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Algebraic Equations Warm-Up Problem of the Day and Lesson Quizzes

This lesson covers solving linear equations, combining like terms, and using the distributive property. Includes warm-up problems, lesson presentation, and quizzes.

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Algebraic Equations Warm-Up Problem of the Day and Lesson Quizzes

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  1. Warm Up Problem of the Day Lesson Presentation Lesson Quizzes

  2. Warm Up Solve. 1. –8p – 8 = 56 2. 13d – 5 = 60 3. 9x + 24 = 60 4. p = –8 d = 5 x = 4 k 7 k = 49 + 4 = 11 z 4 z = 20 5. 19 + = 24

  3. Problem of the Day Without using a calculator, multiply 2.637455 by 6, add 12, divide the result by 3, subtract 4, and then multiply by 0.5. What number will you end with? (Hint: If you start with x, what do you end with?) 2.637455 (the number you started with)‏

  4. Sunshine State Standards Preview of MA.912.A.3.1 Solve linear equations in one variable that include simplifying algebraic equations.

  5. Additional Example 1: Combining Like Terms to Solve Equations Solve 12 – 7b + 10b = 18. 12 – 7b + 10b = 18 12 + 3b = 18 Combine like terms. – 12 – 12 Subtract 12 from both sides. 3b = 6 3b = 6 Divide both sides by 3. 3 3 b = 2

  6. Check It Out: Example 1 Solve 14 – 8b + 12b = 62. 14 – 8b + 12b = 62 14 + 4b = 62 Combine like terms. – 14 – 14 Subtract 14 from both sides. 4b = 48 4b = 48 Divide both sides by 4. 4 4 b = 12

  7. You may need to use the Distributive Property to solve an equation that has parentheses. Multiply each term inside the parentheses by the factor that is outside the parentheses. Then combine like terms.

  8. Remember! The Distributive Property states that a(b + c) = ab + ac. For instance, 2(3 + 5) = 2(3) + 2(5).

  9. Additional Example 2: Using the Distributive Property to Solve Equations Solve 5(y – 2) + 6 = 21 5(y – 2) + 6 = 21 Distribute 5 on the left side. 5(y)– 5(2) + 6 = 21 Simplify and combine like terms. 5y – 4 = 21 + 4 + 4 Add 4 to both sides. 5y = 25 55 Divide both sides by 5. y = 5

  10. Check It Out: Example 2 Solve 3(x – 3) + 4 = 28. 3(x – 3) + 4 = 28 Distribute 3 on the left side. 3(x)– 3(3) + 4 = 28 Simplify and combine like terms. 3x – 5 = 28 + 5 + 5 Add 5 to both sides. 3x = 33 33 Divide both sides by 3. x = 11

  11. Additional Example 3: Problem Solving Application Troy has three times as many trading cards as Hillary. Subtracting 9 from the number of trading cards Troy has and then dividing by 6 gives the number of cards Sean has. If Sean owns 24 trading cards, how many trading cards does Hillary own?

  12. 1 Understand the Problem Additional Example 3 Continued Rewrite the question as a statement. • Find the number of trading cards that Hillary owns. List the important information: • Troy owns 3 times as many trading cards as Hillary has. • Subtracting 9 from the number of trading cards that Troy owns and then dividing by 6 gives the number cards Sean owns. • Sean owns 24 trading cards.

  13. 2 Make a Plan Let c represent the number of trading cards Hillary owns. Then 3c represents the number Troy has, and 3c – 9 6 represents the number Sean owns, which equals 24. 3c – 9 6 Solve the equation = 24 for c to find the number of cards Hillary owns. Additional Example 3 Continued

  14. 3 Solve 3c – 9 6 = 24 3c – 9 6 (6)‏ = (6)24 3c = 153 3 3 Additional Example 3 Continued Multiply both sides by 6. 3c – 9 = 144 3c – 9 + 9 = 144 + 9 Add 9 to both sides. 3c = 153 Divide both sides by 3. c = 51 Hillary owns 51 cards.

  15. 4 Additional Example 3 Continued Look Back If Hillary owns 51 cards, then Troy owns 153 cards. When you subtract 9 from 153, you get 144. And 144 divided by 6 is 24, which is the number of cards that Sean owns. So the answer is correct.

  16. Check It Out: Example 3 John is twice as old as Helen. Subtracting 4 from John’s age and then dividing by 2 gives William’s age. If William is 24, how old is Helen?

  17. 1 Understand the Problem Check It Out: Example 3 Continued Rewrite the question as a statement. • Find Helen’s age. List the important information: • John is 2 times as old as Helen. • Subtracting 4 from John’s age and then dividing by 2 gives William’s age. • William is 24 years old.

  18. 2 Make a Plan Check It Out: Example 3 Continued Let h stand for Helen’s age. Then 2h represents John’s age, and 2h – 4 2 represents William’s age, which equals 24. 2h – 4 2 Solve the equation = 24 for h to find Helen’s age.

  19. 3 Solve Check It Out: Example 3 Continued 2h – 4 2 = 24 2h – 4 2 = (2)24 Multiply both sides by 2. (2)‏ 2h – 4 = 48 Add 4 to both sides. 2h – 4 + 4 = 48 + 4 2h = 52 2h = 52 Divide both sides by 2. 2 2 h = 26 Helen is 26 years old.

  20. 4 Check It Out: Example 3 Continued Look Back If Helen is 26 years old, then John is 52 years old. When you subtract 4 from 52, you get 48. And 48 divided by 2 is 24, which is William’s age. So the answer is correct.

  21. Lesson Quizzes Standard Lesson Quiz Lesson Quiz for Student Response Systems

  22. Lesson Quiz Solve. 1.c + 21 + 5c = 63 2. –x – 11 + 17x = 53 3. 59 = w – 16 + 4w 4. 4(k – 3) + 1 = 33 c = 7 x = 4 w= 15 k = 11 5. Kelly swam 4 times as many laps as Kathy. Adding 5 to the number of laps Kelly swam gives you the number of laps Julie swam. If Julie swam 9 laps, how many laps did Kathy swim? 1 lap

  23. Lesson Quiz for Student Response Systems 1. Solve m + 29 + 8m = 110. • A.m = 8 B.m = 9 C.m = 10 D.m = 11

  24. Lesson Quiz for Student Response Systems 2. Solve –f – 13 + 15f = 71. • A.f = 3 B.f = 5 C.f = 6 D.f = 7

  25. Lesson Quiz for Student Response Systems 3. Solve 7(r – 5) + 9 = 72. • A.r = 7 B.r = 9 C.r = 12 D.r = 14

  26. Lesson Quiz for Student Response Systems • 4. Mark has 5 times as many collectibles as Maria has. Subtracting 6 from the number of collectibles Mark has and then dividing by 2 gives the number of collectibles Jim has. If Jim has 112 collectibles, how many collectibles does Maria have? • A.35 collectibles B.40 collectibles C.46 collectibles D.59 collectibles

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