ \ \text{ or }\ \ x\gt k \\ |x| \ge k\ \ \ \text{ is equivalent to }\ \ x\le -k\ \ \text{ or }\ \ x\ge k \end{gather} $$
EXAMPLE (an absolute value equation):
Solve: [beautiful math coming... please be patient] $|2 - 3x| = 7$
Solution:
Write a nice, clean list of equivalent sentences:
[beautiful math coming... please be patient] $|2 - 3x| = 7$ (original equation)
[beautiful math coming... please be patient] $2-3x = \pm 7$ (check that $\,k\ge 0\,$; use the theorem)
[beautiful math coming... please be patient] $2-3x = 7\ \text{ or }\ 2-3x = -7$ (expand the plus/minus)
[beautiful math coming... please be patient] $-3x = 5\ \text{ or }\ -3x = -9$ (subtract $\,2\,$ from both sides of both equations)
[beautiful math coming... please be patient] $\displaystyle x = -\frac{5}{3}\ \text{ or } x = 3$ (divide both sides of both equations by $\,-3\,$)
It's a good idea to check your solutions:

[beautiful math coming... please be patient] $|2 - 3(-\frac{5}{3})|\ \overset{\text{?}}{=}\ 7$,     $|2 + 5| = 7$     Check!

[beautiful math coming... please be patient] $|2 - 3(3)|\ \overset{\text{?}}{=}\ 7$,     $|2 - 9| = 7$     Check!
EXAMPLE (an absolute value inequality involving ‘less than’):
Solve: [beautiful math coming... please be patient] $3|-6x + 7| \le 9$
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:
[beautiful math coming... please be patient] $3|-6x + 7| \le 9$ (original sentence)
[beautiful math coming... please be patient] $|-6x + 7| \le 3$ (divide both sides by $\,3$)
[beautiful math coming... please be patient] $-3 \le -6x + 7 \le 3$ (check that $\,k \ge 0\,$; use the theorem)
[beautiful math coming... please be patient] $-10 \le -6x \le -4$ (subtract $\,7\,$ from all three parts of the compound inequality)
[beautiful math coming... please be patient] $\displaystyle \frac{10}{6} \ge x \ge \frac{4}{6}$ (divide all three parts by $\,-6\,$; change direction of inequality symbols)
[beautiful math coming... please be patient] $\displaystyle \frac{2}{3} \le x \le \frac{5}{3}$ (simplify fractions; write in the conventional way)
Check the ‘boundaries’ of the solution set:
[beautiful math coming... please be patient] $3|-6(\frac{2}{3}) + 7| = 3|-4 + 7| = 3|3| = 9$     Check!
[beautiful math coming... please be patient] $3|-6(\frac{5}{3}) + 7| = 3|-10 + 7| = 3|-3| = 9$     Check!
EXAMPLE (an absolute value inequality involving ‘greater than’):
Solve: [beautiful math coming... please be patient] $3|-6x + 7| \ge 9$
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:
[beautiful math coming... please be patient] $3|-6x + 7| \ge 9$ (original sentence)
[beautiful math coming... please be patient] $|-6x + 7| \ge 3$ (divide both sides by $\,3$)
[beautiful math coming... please be patient] $-6x + 7 \le -3\ \ \text{or}\ \ -6x + 7\ge 3$ (check that $\,k \ge 0\,$; use the theorem)
[beautiful math coming... please be patient] $-6x\le -10\ \ \text{or}\ \ -6x\ge -4$ (subtract $\,7\,$ from both sides of both subsentences)
[beautiful math coming... please be patient] $\displaystyle x\ge\frac{10}{6}\ \ \text{or}\ \ x\le \frac{4}{6}$ (divide by $\,-6\,$; change direction of inequality symbols)
[beautiful math coming... please be patient] $\displaystyle x\ge\frac{5}{3}\ \ \text{or}\ \ x\le \frac{2}{3}$ (simplify fractions)
[beautiful math coming... please be patient] $\displaystyle x\le \frac{2}{3}\ \ \text{or}\ \ x\ge\frac{5}{3}$ (in the web exercise, the ‘less than’ part is always reported first)
Check the ‘boundaries’ of the solution set:
[beautiful math coming... please be patient] $3|-6(\frac{2}{3}) + 7| = 3|-4 + 7| = 3|3| = 9$     Check!
[beautiful math coming... please be patient] $3|-6(\frac{5}{3}) + 7| = 3|-10 + 7| = 3|-3| = 9$     Check!
EXAMPLE (an absolute value equation that is always false):
Solve: [beautiful math coming... please be patient] $|3x + 1| = -5$
Solution:
The theorem can't be used here, since $\,k\,$ is negative.
In this case, you need to stop and think.
Can absolute value ever be negative?   NO!
No matter what number you substitute for $\,x\,$,
the left-hand side of the equation will always be a number that is greater than or equal to zero.
Therefore, this sentence has no solutions. It is always false.
EXAMPLE (an absolute value inequality that is always false):
Solve: [beautiful math coming... please be patient] $|5 - 2x| \lt -3$
Solution:
The theorem can't be used here, since $\,k\,$ is negative.
In this case, you need to stop and think.
Can absolute value ever be negative?   NO!
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 can't possibly be less than $\,-3\,$.
Therefore, this sentence has no solutions. It is always false.
EXAMPLE (an absolute value inequality that is always true):
Solve: [beautiful math coming... please be patient] $|5 - 2x| \gt -3$
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:
Identifying Perfect Squares

 
 
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 sentence.
Write the result in the most conventional way.

For more advanced students, a graph is displayed.
For example, the inequality $\,|2 - 3x| \lt 7\,$
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 below the purple graph.
Click the “show/hide graph” button if you prefer not to see the graph.

Solve: