Holt Algebra 2 2-8 Solving Absolute-Value Equations and Inequalities 2-8 Solving Absolute-Value Equations and Inequalities Holt Algebra 2 Warm Up Warm

Holt Algebra 2

2-8Solving Absolute-Value Equations and Inequalities 2-8Solving Absolute-Value Equations and Inequalities

Holt Algebra 2

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz

Holt Algebra 2

2-8Solving Absolute-Value Equations and Inequalities

Warm UpSolve.1. y + 7 < –11

2. 4m ≥ –12

3. 5 – 2x ≤ 17

y < –18

m ≥ –3

x ≥ –6

Use interval notation to indicate the graphed numbers.

4.

5.

(-2, 3]

(-, 1]

Holt Algebra 2


Solve compound inequalities.

Write and solve absolute-value equations and inequalities.

Objectives

Holt Algebra 2


disjunctionconjunctionabsolute-value

Vocabulary

Holt Algebra 2


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.

Holt Algebra 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< 2 –3 ≤ x < 2

U

Holt Algebra 2


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

Reading Math

Holt Algebra 2


Example 1A: Solving Compound Inequalities

Solve the compound inequality. Then graph the solution set.

Solve both inequalities for y.

The solution set is all points that satisfy {y|y < –4 or y ≥ –2}.

6y < –24 OR y +5 ≥ 3

6y < –24 y + 5 ≥3

y < –4 y ≥ –2or

–6 –5 –4 –3 –2 –1 0 1 2 3 (–∞, –4) U [–2, ∞)

Holt Algebra 2


Solve both inequalities for c.

The solution set is the set of points that satisfy both c ≥ –4 and c < 0.

c ≥ –4 c < 0

–6 –5 –4 –3 –2 –1 0 1 2 3[–4, 0)

Example 1B: Solving Compound Inequalities


and 2c + 1 < 1

Holt Algebra 2


Example 1C: Solving Compound Inequalities


Solve both inequalities for x.

The solution set is the set of all points that satisfy {x|x < 3 or x ≥ 5}.

–3 –2 –1 0 1 2 3 4 5 6

x – 5 < –2 OR –2x ≤ –10

x < 3 x ≥ 5

x – 5 < –2 or –2x ≤ –10

(–∞, 3) U [5, ∞)

Holt Algebra 2




The solution set is all points that satisfy {x|x < 3 U x ≥ 6}.

x – 2 < 1 OR 5x ≥ 30

x – 2 < 1 5x ≥ 30

x < 3 x ≥ 6or

–1 0 1 2 3 4 5 6 7 8 (–∞, 3) U [6, ∞)

Check It Out! Example 1a

Holt Algebra 2



2x ≥ –6 and –x > –4

–4 –3 –2 –1 0 1 2 3 4 5

2x ≥ –6 AND –x > –4

x ≥ –3 x < 4

The solution set is the set of points that satisfy both {x|x ≥ –3 x < 4}.

U

[–3, 4)


Check It Out! Example 1b

Holt Algebra 2



x –5 < 12 or 6x ≤ 12

2 4 6 8 10 12 14 16 18 20

x –5 < 12 OR 6x ≤ 12

x < 17 x ≤ 2

Because every point that satisfies x < 2 also satisfies x < 2, the solution set is {x|x < 17}.

(-∞, 17)


Check It Out! Example 1c

Holt Algebra 2



–3x < –12 and x + 4 ≤ 12

2 3 4 5 6 7 8 9 10 11

–3x < –12 AND x + 4 ≤ 12

x < –4 x ≤ 8

The solution set is the set of points that satisfy both {x|4 < x ≤ 8}.

(4, 8]


Check It Out! Example 1d

Holt Algebra 2


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.

Holt Algebra 2


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.

Holt Algebra 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.

Holt Algebra 2


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.

Holt Algebra 2


Think: Greator inequalities involving > or ≥ symbols are disjunctions.

Think: Less thand inequalities involving < or ≤ symbols are conjunctions.

Helpful Hint

Holt Algebra 2


Note: The symbol ≤ can replace <, and the rules still apply. The symbol ≥ can replace >, and the rules still apply.

Holt Algebra 2


Solve the equation.

Example 2A: Solving Absolute-Value Equations

Rewrite the absolute value as a disjunction.

This can be read as “the distance from k to –3 is 10.”

Add 3 to both sides of each equation.

|–3 + k| = 10

–3 + k = 10 or –3 + k = –10

k = 13 or k = –7

Holt Algebra 2


Solve the equation.

Example 2B: Solving Absolute-Value Equations

x = 16 or x = –16

Isolate the absolute-value expression.


Multiply both sides of each equation by 4.

Holt Algebra 2



|x + 9| = 13

Solve the equation.


This can be read as “the distance from x to +9 is 4.”

