# Solving Absolute Value Sentences, All Types

This web exercise mixes up problems from the previous three web exercises:

Visit these earlier sections for a thorough discussion of the concepts.

THEOREM solving absolute value sentences
Let $\,x\in\mathbb{R}\,,$ and let $\,k\ge 0\,.$   Then: $$\begin{gather} \cssId{s13}{|x| = k\ \ \text{ is equivalent to }\ \ x = \pm k} \\ \\ \cssId{s14}{|x| \lt k\ \ \ \text{ is equivalent to }\ \ -k \lt x \lt k} \\ \cssId{s15}{|x| \le k\ \ \ \text{ is equivalent to }\ \ -k \le x \le k} \\ \\ \cssId{s16}{|x| \gt k\ \ \ \text{ is equivalent to }\ \ x\lt -k\ \ \text{ or }\ \ x\gt k} \\ \cssId{s17}{|x| \ge k\ \ \ \text{ is equivalent to }\ \ x\le -k\ \ \text{ or }\ \ x\ge k} \end{gather}$$

## 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 $2-3x = 7$       or $\,2-3x = -7$ expand the plus/minus $-3x = 5$       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 sentence. 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 sentence. 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 $\,3$ $-6x + 7 \le -3$       or $\,-6x + 7\ge 3$ check that $\,k \ge 0\,$; use the theorem $-6x\le -10$       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.

## Concept Practice

Solve the given absolute value sentence. Write the result in the most conventional way.

For more advanced students, a graph is available. 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 to toggle.

Solve: