1. A compound statement is made up of more than one equation or inequality.
A disjunction is a compound statement that uses the word or.
Disjunction: x ≤ –3 OR x > 2 Set builder notation: {x|x ≤ –3 U x > 2}
A disjunction is true if and only if at least one of its parts is true.

2. A conjunction is a compound statement that uses the word and.
Conjunction: x ≥ –3 AND x < 2 Set builder notation: {x|x ≥ –3 x < 2}.
A conjunction is true if and only if all of its parts are true. Conjunctions can be written as a single statement as shown.
x ≥ –3 and x< –3 ≤ x < 2 U

3. Dis- means “apart. ” Disjunctions have two separate pieces
Dis- means “apart.” Disjunctions have two separate pieces. Con- means “together” Conjunctions represent one piece.
Reading Math

4. Example 1: Solving Compound Inequalities
Solve the compound inequality. Then graph the solution set.
6y < –24 OR y +5 ≥ 3
Solve both inequalities for y.
6y < –24
y + 5 ≥3 or
y < –4
y ≥ –2
The solution set is all points that satisfy {y|y < –4 or y ≥ –2}.
–6 –5 –4 –3 –2 –
(–∞, –4) U [–2, ∞)

5. Solve the compound inequality. Then graph the solution set.
Example 2
Solve the compound inequality. Then graph the solution set.
–3x < –12 AND x + 4 ≤ 12
Solve both inequalities for x.
–3x < – and x + 4 ≤ 12
x > x ≤ 8
The solution set is the set of points that satisfy both {x|4 < x ≤ 8}.
(4, 8]

6. Example 3: Solving Compound Inequalities
Solve the compound inequality. Then graph the solution set.
X +3 < 7 OR 3x > 18
Solve both inequalities for x.
x + 3 < 7 or x > 18
x < x > 6
The solution set is the set of all points that satisfy {x|x < 4 or x > 6}.
–3 –2 –
(–∞, 4) U (6, ∞)

7. 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.

8. 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.

9. Example 2A: Solving Absolute-Value Equations
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
Add 3 to both sides of each equation.
k = 13 or k = –7

10. Check It Out! Example 2b
Solve the equation.
|6x| – 8 = 22
Isolate the absolute-value expression.
|6x| = 30
Rewrite the absolute value as a disjunction.
6x = 30 or 6x = –30
Divide both sides of each equation by 6.
x = 5 or x = –5

11. You can solve absolute-value inequalities using the same methods that are used to solve an absolute-value equation.

12. 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.

13. The solutions of |x| < 3 are the points that are less than 3 units from zero. The solution is a conjunction: –3 < x < 3.

14. Check It Out! Example 4a
Solve the compound inequality. Then graph the solution set.
|x – 5| ≤ 8
Multiply both sides by 2.
Rewrite the absolute value as a conjunction.
x – 5 ≤ 8 and x – 5 ≥ –8
Add 5 to both sides of each inequality.
x ≤ 13 and x ≥ –3

15. The solution set is {x|–3 ≤ x ≤ 13}.
Check It Out! Example 4
The solution set is {x|–3 ≤ x ≤ 13}.
–10 –

16. 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.

17. 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

18. 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 

19. Check It Out! Example 3a
Solve the inequality. Then graph the solution.
|4x – 8| > 12
Rewrite the absolute value as a disjunction.
4x – 8 > 12 or 4x – 8 < –12
Add 8 to both sides of each inequality.
4x > 20 or 4x < –4
Divide both sides of each inequality by 4.
x > 5 or x < –1

20. Check It Out! Example 3a Continued
{x|x < –1 or x > 5}
–3 –2 –
(–∞, –1) U (5, ∞)
To check, you can test a point in each of the three region.
|4(–2) + 8| > 12
|–16| > 12 
|4(0) + 8| > 12
|8| > 12 x
|4(6) + 8| > 12
|32| > 12 

21. Solve the inequality. Then graph the solution.
Check It Out! Example 3b
Solve the inequality. Then graph the solution.
|3x| + 36 > 12
Isolate the absolute value as a disjunction.
|3x| > –24
Rewrite the absolute value as a disjunction.
3x > –24 or 3x < 24
Divide both sides of each inequality by 3.
x > –8 or x < 8
The solution is all real numbers, R.
–3 –2 –
(–∞, ∞)

22. Check It Out! Example 4b
Solve the compound inequality. Then graph the solution set.
–2|x +5| > 10
Divide both sides by –2, and reverse the inequality symbol.
|x + 5| < –5
Rewrite the absolute value as a conjunction.
x + 5 < –5 and x + 5 > 5
Subtract 5 from both sides of each inequality.
x < –10 and x > 0
Because no real number satisfies both x < –10 and x > 0, there is no solution. The solution set is ø.