THEOREM
solving absolute value sentences
Let
$\,x\in\mathbb{R}\,$,
and let
$\,k\ge 0\,$.
Then,
$$
\begin{gather}
\cssId{s13}{x = k\ \ \text{ is equivalent to }\ \ x = \pm k} \\
\\
\cssId{s14}{x \lt k\ \ \ \text{ is equivalent to }\ \ k \lt x \lt k} \\
\cssId{s15}{x \le k\ \ \ \text{ is equivalent to }\ \ k \le x \le k} \\
\\
\cssId{s16}{x \gt k\ \ \ \text{ is equivalent to }\ \ x\lt k\ \ \text{ or }\ \ x\gt k} \\
\cssId{s17}{x \ge k\ \ \ \text{ is equivalent to }\ \ x\le k\ \ \text{ or }\ \ x\ge k}
\end{gather}
$$
Solve:
$2  3x = 7$
Solution:
Write a nice, clean list of equivalent sentences:
$2  3x = 7$

(original equation) 
$23x = \pm 7$

(check that $\,k\ge 0\,$; use the theorem) 
$23x = 7\ \text{ or }\ 23x = 7$

(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:
$36x + 7 \le 9$
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:
$36x + 7 \le 9$

(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:
$36(\frac{2}{3}) + 7 = 34 + 7 = 33 = 9$ Check!
$36(\frac{5}{3}) + 7 = 310 + 7 = 33 = 9$ Check!
EXAMPLE (an absolute value inequality involving ‘greater than’):
Solve:
$36x + 7 \ge 9$
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:
$36x + 7 \ge 9$

(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) 
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 (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 lefthand 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 lefthand 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 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.