Lesson 27
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Lesson 27 NYS COMMON CORE MATHEMATICS CURRICULUM 6•4 Lesson 27: One-Step EquationsβMultiplication and Division Student Outcomes ο§ Students solve one-step equations by relating an equation to a diagram. ο§ Students check to determine if their solutions make the equations true. Lesson Notes This lesson teaches students to solve one-step equations using tape diagrams. Through the construction of tape diagrams, students create algebraic equations and solve for one variable. This lesson not only allows students to continue studying the properties of operations and identity but also allows students to develop intuition of the properties of equality. This lesson continues the informal study of the properties of equality students have practiced since Grade 1 and also serves as a springboard to the formal study, use, and application of the properties of equality seen in Grade 7. Understand that, while students intuitively use the properties of equality, diagrams are the focus of this lesson. This lesson purposefully omits focusing on the actual properties of equality, which are covered in Grade 7. Students relate an equation directly to diagrams and verbalize what they do with diagrams to construct and solve algebraic equations. Poster paper is needed for this lesson. Posters need to be prepared ahead of time, one set of questions per poster. Classwork Example 1 (5 minutes) Example 1 Solve ππ = π using tape diagrams and algebraically. Then, check your answer. First, draw two tape diagrams, one to represent each side of the equation. π π π π If π had to be split into three groups, how big would each group be? π Demonstrate the value of π using tape diagrams. π π π π Lesson 27: One-Step EquationsβMultiplication and Division This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M4-TE-1.3.0-09.2015 288 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 27 NYS COMMON CORE MATHEMATICS CURRICULUM 6•4 How can we demonstrate this algebraically? We know we have to split π into three equal groups, so we have to divide by π to show this algebraically. ππ ÷ π = π ÷ π How does this get us the value of π? The left side of the equation will equal π because we know the identity property, where π β π ÷ π = π, so we can use this identity here. The right side of the equation will be π because π ÷ π = π. Therefore, the value of π is π. How can we check our answer? We can substitute the value of π into the original equation to see if the number sentence is true. π(π) = π; π = π. This number sentence is true, so our answer is correct. Example 2 (5 minutes) Example 2 Solve π π = π using tape diagrams and algebraically. Then, check your answer. First, draw two tape diagrams, one to represent each side of the equation. π÷π π If the first tape diagram shows the size of π ÷ π, how can we draw a tape diagram to represent π? The tape diagram to represent π should be four sections of the size π ÷ π. Draw this tape diagram. π π÷π π÷π π÷π π÷π What value does each π ÷ π section represent? How do you know? Each π ÷ π section represents a value of π. We know this from our original tape diagram. How can you use a tape diagram to show the value of π? Draw four equal sections of π, which will give π the value of π. π Lesson 27: π π One-Step EquationsβMultiplication and Division This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M4-TE-1.3.0-09.2015 π 289 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 27 NYS COMMON CORE MATHEMATICS CURRICULUM 6•4 How can we demonstrate this algebraically? π π π π β π = π β π. Because we multiplied the number of sections in the original equation by π, we know the identity β π = π can be used here. How does this help us find the value of π? The left side of the equation will equal π, and the right side will equal π. Therefore, the value of π is π. How can we check our answer? Substitute π into the equation for π, and then check to see if the number sentence