1 / 34

Chapter 6

Chapter 6. Linear Equations and Inequalities. 6.1 – Solving equations by using inverse operations. Chapter 6. Fixing a flat tire. Listed are the steps to remove a flat tire. How can you use those steps to figure out how to put on a spare?. equations. Take out a spare sheet of paper.

wallis
Télécharger la présentation

Chapter 6

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 6 Linear Equations and Inequalities

  2. 6.1 – Solving equations by using inverse operations Chapter 6

  3. Fixing a flat tire Listed are the steps to remove a flat tire. How can you use those steps to figure out how to put on a spare?

  4. equations

  5. Take out a spare sheet of paper • Write x = ___. Pick any number you want to go into the blank. • Multiply or divide each side by another number of your choice. • Now add or subtract something from each side. • This is your final equation. • Trade equations with the person next to you, and try to figure out what number the other person started with. What are some of the strategies that you used?

  6. Inverse operations • When we are solving linear equations, we use inverse operations, which means that we do the opposite of what’s in the equation. • What’s the opposite of adding (+)? • What’s the opposite of subtracting (–)? • What’s the opposite of multiplication (×)? • What’s the opposite of division (÷)? • By using inverse operations we can cancel out what is in the equation. • When we do something to one side of the equation, we need to do it to the other side as well. Think of it as a scale, and you need to keep it balanced.

  7. example 3x = –3.6  x = –3.6 ÷ 3  x = –1.2 Verify: (3)(–1.2) = –3.6 it works b) x ÷ 4 = 1.5  x = 4(1.5)  x = 6 Verify: 6 ÷ 4 = 1.5  it works

  8. Example • How do we represent 7 percent? •  7 percent = 7/100 = 0.07 • What does “of” mean in an equation •  it means multiply, so to find 7 percent of a number we need to multiply. • 0.07x = 56.7 • 0.07x ÷ 0.07 = 56.7 ÷ 0.07 • x = 810 b) (0.07)(810) = 56.7 So, the solution is correct.

  9. Try it • L = 3.7 • P = 13.2 • W = ? • P = L + L + W + W • P = 2L + 2W • 13.2 = 2(3.7) + 2W • 13.2 = 7.4 + 2W • b) 13.2 = 7.4 + 2W • 2W + 7.4 – 7.4 = 13.2 – 7.4 • 2W = 5.8 • 2W ÷ 2 = 5.8 ÷ 2 • W = 2.9 cm • c) 7.4 + 2(2.9) = 7.4 + 5.8 = 13.2 • The solution works.

  10. Translating English into algebra In math, one of the most important skills is understanding how to turn what we see in word problems into equations. Write each phrase as an algebraic expression to practice for word problems.

  11. pg. 272-274, # 8, 9, 10, 13, 16, 18, 21, 24. Independent Practice

  12. 6.2 – solving equations by using balance strategies Chapter 6

  13. Example: Balance model Solve: 6x + 2 = 10 + 4x • Algebraically: • 6x + 2 = 10 + 4x • 6x + 2 – 4x = 10 + 4x – 4x • 2x + 2 = 10 • 2x + 2 – 2 = 10 – 2 • 2x = 8 • 2x ÷ 2 = 8 ÷ 2 • x = 4

  14. example Solve:

  15. A cell phone company offers two plans. Plan A: 120 free minutes, $0.75 per additional minute Plan B: 30 free minutes, $0.25 per additional minute Which time for calls will result in the same cost for both plans? Model the problem with an equation. Solve the problem. a) Let t be time, in minutes. Plan A: You will pay for every minute after 120 minutes. So, you will pay for (t – 120) minutes. Cost of Plan A = 0.75(t – 120) Similarly, Cost of Plan B = 0.25(t – 30)  0.75(t – 120) = 0.25(t – 30) • b) 0.75(t – 120) = 0.25(t – 30) • 0.75(t – 120) ÷ 0.25 = 0.25(t – 30) ÷ 0.25 • 3(t – 120) = t – 30 • 3t – 360 = t – 30 • 3t – 360 + 360 = t – 30 + 360 • 3t = t + 330 • 3t – t = t + 330 – t • 2t = 330 • 2t ÷ 2 = 330 ÷ 2 • t = 165 minutes They will be the same cost if there are 165 minutes of calls.

