Holt Algebra 2 2-8 Solving Absolute-Value Equations and Inequalities 2-8,9 Solving and Graphing...

Holt Algebra 2

2-8Solving Absolute-Value Equations and Inequalities 2-8,9 Solving and Graphing Absolute-Value

Equations and Inequalities

Holt Algebra 2

VocabVocab

Lesson PresentationLesson Presentation

Text QuestionsText Questions

Objective and PA Math StandardsObjective and PA Math Standards

Warm UpWarm UpLesson PlanLesson Plan

WorksheetWorksheet

Lesson QuizLesson Quiz

Holt Algebra 2

2-8Solving Absolute-Value Equations and Inequalities

Warm UpSolve.1. y + 7 < –11 2. 5 – 2x ≤ 17y < –18 x ≥ –6

Use interval notation to show the graphed numbers

6.

7.

(-2, 3]

(-, 1]

3. f(x) = –|x + 1|

4. f(x) = 2|x| – 1

5. f(x) = |x + 1| + 2

–5; –2

7; 5

7; 4

Evaluate each expression in #3-5 for f(4) and f(-3).

Solving and Graphing Absolute-Value Equations and Inequalities

Holt Algebra 2


Solve and graph compound inequalities.

Write, solve and graph absolute-value equations and inequalities.

Objectives


Holt Algebra 2


compound statementdisjunctionconjunctionabsolute-valueabsolute value function

Vocabulary


Holt Algebra 2


A compound statement is made up of more than one equation or inequality.

A disjunction is a compound statement using or.

Disjunction: x ≤ –3 OR x > 2 Set builder notation: {x|x ≤ –3 U x > 2}Interval Notation: (-∞, -2] U (2, ∞)

A disjunction is true if at least one of its parts is true.


Holt Algebra 2


A conjunction is a compound statement using and.

Conjunction: x ≥ –3 AND x < 2 Set builder notation: {x|x ≥ –3 x < 2}

Interval Notation: [-3, 2)

A conjunction is true 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 pieces.

Con- means “together” Conjunctions have one piece.

Reading Math


Holt Algebra 2


Absolute value of a number x, written |x|, is the distance from x to zero on the number line. Because it represents distance without regard to direction, the absolute value of any real number is always positiveabsolute value of any real number is always positive.


Holt Algebra 2



Holt Algebra 2


Example 1A: Solving Compound InequalitiesSolve and graph the compound inequality.

{y|y < –4 or y ≥ –2}

6y < –24 OR y +5 ≥ 3

y < –4 y ≥ –2

–6 –5 –4 –3 –2 –1 0 1 2 3

/6 /6 -5 -5

orInequality Notation

Set Builder Notation

Interval Notation (–∞,-4) U [-2, ∞)


Holt Algebra 2


Example 1C: Solving Compound Inequalities

Solve and graph the compound inequality.

{x|x < 3 or x ≥ 5}.

–3 –2 –1 0 1 2 3 4 5 6

x – 5 < –2 OR –2x ≤ –10

x < 3 or x ≥ 5

+5 +5 /–2 /-2

(–∞, 3) U [5, ∞)

Inequality Notation

Set Builder Notation

Interval Notation


Holt Algebra 2


–4 –3 –2 –1 0 1 2 3 4 5

2x ≥ –6 AND –x > –4

x ≥ –3 x < 4

{x|x ≥ –3 x < 4}. U

[–3, 4)


Check It Out! Example 1b


Holt Algebra 2


2 4 6 8 10 12 14 16 18 20

x – 5 < 12 OR 6x ≤ 12

x < 17 x ≤ 2

Note: Every point that satisfies x < 17 also satisfies x < 2 {x|x < 17}.

(-∞, 17)


Check It Out! Example 1c


Holt Algebra 2


Example 1A Continued

The graph of g(x) = |x|– 5 is the graph of f(x) = |x| after a vertical shift of 5 units down. The vertex of g(x) is (0, –5).

f(x)

g(x)


Holt Algebra 2


Example 1B Continued

f(x)

g(x)

The graph of g(x) = |x + 1| is the graph of f(x) = |x| after a horizontal shift of 1 unit left. The vertex of g(x) is (–1, 0).


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


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

|–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


Holt Algebra 2


Check It Out! Example 2a

|x + 9| = 13

Solve the equation.

x + 9 = 13 or x + 9 = –13

x = 4 or x = –22


Holt Algebra 2



|6x| – 8 = 22

Solve the equation.

|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.

Remember this:“LESS THAN AND…

GREATER THAN OR.”

When the abs. value inequality is <, it’s an “AND”When the abs. value inequality is >, it’s an “OR”


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

q ≤ –2 or q ≥ 3

{q|q ≤ –2 or q ≥ 3}

(–∞, –2] U [3, ∞)


Holt Algebra 2


Solve the inequality. Then graph the solution.

|4x – 8| > 12

4x – 8 > 12 or 4x – 8 < –12

4x > 20 or 4x < –4

x > 5 or x < –1


x < -1 or x > 5

{x|x < –1 or x > 5}

(–∞, –1) U (5, ∞)


Holt Algebra 2


Solve the compound inequality. Then graph the solution set.

Example 4A: Solving Absolute-Value Inequalities with Conjunctions

|2x +7| ≤ 3

2x + 7 ≤ 3 and 2x + 7 ≥ –3

2x ≤ –4 and 2x ≥ –10

x ≤ –2 and x ≥ –5


Holt Algebra 2



Example 4B: Solving Absolute-Value Inequalities with Conjunctions

|p – 2| ≤ –6

|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

x – 5 ≤ 8 and x – 5 ≥ –8

x ≤ 13 and x ≥ –3



Holt Algebra 2



–2|x +5| > 10

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


Holt Algebra 2 2-8 Solving Absolute-Value Equations and Inequalities 2-8,9 Solving and Graphing...

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