is true. π π = π. This is a true number sentence, so π is the correct answer. Exploratory Challenge (15 minutes) Each group (two or three) of students receives one set of problems. Have students solve both problems on poster paper with tape diagrams and algebraically. Students should also check their answers on the poster paper. More than one group may have each set of problems. Scaffolding: If students are struggling, model one set of problems before continuing with the Exploratory Challenge. Set 1 On poster paper, solve each problem below algebraically and using tape diagrams. Check each answer to show that you solved the equation correctly (algebraic and tape diagram sample responses are below). 1. 2π = 16 Tape Diagrams: ππ π π π π MP.1 π Algebraically: 2π = 16 2π ÷ 2 = 16 ÷ 2 π=8 Check: 2 β 8 = 16; 16 = 16. This is a true number sentence, so 8 is the correct solution. Lesson 27: One-Step EquationsβMultiplication and Division This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M4-TE-1.3.0-09.2015 290 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 27 NYS COMMON CORE MATHEMATICS CURRICULUM 2. π 3 6•4 =4 Tape Diagrams: π÷π π π π÷π π÷π π÷π π π π π ππ Algebraically: π =4 3 MP.1 π β3= 4β3 3 π = 12 Check: 12 3 = 4; 4 = 4. This number sentence is true, so 12 is the correct solution. Set 2 On poster paper, solve each problem below algebraically and using tape diagrams. Check each answer to show that you solved the equation correctly (algebraic and tape diagram sample responses are below). 1. 4 β π = 24 Tape Diagrams: ππ π π π π π π π π π Lesson 27: One-Step EquationsβMultiplication and Division This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M4-TE-1.3.0-09.2015 291 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 27 NYS COMMON CORE MATHEMATICS CURRICULUM 6•4 Algebraically: 4 β π = 24 4 β π ÷ 4 = 24 ÷ 4 π=6 Check: 4 β 6 = 24; 24 = 24. This number sentence is true, so 6 is the correct solution. 2. π 7 =1 Tape Diagrams: π ÷π π π MP.1 π ÷π π ÷π π ÷π π ÷π π ÷π π ÷π π ÷π π π π π π π π π π Algebraically: π =1 7 π β7= 1β7 7 π=7 Check: 7 7 = 1; 1 = 1. This number sentence is true, so 7 is the correct solution. Lesson 27: One-Step EquationsβMultiplication and Division This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M4-TE-1.3.0-09.2015 292 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 27 NYS COMMON CORE MATHEMATICS CURRICULUM 6•4 Set 3 On poster paper, solve each problem below algebraically and using tape diagrams. Check each answer to show that you solved the equation correctly (algebraic and tape diagram sample responses are below). 1. 5π = 45 Tape Diagrams: ππ π π π π π π π π π π π Algebraically: 5π = 45 5π ÷ 5 = 45 ÷ 5 π=9 Check: 5(9) = 45; 45 = 45. This number sentence is true, so 9 is the correct solution. 2. MP.1 π 3 = 10 Tape Diagrams: π÷π ππ π π÷π π÷π π÷π ππ ππ ππ π ππ Algebraically: π = ππ π π β π = ππ β π π π = ππ Check: 30 3 = 10; 10 = 10. This number sentence is true, so 30 is the correct solution. Lesson 27: One-Step EquationsβMultiplication and Division This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M4-TE-1.3.0-09.2015 293 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 27 NYS COMMON CORE MATHEMATICS CURRICULUM 6•4 Set 4 On poster paper, solve each problem below algebraically and using tape diagrams. Check each answer to show that you solved the equation correctly (algebraic and tape diagram sample responses are below). 1. 9 β π = 54 Tape Diagrams: ππ π π π π π π π π π π π π π π π π π π π Algebraically: 9 β π = 54 9 β π ÷ 9 = 54 ÷ 9 π=6 Check: 9 β 6 = 54; 54 = 54. This number sentence is true, so 6 is the correct solution. MP.1 2. 2= β 7 Tape Diagrams: π÷π π π π÷π π÷π π÷π π÷π π÷π π÷π π÷π π π π π π π π π ππ Lesson 27: One-Step EquationsβMultiplication and Division This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M4-TE-1.3.0-09.2015 294 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 27 NYS COMMON CORE MATHEMATICS CURRICULUM 6•4 Algebraically: β 7 β 2β7= β7 7 14 = β 2= Check: 2 = 14 ; 2 = 2. This number sentence is true, so 14 is the correct solution. 