  16. Magic squares activity Answer all of the questions on the Magic Squares Activity sheet to the best of your ability.

  17. P. 280-283, # 4, 6, 8, 11, 13, 18, 19, 20, 21, 23 Independent Practice

  18. 6.3 – introduction to linear inequalities6.4 – Solving linear inequalities by using addition and subtraction Chapter 6

  19. inequalities What is an inequality? We use inequalities to model a situation that can be described by a range of numbers instead of a single number. < > ≤ ≥ Less than Greater than Less than or equal to Greater than or equal to

  20. t ≤ 30

  21. Try it

  22. example We want to know the numbers that are greater than or equal to –4. –8 is lower than –4 (the more negative, the smaller a number is), so No. –3.5 is a bit higher than –4, so Yes. –4 is equal to –4, so Yes. –4.5 is lower than –4, so No. 0 is higher than –4, so Yes.

  23. Inequalities and number lines We can represent inequalities using number lines. It’s important to differentiate between “greater/less than” and “greater/less than or equal to.” Example: a < 3 Example: b ≤ –5 When it’s greater/less than, you leave the circle open (not coloured in). When it’s greater/less than or equal to, you need to remember to colour in the circle.

  24. Try it a) b) c) d)

  25. Solving inequalities Inequalities are basically just equations, with a different symbol in the place of the equals sign. We can represent inequalities with balances, like we can regular equations. How can represent this using an inequality? 7 grams < 9 grams • What would happen if we added two grams to each side? • What if we subtracted two grams from each side?

  26. Example Solve: 6 ≤ x – 4.5 • 6 ≤ x – 4.5 • 6 + 4.5 ≤ x – 4.5 + 4.5 • 10.5 ≤ x • x ≥ 10.5 Try it: 1 > r + 9 • 1 > r + 9 • 1 – 9 > r + 9 – 9 • –8 > r • r < –8

  27. example • Jake plans to board his dog while he is away on vacation. • Boarding house A charges $90 plus $5 per day. • Boarding house B charges $100 plus $4 per day. • For how many days must Jake board his dog for boarding house A to be less expensive than boarding house B? • Choose a variable and write an equality that can be used to solve the problem. • Solve and graph the problem. • Let d be the number of days. • Boarding house A = 90 + 5d • Boarding house B = 100 + 4d • We want: • Boarding house A < Boarding house B •  90 + 5d < 100 + 4d • b) 90 + 5d < 100 + 4d • 90 + 5d – 90 < 100 + 4d – 90 • 5d < 10 + 4d • 5d – 4d < 10 + 4d – 4d • d < 10 • For boarding house A to be less expensive than B, he needs to board his dog for less than 10 days.

  28. pg. 292-293, # 3, 8, 12, 13, 14, 15. pg. 298-300, # 9, 10, 13, 14, 15, 16. Independent Practice

  29. 6.5 – solving linear inequalities by using multiplication and division Chapter 6

  30. Pattern handout Fill in the blanks with the appropriate inequality symbols. From the patterns, what can we say about multiplying and dividing while solving inequalities? • The properties of inequalities: • When each side of an inequalities is multiplied or divided by a positive number, the resulting inequality is still true. • When each side of an inequality is multiplied or divided by a negative number, the inequality sign must be reversed for the inequality to remain true.

  31. example Solve: a) b) a) b)

  32. example A super-slide charges $1.25 to rent a mat and $0.75 per ride. Haru as $10.25. How many rides can Haru go on? Choose a variable, then write an inequality to solve this problem. Solve the problem and graph the solution. • Let r be the number of rides. • The total cost has to be less than or equal to $10.25. • 1.25 + 0.75r ≤ 10.25 • b) 1.25 + 0.75r ≤ 10.25 • 1.25 + 0.75r – 1.25 ≤ 10.25 – 1.25 • 0.75r ≤ 9 • 0.75r ÷ 0.75 ≤ 9 ÷ 0.75 • r ≤ 12 • Haru has to go on less than or equal to 12 rides.

  33. Inequalities Bingo

  34. pg. 304-306, # 9, 10, 11, 15, 16, 17, 18. Independent Practice

More Related