﻿ Solving Absolute Value Sentences, All Types
SOLVING ABSOLUTE VALUE SENTENCES, ALL TYPES

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

$|2 - 3(-\frac{5}{3})|\ \overset{\text{?}}{=}\ 7$,     $|2 + 5| = 7$     Check!

$|2 - 3(3)|\ \overset{\text{?}}{=}\ 7$,     $|2 - 9| = 7$     Check!
EXAMPLE (an absolute value inequality involving ‘less than’):
Solve: $3|-6x + 7| \le 9$
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:
 $3|-6x + 7| \le 9$ (original sentence) $|-6x + 7| \le 3$ (divide both sides by $\,3$) $-3 \le -6x + 7 \le 3$ (check that $\,k \ge 0\,$; use the theorem) $-10 \le -6x \le -4$ (subtract $\,7\,$ from all three parts of the compound inequality) $\displaystyle \frac{10}{6} \ge x \ge \frac{4}{6}$ (divide all three parts by $\,-6\,$; change direction of inequality symbols) $\displaystyle \frac{2}{3} \le x \le \frac{5}{3}$ (simplify fractions; write in the conventional way)
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 (an absolute value inequality involving ‘greater than’):
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 $\,3$) $-6x + 7 \le -3\ \ \text{or}\ \ -6x + 7\ge 3$ (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)
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 (an absolute value equation that is always false):
Solve: $|3x + 1| = -5$
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: $|5 - 2x| \lt -3$
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: $|5 - 2x| \gt -3$
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

When you're done practicing, move on to:
Identifying Perfect Squares

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.