7 Set 5 On poster paper, solve each problem below algebraically and using tape diagrams. Check each answer to show that you solved the equation correctly (algebraic and tape diagram sample responses are below). 1. MP.1 50 = 10π Tape Diagrams: ππ π π π π π π π π π π π π π π π π π π π π π Algebraically: 50 = 10π 50 ÷ 10 = 10π ÷ 10 5=π Check: 50 = 10(5); 50 = 50. This number sentence is true, so 5 is the correct solution. Lesson 27: One-Step EquationsβMultiplication and Division This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M4-TE-1.3.0-09.2015 295 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 27 NYS COMMON CORE MATHEMATICS CURRICULUM 2. π 8 6•4 =3 Tape Diagrams: π÷π π π π÷π π÷π π÷π π÷π π÷π π÷π π÷π π÷π π π π π π π π π MP.1 π ππ Algebraically: π =3 8 π β8=3β8 8 π = 24 Check: 24 8 = 3; 3 = 3. This number sentence is true, so 24 is the correct solution. Hang completed posters around the room. Students walk around to examine other groups’ posters. Students may MP.3 either write on a piece of paper, write on Post-it notes, or write on the posters any questions or comments they may have. Answer students’ questions after providing time for students to examine posters. Lesson 27: One-Step EquationsβMultiplication and Division This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M4-TE-1.3.0-09.2015 296 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 27 NYS COMMON CORE MATHEMATICS CURRICULUM 6•4 Exercises (10 minutes) Students complete the following problems individually. Remind students to check their solutions. Exercises 1. Use tape diagrams to solve the following problem: ππ = ππ. ππ π π π π π π π Check: π(π) = ππ; ππ = ππ. This number sentence is true, so π is the correct solution. 2. π π Solve the following problem algebraically: ππ = . π π π ππ β π = β π π ππ = π ππ = Check: ππ = 3. ππ ; ππ = ππ. This number sentence is true, so ππ is the correct solution. π Calculate the solution of the equation using the method of your choice: ππ = ππ. Tape Diagrams: ππ π π π π π π π π π Algebraically: ππ = ππ ππ ÷ π = ππ ÷ π π=π Check: π(π) = ππ; ππ = ππ. This number sentence is true, so π is the correct solution. Lesson 27: One-Step EquationsβMultiplication and Division This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M4-TE-1.3.0-09.2015 297 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 27 NYS COMMON CORE MATHEMATICS CURRICULUM 4. 6•4 Examine the tape diagram below, and write an equation it represents. Then, calculate the solution to the equation using the method of your choice. ππ π π π π π π π ππ = ππ or ππ = ππ Tape Diagram: ππ ππ ππ ππ ππ ππ ππ π Algebraically: ππ = ππ ππ ÷ π = ππ ÷ π π = ππ ππ = ππ ππ ÷ π = ππ ÷ π π = ππ Check: π(ππ) = ππ, ππ = π(ππ); ππ = ππ. This number sentence is true, so ππ is the correct answer. Write a multiplication equation that has a solution of ππ. Use tape diagrams to prove that your equation has a solution of ππ. 5. Answers will vary. Write a division equation that has a solution of ππ. Prove that your equation has a solution of ππ using algebraic methods. 6. Answers will vary. Closing (5 minutes) ο§ ο§ How is solving addition and subtraction equations similar to and different from solving multiplication and division equations? οΊ Solving addition and subtraction equations is similar to solving multiplication and division equations because identities are used for all of these equations. οΊ Solving addition and subtraction equations is different from solving multiplication and division equations because they require different identities. What do you know about the pattern in the operations you used to solve the equations today? οΊ We used inverse operations to solve the equations today. Division was used to solve multiplication equations, and multiplication was used to solve division equations. Exit Ticket (5 minutes) Lesson 27: One-Step EquationsβMultiplication and