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- 1. Lesson 27: One-Step EquationsβMultiplicationand Division Date: 4/20/15 277 Β© 2013 Common Core, Inc. Some rightsreserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 6β’4Lesson 27 π π π π Lesson 27: One-Step EquationsβMultiplication and Division Student Outcomes ο§ Students solveone-step equations by relatingan equation to a diagram. ο§ Students check to determine if their solution makes the equation true. Lesson Notes This lesson teaches students to solveone-step equations using tape diagrams. Through the construction of tape diagrams,students will createalgebraic equations and solvefor one variable. This lesson notonly 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 sincefirstgrade,and also serves as a springboard to the formal study, use, and application of the properties of equality seen in Grade 7. Understand that, whilestudents will intuitively usethe properties of equality,diagrams arethe focus of this lesson. This lesson purposefully omits focusingon the actual properties of equality,which will becovered in Grade 7. Students will relatean equation directly to diagrams and verbalizewhatthey do with diagrams to constructand 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 = πusing tape diagramsand algebraically. Then,check your answer. First, draw two tape diagrams, one to represent each side oftheequation. If πhad to be split into threegroups, how big wouldeach group be? π Demonstrate the valueof π using tape diagrams. π π π π
- 2. Lesson 27: One-Step EquationsβMultiplicationand Division Date: 4/20/15 278 Β© 2013 Common Core, Inc. Some rightsreserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 6β’4Lesson 27 π π Γ· π π Γ· ππ Γ· π π Γ· π How can we demonstrate thisalgebraically? Weknow we haveto split πinto threeequal groups, so wehaveto divideby πto show this algebraically. ππΓ· π = π Γ· π How doesthisget usthe value of π? The left sideoftheequation will equal π, because weknow theidentity property, where π β π Γ· π = π, so wecan usethis identity here. The right sideoftheequation will be πbecause π Γ· π = π. Therefore, thevalueof π is π. How can we check our answer? Wecan substitutethevalueof π into theoriginal equation toseeifthenumber sentenceis true. π( π)= π; π = π. This number sentenceis true, so our answer is correct. Example 2 (5 minutes) Example 2 Solve π π = πusing tape diagramsand algebraically. Then, check your answer. First, draw two tape diagrams, one to represent each side oftheequation. If the first tape diagram showsthe size of π Γ· π, how can we draw atape diagram to represent y? The tapediagram torepresent π should be four sections ofthesize π Γ· π. Draw thistape diagram. What value doeseach πΓ· πsectionrepresent? How do you know? Each π Γ· πsection represents a valueof π, weknows this from our originaltapediagram. How can you use atape diagram to show the value of π? Draw four equal sections of π, whichwill give πthevalueof π. π Γ· π π π ππ π
- 3. Lesson 27: One-Step EquationsβMultiplicationand Division Date: 4/20/15 279 Β© 2013 Common Core, Inc. Some rightsreserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 6β’4Lesson 27 How can we demonstrate thisalgebraically? π π β π = π β π. Becausewemultiplied thenumber ofsections in theoriginalequation by π, weknow theidentity π π β π = π and can usethis identity here. How doesthishelp usfind the valueof π? The left sideoftheequation will equal π, andtheright sidewillequal π. Therefore, thevalueof πis π. How can we check our answer? Substitute πinto theequation for π, thencheck toseeifthenumber sentenceis true. π π = π. This is a truenumber sentence, so πis thecorrect 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. Set 1 On poster paper, solve each problem below algebraically and using tapediagrams. Check each answer to show that you solved the equationcorrectly. 1. ππ = ππ TapeDiagrams: Algebraically: ππ = ππ ππ Γ· π = ππ Γ· π π = π Check: πβ π = ππ; ππ= ππ. This is a truenumber sentence, so πis thecorrect solution. Scaffolding: If students are struggling, model one set of problems before continuingwith the Exploratory Challenge. ππ π ππ π π MP.1
