1. 2-8 Solving Absolute-Value Equations and Inequalities Warm Up
Lesson Presentation
Lesson Quiz
Holt Algebra 2

2. Solve. Then graph the solution.
Lesson Quiz: Part I
Solve. Then graph the solution. 1.
y – 4 ≤ –6 or 2y >8
{y|y ≤ –2 ≤ or y > 4}
–4 –3 –2 – 2.
–7x < 21 and x + 7 ≤ 6
{x|–3 < x ≤ –1}
–4 –3 –2 –
Solve each equation.
3. |2v + 5| = 9
4. |5b| – 7 = 13
2 or –7
+ 4

3. Objectives Solve compound inequalities.
Write and solve absolute-value equations and inequalities.

4. Vocabulary
disjunction
conjunction
absolute-value

5. 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.

6. 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

7. 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

8. 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.

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

10. 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.

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

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

14. 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

15. 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 

16. Example 3B: Solving Absolute-Value Inequalities with Disjunctions
Solve the inequality. Then graph the solution.
|0.5r| – 3 ≥ – (always true?? – all reals)
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 –
(–∞, ∞)

17. 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

18. 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 

19. 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 –
(–∞, ∞)

20. 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

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

22. 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.

23. 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

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

25. Check It Out! Example 4b
Solve the compound inequality. Then graph the solution set.
–2|x +5| > (Never True??-no sol.)
Divide both sides by –2, and revrse 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.

26. Solve. Then graph the solution.
Lesson Quiz: Part II
Solve. Then graph the solution. 5.
|1 – 2x| > 7
{x|x < –3 or x > 4}
–4 –3 –2 – 6.
|3k| + 11 > 8 R
–4 –3 –2 – 7.
–2|u + 7| ≥ 16 ø

27. 20-36 extra credit if you want it
Assignment
Page 154 # 8-13 all,
20-36 extra credit if you want it
Due Wednesday
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Assignment
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Due Tuesday
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