Division This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M4-TE-1.3.0-09.2015 298 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 27 NYS COMMON CORE MATHEMATICS CURRICULUM Name 6•4 Date Lesson 27: One-Step Equations—Multiplication and Division Exit Ticket Calculate the solution to each equation below using the indicated method. Remember to check your answers. π 1. Use tape diagrams to find the solution of 2. Find the solution of 64 = 16π’ algebraically. Lesson 27: 10 = 4. One-Step EquationsβMultiplication and Division This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M4-TE-1.3.0-09.2015 299 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 3. Lesson 27 6•4 Use the method of your choice to find the solution of 12 = 3π£. Lesson 27: One-Step EquationsβMultiplication and Division This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M4-TE-1.3.0-09.2015 300 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 27 NYS COMMON CORE MATHEMATICS CURRICULUM 6•4 Exit Ticket Sample Solutions Calculate the solution to each equation below using the indicated method. Remember to check your answers. 1. Use tape diagrams to find the solution of π ππ = π. π ÷ ππ π π π ÷ ππ π ÷ ππ π ÷ ππ π ÷ ππ π ÷ ππ π ÷ ππ π ÷ ππ π ÷ ππ π ÷ ππ π ÷ ππ π π π π π π π π π π π ππ Check: 2. ππ ππ = π; π = π. This number sentence is true, so ππ is the correct solution. Find the solution of ππ = πππ algebraically. ππ = πππ ππ ÷ ππ = πππ ÷ ππ π=π Check: ππ = ππ(π); ππ = ππ. This number sentence is true, so π is the correct solution. 3. Use the method of your choice to find the solution of ππ = ππ. Tape Diagrams: ππ π π π π π π π Algebraically: ππ = ππ ππ ÷ π = ππ ÷ π π=π Check: ππ = π(π); ππ = ππ. This number sentence is true, so π is the correct solution. Lesson 27: One-Step EquationsβMultiplication and Division This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M4-TE-1.3.0-09.2015 301 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 27 NYS COMMON CORE MATHEMATICS CURRICULUM 6•4 Problem Set Sample Solutions 1. Use tape diagrams to calculate the solution of ππ = ππ. Then, check your answer. ππ π π π π π π π π π π π Check: ππ = π(π); ππ = ππ. This number sentence is true, so π is the correct solution. 2. Solve ππ = π algebraically. Then, check your answer. π π π π ππ β π = β π π ππ = π ππ = Check: ππ = 3. ππ ; ππ = ππ. This number sentence is true, so ππ is the correct solution. π Use tape diagrams to calculate the solution of π π = ππ. Then, check your answer. π÷π ππ π π÷π π÷π π÷π π÷π π÷π ππ ππ ππ ππ ππ π ππ Check: ππ π = ππ; ππ = ππ. This number sentence is true, so ππ is the correct solution. Lesson 27: One-Step EquationsβMultiplication and Division This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M4-TE-1.3.0-09.2015 302 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 27 NYS COMMON CORE MATHEMATICS CURRICULUM 4. 6•4 Solve πππ = ππ algebraically. Then, check your answer. πππ = ππ πππ ÷ ππ = ππ ÷ ππ π=π Check: ππ(π) = ππ; ππ = ππ. This number sentence is true, so π is the correct solution. 5. Write a division equation that has a solution of π. Prove that your solution is correct by using tape diagrams. Answers will vary. 6. Write a multiplication equation that has a solution of π. Solve the equation algebraically to prove that your solution is correct. Answers will vary. 7. When solving equations algebraically, Meghan and Meredith each got a different solution. Who is correct? Why did the other person not get the correct answer? Meghan π =π π Meredith π =π π π βπ=πβπ π π ÷π=π÷π π π=π π=π Meghan is correct. Meredith divided by π to solve the equation, which is not correct because she would end up with π π = π. To solve a division equation, Meredith must multiply by π to end up with π because the identity states π ÷ π β π = π. Lesson 27: One-Step EquationsβMultiplication and Division This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M4-TE-1.3.0-09.2015 303 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
