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

2. 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.

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. Example 1A: 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, ∞)

9. Example 1B: Solving Compound Inequalities
Solve the compound inequality. Then graph the solution set.
Solve both inequalities for c.
and
2c + 1 < 1
c ≥ –4
c < 0
The solution set is the set of points that satisfy both c ≥ –4 and c < 0.
–6 –5 –4 –3 –2 –
[–4, 0)

10. Example 1C: Solving Compound Inequalities
Solve the compound inequality. Then graph the solution set.
x – 5 < –2 OR –2x ≤ –10
Solve both inequalities for x.
x – 5 < –2 or –2x ≤ –10
x < x ≥ 5
The solution set is the set of all points that satisfy {x|x < 3 or x ≥ 5}.
–3 –2 –
(–∞, 3) U [5, ∞)

11. Solve the compound inequality. Then graph the solution set.
Check It Out! Example 1a
Solve the compound inequality. Then graph the solution set.
x – 2 < 1 OR 5x ≥ 30
Solve both inequalities for x.
x – 2 < 1
5x ≥ 30 or
x ≥ 6
x < 3
The solution set is all points that satisfy {x|x < 3 U x ≥ 6}. –
(–∞, 3) U [6, ∞)

12. Solve the compound inequality. Then graph the solution set.
Check It Out! 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
The solution set is the set of points that satisfy both {x|x ≥ –3 x < 4}. U
–4 –3 –2 –
[–3, 4)

13. Solve the compound inequality. Then graph the solution set.
Check It Out! Example 1c
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
Because every point that satisfies x < 2 also satisfies x < 2, the solution set is {x|x < 17}.
(-∞, 17)

14. Solve the compound inequality. Then graph the solution set.
Check It Out! Example 1d
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]

15. 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.

16. 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.

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

18. 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.

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

20. 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.

21. 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

22. Example 2B: Solving Absolute-Value Equations
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

23. Check It Out! Example 2a
Solve the equation.
This can be read as “the distance from x to +9 is 4.”
|x + 9| = 13
Rewrite the absolute value as a disjunction.
x + 9 = 13 or x + 9 = –13
Subtract 9 from both sides of each equation.
x = 4 or x = –22

24. 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

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

26. 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

27. 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 

28. 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 –
(–∞, ∞)

29. 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

30. 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 

31. 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 –
(–∞, ∞)

32. 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

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

34. 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.

35. 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

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

37. 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.

38. 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

39. 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