- 4. Lesson 27: One-Step EquationsβMultiplicationand Division Date: 4/20/15 280 Β© 2013 Common Core, Inc. Some rightsreserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 6β’4Lesson 27 ππ π π Γ· π π π Γ· ππ Γ· π π Γ· π ππ π π 2. π π = π TapeDiagrams: Algebraically: π π = π π π β π = πβ π π = ππ Check: ππ π = π; π = π. This number sentenceis true, so ππis thecorrect solution. Set 2 On poster paper, solve each problem below algebraically and using tapediagrams. Check each answerto show that you solved the equationcorrectly. 1. πβ π = ππ Tape Diagram: Algebraically: π β π = ππ π β π Γ· π = ππΓ· π π = π Check: πβ π = ππ; ππ = ππ. This number sentenceis true, so πis thecorrect solution. π ππ ππ ππ ππ ππ MP.1
- 5. Lesson 27: One-Step EquationsβMultiplicationand Division Date: 4/20/15 281 Β© 2013 Common Core, Inc. Some rightsreserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 6β’4Lesson 27 2. π π = π TapeDiagrams: Algebraically: π π = π π π β π = π β π π = π Check: π π = π; π = π. This number sentenceis true, so πis thecorrect solution. Set 3 On poster paper, solve each problem below algebraically and using tapediagrams. Check each answerto show that you solved the equationcorrectly. 1. ππ = ππ TapeDiagram: Algebraically: ππ = ππ ππΓ· π = ππΓ· π π = π Check: π( π) = ππ; ππ = ππ. This number sentenceis true, so πis thecorrect solution. π π π Γ· π π π π Γ· ππ Γ· π π Γ· ππ Γ· ππ Γ· ππ Γ· π π Γ· π ππ ππππ π ππ ππ ππ π π ππ ππ π MP.1
- 6. Lesson 27: One-Step EquationsβMultiplicationand Division Date: 4/20/15 282 Β© 2013 Common Core, Inc. Some rightsreserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 6β’4Lesson 27 ππ π π Γ· π ππ π Γ· ππ Γ· π π Γ· π ππππ ππ π 2. π π = ππ TapeDiagrams: Algebraically: π π = ππ π π β π = ππβ π π = ππ Check: ππ π = ππ; ππ= ππ. This number sentenceis true, so ππis thecorrect solution. Set 4 On poster paper, solve each problem below algebraically and using tapediagrams. Check each answer to show that you solved the equationcorrectly. 1. πβ π = ππ TapeDiagrams: Algebraically: πβ π = ππ π β π Γ· π = ππ Γ· π π = π Check: πβ π = ππ; ππ = ππ. This number sentenceis true, so πis thecorrect solution. ππ ππ ππ πππ ππ ππ ππ πππ ππ π MP.1
- 7. Lesson 27: One-Step EquationsβMultiplicationand Division Date: 4/20/15 283 Β© 2013 Common Core, Inc. Some rightsreserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 6β’4Lesson 27 2. π = π π TapeDiagrams: Algebraically: π = π π π β π = π π β π ππ = π Check: π = ππ π ; π/π. This number sentenceis true, so ππis thecorrect solution. Set 5 On poster paper, solve each problem below algebraically and using tapediagrams. Check each answerto show that you solved the equationcorrectly. 1. ππ= πππ TapeDiagrams: Algebraically: ππ = πππ ππΓ· ππ = πππΓ· ππ π = π Check: ππ= ππ(π); ππ = ππ. This number sentenceis true, so πis thecorrect solution. ππ π π Γ· π π π π Γ· ππ Γ· π π Γ· ππ Γ· ππ Γ· ππ Γ· π π Γ· π ππ ππππ π ππ ππ ππ πππ ππ π ππ ππ πππ ππ π π MP.1
- 8. Lesson 27: One-Step EquationsβMultiplicationand Division Date: 4/20/15 284 Β© 2013 Common Core, Inc. Some rightsreserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 6β’4Lesson 27 ππ π ππ π ππ π 2. π π = π TapeDiagrams: Algebraically: π π = π π π β π = π β π π = ππ Check: ππ π = π; π = π. This number sentenceis true, so ππis thecorrect solution. Hang completed posters around the room. Students walk around to examine other groupsβ posters. Students may either write on a piece of paper or write on the posters any questions or comments they may have. Answer studentsβ questions after providingtime for students to examine posters. Exercises(10 minutes) Students complete the followingproblems individually. Remind students to check their solutions. Exercises 1. Use tape diagramsto solve the following problem: ππ = ππ. Check: π( π) = ππ; ππ = ππ. This number sentenceis true, so πis thecorrect solution. π π Γ· π π ππ π π Γ· ππ Γ· π π Γ· ππ Γ· ππ Γ· ππ Γ· π π Γ· π π Γ· π ππ ππππ π π MP.3 MP.1
- 9. Lesson 27: One-Step EquationsβMultiplicationand Division Date: 4/20/15 285 Β© 2013 Common Core, Inc. Some rightsreserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 6β’4Lesson 27 2. Solve the following problem algebraically: ππ = π π . ππ = π π ππ β π = π π β π ππ = π Check: ππ = ππ π ; ππ= ππ. This number sentence is true, so ππis thecorrect solution. 3. Calculate the solutionofthe equationusing themethod ofyour choice: ππ = ππ. TapeDiagrams: Algebraically: ππ = ππ ππΓ· π = ππΓ· π π = π Check: π( π) = ππ; ππ = ππ. This number sentenceis true, so πis thecorrect solution. 4. Examine the tape diagram below and writean equation it represents. Then, calculate thesolution tothe equation using the method ofyour choice. ππ = ππor ππ= ππ TapeDiagram: Algebraically: ππ = ππ ππΓ· π = ππΓ· π π = ππ ππ = ππ ππΓ· π = ππΓ· π π = ππ Check: π( ππ)= ππ, ππ = π( ππ); ππ = ππ. This number sentenceis true, so ππis thecorrect answer. 5. Write a multiplication equationthat hasasolution of ππ. Use tape diagramsto prove that yourequation hasa solution of ππ. Answers will vary. 6. Write a division equation that hasasolution of ππ. Prove that yourequation hasasolution of ππusing algebraic methods. Answers will vary. ππ ππ π π π ππ π π ππ ππ ππππ π π ππππ ππππππππ ππ
