1. Warm Up Solve. 1. y + 7 < –11 y < –18 2. 4m ≥ –12 m ≥ –3
3. 5 – 2x ≤ 17
x ≥ –6
Use interval notation to indicate the graphed numbers. 4.
(-2, 3]
(-, 1] 5.

2. Do Now 10/15/2015
Solve each equation. 1.
|2v + 5| = 9
2 or –7 2.
|5b| – 7 = 13
+ 4

3. 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

4. A conjunction is a compound statement that uses the word and.
Conjunction: x ≥ –3 AND x < 2
Conjunctions can be written as a single statement as shown.
x ≥ –3 and x< –3 ≤ x < 2

5. 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

6. Solve both inequalities for x.
Example 1a
Solve the compound inequality. Then graph the solution set and write the interval notation.
x – 2 < 1 OR 5x ≥ 30
Solve both inequalities for x.
x – 2 < 1
5x ≥ 30 or
x ≥ 6
x < 3 –
(–∞, 3) U [6, ∞)

7. Solve the compound inequality. Then graph the solution set.
Example 1b
Solve the compound inequality. Then graph the solution set.
2x ≥ –6 AND –x > –4
Solve both inequalities for x.
2x ≥ – and –x > –4
x ≥ – x < 4
–4 –3 –2 –
[–3, 4)

8. Solve the compound inequality. Then graph the solution set.
Try it out!
Solve the compound inequality. Then graph the solution set.
x –5 < 12 OR 6x ≤ 12
Solve both inequalities for x.
x –5 < or x ≤ 12
x < x ≤ 2
(-∞, 17)

9. Solve the compound inequality. Then graph the solution set.
Try It Out!
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
(4, 8]

10. 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.

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

12. 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.

13. Helpful Hint
Think: Greator inequalities involving > or ≥ symbols are disjunctions.
Think: Less thand inequalities involving < or ≤ symbols are conjunctions.

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

16. 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

17. To check, you can test a point in each of the three region.
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 

18. Solve the inequality. Then graph the solution.
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 –
(–∞, ∞)

19. 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

20. Rewrite as compound inequality –3 ≤ x ≤ 13
Example 4a Continued
Rewrite as compound inequality
–3 ≤ x ≤ 13
–10 –
[-3, 13]

21. Try it out!
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.