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1-5. Solving Simple Inequalities. Warm Up. Problem of the Day. Lesson Presentation. Course 3. 1-5. Solving Simple Inequalities. z. 6. Course 3. Warm Up Solve. 1. x + 6 = 13 2. 8 n = 48 3. t  2 = 56 4. 6 =. x = 7. n = 6. t = 58. z = 36. 1-5.

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  1. 1-5 Solving Simple Inequalities Warm Up Problem of the Day Lesson Presentation Course 3

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

  3. 1-5 Solving Simple Inequalities Course 3 Problem of the Day Bill 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

  4. 1-5 Solving Simple Inequalities Course 3 Learn to solve and graph inequalities.

  5. 1-5 Solving Simple Inequalities Course 3 Vocabulary inequality algebraic inequality solution of an inequality solution set

  6. 1-5 Solving Simple Inequalities Course 3 An inequality compares two quantities and typically uses one of these symbols:  < is greater than is less than   is greater than or equal to is less than or equal to

  7. 1-5 Solving Simple Inequalities A. 23 – 14 6 9 6 B. 5(12) 70 60 70 Course 3 Additional Example 1: Completing an Inequality Compare. Write < or >. > <

  8. 1-5 Solving Simple Inequalities A. 19 – 3 17 16 17 B. 4(15) 50 60 50 Course 3 Try This: Example 1 Compare. Write < or >. < >

  9. 1-5 Solving Simple Inequalities Course 3 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.

  10. 1-5 Solving Simple Inequalities Word Phrase Inequality Sample Solutions Solution Set 1 2 3 4 5 6 7 Course 3 x is less than 5 x < 5 x = 4 4 < 5 x = 2.1 2.1 < 5

  11. 1-5 Solving Simple Inequalities Word Phrase Inequality Sample Solutions Solution Set –3 –2 –1 0 1 2 3 Course 3 a is greater than 0 a is more than 0 a > 0 a = 7 7 > 0 a = 25 25 > 0

  12. 1-5 Solving Simple Inequalities Word Phrase Inequality Sample Solutions Solution Set –3 –2 –1 0 1 2 3 Course 3 y is less than or equal to 2 y is at most 2 y 2 y = 0 0  2 y = 1.5 1.5  2

  13. 1-5 Solving Simple Inequalities Word Phrase Inequality Sample Solutions Solution Set –1 0 1 2 3 4 5 Course 3 m is greater than or equal to 3 m is at least 3 m 3 m = 17 17  3 m = 3 3  3

  14. 1-5 Solving Simple Inequalities Course 3 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.

  15. 1-5 Solving Simple Inequalities 1 2 3 4 5 6 7 Course 3 Additional Example 2A: Solving and Graphing Inequalities Solve and graph the inequality. A. x + 2.5  8 –2.5 –2.5 Subtract 2.5 from both sides. x 5.5 According to the graph, 5.4 is a solution, since 5.4 < 5.5, and 6 should not be solution because 6 > 5.5.

  16. 1-5 Solving Simple Inequalities 1 2 3 4 5 6 7 Course 3 Additional Example 2B: Solving and Graphing Inequalities Solve and graph the inequality. B. 5t> 15 5t> 15 Divide both sides by 5. 5 5 t> 3

  17. 1-5 Solving Simple Inequalities –3 0 3 6 9 12 15 Course 3 Additional Example 2C: Solving and Graphing Inequalities Solve and graph the inequality. C. w – 1 < 8 + 1+ 1 Add 1 to both sides. w< 9

  18. 1-5 Solving Simple Inequalities p p 4 4 3  0 3 6 9 12 15 18 Course 3 Additional Example 2D: Solving and Graphing Inequalities Solve and graph the inequality. D. 3  Multiply both sides by 4. 4 • 4 • 12 p

  19. 1-5 Solving Simple Inequalities 3 6 9 12 15 18 21 1 2 3 4 5 6 7 Course 3 Try This: Examples 2A and 2B Solve and graph each inequality. A. x + 2  3.5 –2 –2 Subtract 2 from both sides. x 1.5 B. 6u> 72 6u> 72 Divide both sides by 6. 6 6 u> 12

  20. 1-5 Solving Simple Inequalities b b 9 9 2  D. 2  –21 –14 –7 0 7 14 21 0 3 6 9 12 15 18 Course 3 Try This: Examples 2C and 2D Solve and graph each inequality. C. z – 6 < 15 + 6+ 6 Add 6 to both sides. z< 21 Multiply both sides by 9. 9 • 9 • 18 b

  21. 1-5 Solving Simple Inequalities Course 3 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?

  22. 1-5 Solving Simple Inequalities 1 Understand the Problem Distance around the room Course 3 Additional Example 3 Continued 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 length of one package of border the number of packages of border  •

  23. 1-5 Solving Simple Inequalities Make a Plan 2 Course 3 Additional Example 3 Continued Use the relationship to write an inequality. Let x represent the number of packages of border.  • x 88 ft 16 ft

  24. 1-5 Solving Simple Inequalities 3 Solve Course 3 Additional Example 3 Continued 16x 88 16x 88 Divide both sides by 16. 16 16 x 5.5 At least 5.5 packages of border must be used to complete the room.

  25. 1-5 Solving Simple Inequalities 4 Course 3 Additional Example 3 Continued Look Back 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.

  26. 1-5 Solving Simple Inequalities Course 3 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?

  27. 1-5 Solving Simple Inequalities 1 Understand the Problem Course 3 Try This: Example 3 Continued 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 cookies in one package the number of packages of cookies to be purchased 130 cookies  •

  28. 1-5 Solving Simple Inequalities Make a Plan 2 Course 3 Try This: Example 3 Continued Use the relationship to write an inequality. Let x represent the number of packages of cookies.  20 cookies 130 cookies • x

  29. 1-5 Solving Simple Inequalities 3 Solve Course 3 Try This: Example 3 Continued 20x 130 20x 130 Divide both sides by 20. 20 20 x 6.5 At least 6.5 packages of cookies need to be purchased.

  30. 1-5 Solving Simple Inequalities 4 Course 3 Try This: Example 3 Continued Look Back Because whole packages of cookies must be purchased, at least 7 packages of cookies must be purchased for the party.

  31. 1-5 Solving Simple Inequalities m 2 ? –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 Course 3 Lesson Quiz Use < or > to compare each inequality. 1.135(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. > > k< 3 6  m

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