- 10. Lesson 27: One-Step EquationsβMultiplicationand Division Date: 4/20/15 286 Β© 2013 Common Core, Inc. Some rightsreserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 6β’4Lesson 27 Closing(5 minutes) ο§ How is solvingaddition and subtraction equations similarand differentto s olvingmultiplication and division equations? ο§ What do you know aboutthe pattern in the operations you used to solvethe 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)
- 11. Lesson 27: One-Step EquationsβMultiplicationand Division Date: 4/20/15 287 Β© 2013 Common Core, Inc. Some rightsreserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 6β’4Lesson 27 Name Date Lesson 27: One-Step EquationsβMultiplication and Division Exit Ticket Calculatethe solution to each equation below usingthe indicated method. Remember to check your answers. 1. Use tape diagrams to find the solution of π 10 = 4. 2. Find the solution of 64 = 16π’ algebraically. 3. Use the method of your choice to find the solution of 12 = 3π£.
- 12. Lesson 27: One-Step EquationsβMultiplicationand Division Date: 4/20/15 288 Β© 2013 Common Core, Inc. Some rightsreserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 6β’4Lesson 27 Exit Ticket Sample Solutions Calculate the solutionto each equation below using the indicatedmethod. Rememberto check youranswers. 1. Use tape diagramsto find the solution of π ππ = π. Check: ππ ππ = π; π = π. This number sentenceis true, so ππis thecorrect solution. 2. Find the solution of ππ = πππalgebraically. ππ = πππ ππΓ· ππ = πππ Γ· ππ π = π Check: ππ = ππ( π); ππ= ππ. This number sentenceis true, so πis thecorrect solution. 3. Use the method ofyour choiceto find thesolution of ππ = ππ. TapeDiagrams: Algebraically: ππ= ππ ππΓ· π = ππΓ· π π = π Check: ππ = π(π); ππ = ππ. This number sentenceis true, so πis thecorrect solution. ππ π π Γ· ππ π π π Γ· πππ Γ· ππ π Γ· πππ Γ· πππ Γ· ππ π Γ· πππ Γ· ππ π Γ· ππ π Γ· πππ Γ· ππ ππ ππππ π π ππ ππ π ππ π ππ π
- 13. Lesson 27: One-Step EquationsβMultiplicationand Division Date: 4/20/15 289 Β© 2013 Common Core, Inc. Some rightsreserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 6β’4Lesson 27 Problem Set Sample Solutions 1. Use tape diagramsto calculate the solution of ππ = ππ. Then, check youranswer. Check: ππ = π( π); ππ = ππ. This number sentenceis true, so πis thecorrect solution. 2. Solve ππ = π π algebraically. Then, check your answer. ππ = π π ππβ π = π π β π ππ = π Check: ππ = ππ π ; ππ= ππ. This number sentenceis true, so ππis thecorrect solution. 3. Use tape diagramsto calculate thesolution of π π = ππ. Then, check your answer. π Γ· π ππ π π Γ· π π Γ· π π Γ· π π Γ· π π Γ· π ππ ππ ππ ππ ππ π ππ Check: ππ π = ππ; ππ= ππ. This number sentenceis true, so thesolutionis correct. ππ ππ π π π π ππ π π π
- 14. Lesson 27: One-Step EquationsβMultiplicationand Division Date: 4/20/15 290 Β© 2013 Common Core, Inc. Some rightsreserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 6β’4Lesson 27 4. Solve πππ= ππalgebraically. Then, check your solution. πππ = ππ πππΓ· ππ = ππΓ· ππ π = π Check: ππ( π)= ππ; ππ= ππ. This number sentenceis true, so thesolutionis correct. 5. Write adivision equation that hasasolution of π. Prove that yoursolution iscorrect by using tapediagrams. Answers will vary. 6. Write amultiplication equationthat hasasolution of π. Solve the equation algebraically to provethat your solution iscorrect. Answers will vary. 7. When solving equationsalgebraically,Meghan and Meredith each got adifferent solution. Who iscorrect? Why did the other person not get thecorrect answer? Meghan Meredith π π = π π π = π π π β π = πβ π π π Γ· π = πΓ· π π = π π = π Meghan is correct.Meredith divided by πto solvetheequation, which is notcorrect because she would end up with π π = π. To solvea division equation, Meredith must multiply by π to end upwith ππ,which is thesameas π.