1. Recall that the absolute value of a number x, written |x|, is the distance from x to zero on the number line. Because absolute value represents distance without regard to direction, the absolute value of any real number is nonnegative.
Absolute-value equations and inequalities can be represented by compound statements. Consider the equation |x| = 3.
The solutions of |x| = 3 are the two points that are 3 units from zero. The solution is a disjunction: x = –3 or x = 3.

2. The solutions of |x| < 3 are the points that are less than 3 units from zero. The solution is a conjunction: –3 < x < 3.
The solutions of |x| > 3 are the points that are more than 3 units from zero. The solution is a disjunction:
x < –3 or x > 3.
Think: Greator inequalities involving > or ≥ symbols are disjunctions.
Think: Less thand inequalities involving < or ≤ symbols are conjunctions.
Helpful Hint

3. Note: The symbol ≤ can replace <, and the rules still apply
Note: The symbol ≤ can replace <, and the rules still apply. The symbol ≥ can replace >, and the rules still apply.
Ex 2A:
Solve the equation.
This can be read as “the distance from k to –3 is 10.”
|–3 + k| = 10
Rewrite the absolute value as a disjunction.
–3 + k = 10 or –3 + k = –10
k = 13 or k = –7
Add 3 to both sides of each equation.

4. Ex 2B:
Solve the equation.
Isolate the absolute-value expression.
Rewrite the absolute value as a disjunction.
Multiply both sides of each equation
by 4.
x = 16 or x = –16
You can solve absolute-value inequalities using the same methods that are used to solve an absolute-value equation.

5. Example 3A: Solving Absolute-Value Inequalities with Disjunctions
Solve the inequality. Then graph the solution.
|–4q + 2| ≥ 10
Rewrite the absolute value as a disjunction.
–4q + 2 ≥ 10 or –4q + 2 ≤ –10
Subtract 2 from both sides of each inequality.
–4q ≥ 8 or –4q ≤ –12
Divide both sides of each inequality by –4 and reverse the inequality symbols.
q ≤ –2 or q ≥ 3

6. To check, you can test a point in each of the three region.
Example 3A Continued
{q|q ≤ –2 or q ≥ 3}
–3 –2 –
(–∞, –2] U [3, ∞)
To check, you can test a point in each of the three region.
|–4(–3) + 2| ≥ 10
|14| ≥ 10 
|–4(0) + 2| ≥ 10
|2| ≥ 10 x
|–4(4) + 2| ≥ 10
|–14| ≥ 10 

7. Example 3B: Solving Absolute-Value Inequalities with Disjunctions
Solve the inequality. Then graph the solution.
|0.5r| – 3 ≥ –3
Isolate the absolute value as a disjunction.
|0.5r| ≥ 0
Rewrite the absolute value as a disjunction.
0.5r ≥ 0 or 0.5r ≤ 0
Divide both sides of each inequality by 0.5.
r ≤ 0 or r ≥ 0
The solution is all real numbers, R.
–3 –2 –
(–∞, ∞)

8. Example 4A: Solving Absolute-Value Inequalities with Conjunctions
Solve the compound inequality. Then graph the solution set.
|2x +7| ≤ 3
Multiply both sides by 3.
Rewrite the absolute value as a conjunction.
2x + 7 ≤ 3 and 2x + 7 ≥ –3
Subtract 7 from both sides of each inequality.
2x ≤ –4 and 2x ≥ –10
Divide both sides of each inequality by 2.
x ≤ –2 and x ≥ –5

9. The solution set is {x|–5 ≤ x ≤ 2}.
Example 4A Continued
The solution set is {x|–5 ≤ x ≤ 2}.
–6 –5 –3 –2 –

10. Example 4B: Solving Absolute-Value Inequalities with Conjunctions
Solve the compound inequality. Then graph the solution set.
Multiply both sides by –2, and reverse the inequality symbol.
|p – 2| ≤ –6
Rewrite the absolute value as a conjunction.
|p – 2| ≤ –6 and p – 2 ≥ 6
Add 2 to both sides of each inequality.
p ≤ –4 and p ≥ 8
Because no real number satisfies both p ≤ –4 and p ≥ 8, there is no solution. The solution set is ø.