Subtract 9 from both sides of each equation.

x + 9 = 13 or x + 9 = –13

x = 4 or x = –22

Holt Algebra 2



|6x| – 8 = 22

Solve the equation.

Isolate the absolute-value expression.

Rewrite the absolute value as a disjunction. Divide both sides of each equation by 6.

|6x| = 30

6x = 30 or 6x = –30

x = 5 or x = –5

Holt Algebra 2


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

Holt Algebra 2


Example 3A: Solving Absolute-Value Inequalities with Disjunctions

Solve the inequality. Then graph the solution.


|–4q + 2| ≥ 10

–4q + 2 ≥ 10 or –4q + 2 ≤ –10

–4q ≥ 8 or –4q ≤ –12

Divide both sides of each inequality by –4 and reverse the inequality symbols.

Subtract 2 from both sides of each inequality.

q ≤ –2 or q ≥ 3

Holt Algebra 2


Example 3A Continued

{q|q ≤ –2 or q ≥ 3}

–3 –2 –1 0 1 2 3 4 5 6

(–∞, –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

Holt Algebra 2



|0.5r| – 3 ≥ –3

0.5r ≥ 0 or 0.5r ≤ 0

Divide both sides of each inequality by 0.5.

Isolate the absolute value as a disjunction.

r ≤ 0 or r ≥ 0

|0.5r| ≥ 0


Example 3B: Solving Absolute-Value Inequalities with Disjunctions

–3 –2 –1 0 1 2 3 4 5 6

(–∞, ∞)

The solution is all real numbers, R.

Holt Algebra 2




|4x – 8| > 12

4x – 8 > 12 or 4x – 8 < –12

4x > 20 or 4x < –4

Divide both sides of each inequality by 4.

Add 8 to both sides of each inequality.

x > 5 or x < –1


Holt Algebra 2


{x|x < –1 or x > 5}

–3 –2 –1 0 1 2 3 4 5 6

(–∞, –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

Check It Out! Example 3a Continued

Holt Algebra 2



|3x| + 36 > 12


Isolate the absolute value as a disjunction.



3x > –24 or 3x < 24

x > –8 or x < 8

|3x| > –24

–3 –2 –1 0 1 2 3 4 5 6

(–∞, ∞)

The solution is all real numbers, R.

Holt Algebra 2



Example 4A: Solving Absolute-Value Inequalities with Conjunctions

|2x +7| ≤ 3 Multiply both sides by 3.



Rewrite the absolute value as a conjunction.

2x + 7 ≤ 3 and 2x + 7 ≥ –3

2x ≤ –4 and 2x ≥ –10

x ≤ –2 and x ≥ –5

Holt Algebra 2


Example 4A Continued

The solution set is {x|–5 ≤ x ≤ 2}.

–6 –5 –3 –2 –1 0 1 2 3 4

Holt Algebra 2



Example 4B: Solving Absolute-Value Inequalities with Conjunctions

|p – 2| ≤ –6Multiply both sides by –2, and reverse the inequality symbol.



|p – 2| ≤ –6 and p – 2 ≥ 6

p ≤ –4 and p ≥ 8

Because no real number satisfies both p ≤ –4 andp ≥ 8, there is no solution. The solution set is ø.

Holt Algebra 2



|x – 5| ≤ 8 Multiply both sides by 2.


Rewrite the absolute value as a conjunction.x – 5 ≤ 8 and x – 5 ≥ –8

x ≤ 13 and x ≥ –3


Holt Algebra 2


The solution set is {x|–3 ≤ x ≤ 13}.

Check It Out! Example 4

–10 –5 0 5 10 15 20 25

Holt Algebra 2



–2|x +5| > 10

Divide both sides by –2, and reverse the inequality symbol.



x + 5 < –5 and x + 5 > 5

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.


|x + 5| < –5

Holt Algebra 2


Lesson Quiz: Part I

Solve. Then graph the solution.

1.

2.

y – 4 ≤ –6 or 2y >8

–7x < 21 and x + 7 ≤ 6

{y|y ≤ –2 ≤ or y > 4}

{x|–3 < x ≤ –1}

–4 –3 –2 –1 0 1 2 3 4 5

–4 –3 –2 –1 0 1 2 3 4 5

Solve each equation.

3. |2v + 5| = 9 4. |5b| – 7 = 13

2 or –7 + 4

Holt Algebra 2


Lesson Quiz: Part II

Solve. Then graph the solution.

5.

6.

7.

|1 – 2x| > 7 {x|x < –3 or x > 4}

|3k| + 11 > 8 R

–2|u + 7| ≥ 16 ø

–4 –3 –2 –1 0 1 2 3 4 5

–4 –3 –2 –1 0 1 2 3 4 5

Documents

Holt Algebra 2 2-8 Solving Absolute-Value Equations and Inequalities 2-8 Solving Absolute-Value Equations and Inequalities Holt Algebra 2 Warm Up Warm