Solving Simple Inequalities

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1-5Solving Simple Inequalities Pre-Algebra 1-5 Solving Simple Inequalities Pre-Algebra Warm Up Warm Up Problem of the Day Problem of the Day Lesson Lesson Presentation Presentation

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Solving Simple Inequalities 1-5 Solving Simple Inequalities Pre-Algebra Warm Up Solve. 1. x + 6 = 13 2. 8n = 48 3. t  2 = 56 4. 6 = x = 7 n = 6 t = 58 z 6 z = 36

Transcript of Solving Simple Inequalities

Page 1: Solving Simple Inequalities

1-5 Solving Simple Inequalities

Pre-Algebra

1-5 Solving Simple Inequalities

Pre-Algebra

Warm UpWarm UpProblem of the DayProblem of the DayLesson PresentationLesson Presentation

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1-5 Solving Simple Inequalities

Pre-Algebra

Warm UpSolve.

1. x + 6 = 132. 8n = 483. t 2 = 564. 6 =

x = 7n = 6t = 58z = 36z

6

1-5 Solving Simple Inequalities

Pre-Algebra

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1-5 Solving Simple Inequalities

Pre-Algebra

Problem of the DayBill and Brad are taking drivers education. Bill drives with his instructor for one and a half hours three times a week. He needs a total of 27 hours. Brad drives two times a week, two hours each time. He needs 26 hours. Who will finish his hours first? Bill

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1-5 Solving Simple Inequalities

Pre-Algebra

Learn to solve and graph inequalities.

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1-5 Solving Simple Inequalities

Pre-Algebra

Vocabularyinequalityalgebraic inequalitysolution of an inequalitysolution set

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1-5 Solving Simple Inequalities

Pre-Algebra

An inequality compares two quantities and typically uses one of these symbols:

<<is less than

is greater than

is less than or equal to

is greater than or equal to

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1-5 Solving Simple Inequalities

Pre-Algebra

Additional Example 1: Completing an InequalityCompare. Write < or >.

A. 23 – 14 6

9 6>

B. 5(12) 70 60 70<

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1-5 Solving Simple Inequalities

Pre-Algebra

Try This: Example 1Compare. Write < or >.

A. 19 – 3 17

16 17<

B. 4(15) 50 60 50>

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1-5 Solving Simple Inequalities

Pre-Algebra

An inequality that contains a variable is an algebraic inequality.

A number that makes an inequality true is a solution of the inequality.

The set of all solutions is called the solution set. The solution set can be shown by graphing it on a number line.

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1-5 Solving Simple Inequalities

Pre-Algebra

x < 54 < 5

x = 2.1 2.1 < 5

x is less than 5Word Phrase

Inequality

Sample Solutions

Solution Set 1 2 3 4 5 6 7

x = 4

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Pre-Algebra

a > 07 > 0

a = 25 25 > 0

a is greater than 0a is more than 0

Word Phrase

Inequality

Sample Solutions

Solution Set–3 –2 –1 0 1 2 3

a = 7

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1-5 Solving Simple Inequalities

Pre-Algebra

y 20 2

y = 1.5 1.5 2

y is less than or equal to 2y is at most 2

Word Phrase

Inequality

Sample Solutions

Solution Set–3 –2 –1 0 1 2 3

y = 0

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1-5 Solving Simple Inequalities

Pre-Algebra

m 3

17 3m = 3 3 3

m is greater than or equal to 3m is at least 3

Word Phrase

Inequality

Sample Solutions

Solution Set–1 0 1 2 3 4 5

m = 17

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1-5 Solving Simple Inequalities

Pre-Algebra

Most inequalities can be solved the same way equations are solved. Use inverse operations on both sides of the inequality to isolate the variable.

There are special rules when multiplying or dividing by a negative number, which you will learn in the next chapter.

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Pre-Algebra

Additional Example 2A: Solving and Graphing Inequalities

Solve and graph the inequality.A. x + 2.5 8

–2.5 –2.5x 5.5

1 2 3 4 5 6 7

Subtract 2.5 from both sides.

According to the graph, 5.4 is a solution, since 5.4 < 5.5, and 6 should not be solution because 6 > 5.5.

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1-5 Solving Simple Inequalities

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Additional Example 2B: Solving and Graphing Inequalities

Solve and graph the inequality.

B. 5t > 15

5 5t > 3

1 2 3 4 5 6 7

5t > 15 Divide both sides by 5.

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1-5 Solving Simple Inequalities

Pre-Algebra

Additional Example 2C: Solving and Graphing InequalitiesSolve and graph the inequality.

C. w – 1 < 8

w < 9

–3 0 3 6 9 12 15

+ 1 + 1 Add 1 to both sides.

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Additional Example 2D: Solving and Graphing InequalitiesSolve and graph the inequality.

D. 3

12 p

0 3 6 9 12 15 18

Multiply both sides by 4.

p4

3 p44 • 4 •

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1-5 Solving Simple Inequalities

Pre-Algebra

Try This: Examples 2A and 2BSolve and graph each inequality.

A. x + 2 3.5 –2 –2x 1.5

1 2 3 4 5 6 7

Subtract 2 from both sides.

B. 6u > 72

6 6u > 12

3 6 9 12 15 18 21

6u > 72 Divide both sides by 6.

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Try This: Examples 2C and 2DSolve and graph each inequality.

C. z – 6 < 15

z < 21 –21 –14 –7 0 7 14 21

+ 6 + 6 Add 6 to both sides.

18 b 0 3 6 9 12 15 18

Multiply both sides by 9.

D. 2 b9

2 b99 • 9 •

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Additional Example 3: Problem Solving Application

An interior designer is planning to place a wallpaper border along the edges of all four walls of a room. The total distance around the room is 88 feet. The border comes in packages of 16 feet. What is the least number of packages that must be purchased to be sure that there is enough border to complete the room?

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Additional Example 3 Continued11 Understand the Problem

The answer will be the least number of packages of border needed to wallpaper a room.

List the important information:• The total distance around the room is 88 feet.• The border comes in packages of 16 feet.

Show the relationship of the information:

the number of packages of border

the length of one package of border • Distance around

the room

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Pre-Algebra

Additional Example 3 Continued

22 Make a Plan

Use the relationship to write an inequality. Let x represent the number of packages of border.

x 16 ft 88 ft•

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Additional Example 3 Continued

16x 8816x 88

Solve33

16 16x 5.5

At least 5.5 packages of border must be used to complete the room.

Divide both sides by 16.

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Additional Example 3 Continued

Look Back44

Because whole packages of border must be purchased, at least 6 packages of border must be purchased to ensure that there is enough to complete the room.

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1-5 Solving Simple Inequalities

Pre-Algebra

Try This: Example 3

Ron will provide 130 cookies for the school fundraiser. He has to buy the cookies in packages of 20. What is the least number of packages Ron must buy to be sure to have enough cookies?

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Try This: Example 311 Understand the Problem

The answer will be the number of packages of cookies a customer needs to purchase.List the important information:

• Cookies are sold in packages of 20 cookies.• A customer needs to purchase 130 cookies.

Show the relationship of the information:the number of

packages of cookies to be purchased

the number of cookies in one

package 130

cookies•

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Try This: Example 3 Continued

22 Make a Plan

Use the relationship to write an inequality. Let x represent the number of packages of cookies.

x 20 cookies 130 cookies•

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Try This: Example 3 Continued

20x 130 20x 130

Solve33

20 20x 6.5

At least 6.5 packages of cookies need to be purchased.

Divide both sides by 20.

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Try This: Example 3 Continued

Look Back44

Because whole packages of cookies must be purchased, at least 7 packages of cookies must be purchased for the party.

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Lesson QuizUse < or > to compare each inequality.

1. 13 5(2) 2. 14 – 2 11

Solve and graph each inequality.

3. k + 9 < 12

4. 3 5. A school bus can hold 64 passengers. Three classes would like to use the bus for a field trip. Each class has 21 students. Write and solve an inequality to determine whether all three classes will fit on the bus.

>

6 m

>

k< 3

m2

–5 –4–3–2–1 0 1 2 3 4 5

–4 –3–2–1 0 1 2 3 4 5 6

3(21) 64; 63 64; yes?