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Coach Bridges’ Lesson Plans

1/14/19-1/18/19

Monday 1/14:

Unit 6, Lesson 7: Reasoning about Solving Equations (Part 1)

Lesson Goals• Make connections between solving an equation of the form px+q=r and finding an

unknown weight on a balanced hanger diagram.• Describe balanced hangers with equations of the form px+q=r

Required Materials

7.1: Hanger Diagrams (10 minutes)

Setup:Notice and wonder about the photos of socks. 3 minutes of quiet work time followed by a whole-class discussion.

Student task statement

In the two diagrams, all the triangles weigh the same and all the squares weigh the same.

For each diagram, come up with . . .

1. One thing that must be true 2. One thing that could be true3. One thing that cannot possibly be true

Possible responses

Answers vary. Possible responses:

1. Triangle is heavier than square; 1 triangle weighs same as 3 squares and a circle.2. Triangle weighs 32 ounces, square weighs 10 ounces, and circle weighs 2 ounces.3. Triangle and square weigh the same.

7.2: Hanger and Equation Matching (15 minutes)

Setup:Groups of 2. 5–10 minutes of quiet work time and share with a partner followed by a whole-class discussion.


On each balanced hanger, figures with the same letter have the same weight.

• 2+3=5• 3+2=3• 6=2+3• 7=3+1

1. Match each hanger to an equation. Complete the equation by writing x, y, z, or w in the empty box.

2. Find the solution to each equation. Use the hanger to explain what the solution means.

Possible responses1. 7=3w+12. 2 z+3=53. 3 x+2=34. 6=2 y+31. w=22. z=1

3. x=13

4. y=32

7.3: Use Hangers to Understand Equation Solving (15 minutes)

Setup:Keep students in the same groups. 5–10 minutes of quiet work time and share with a partner followed by a whole-class discussion.


Here are some balanced hangers where each piece is labeled with its weight. For each diagram:

1. Write an equation.2. Explain how to figure out the weight of a piece labeled with a letter by reasoning about the

diagram.

3. Explain how to figure out the weight of a piece labeled with a letter by reasoning about the equation.

Possible responses

1. A: 7=3 x+1 B: 2 y+10=31 C: 6.8=2 z+2.2 D: 4w+ 32=172

2. Sample reasoning for diagram A: remove 1 unit of weight from each side of the hanger, leaving 6 units on the left and 3 x’s on the right. Split each side into three equal groups, showing that x=2.

3. Sample reasoning for 7=3 x+1: Subtract 1 from each side, leaving 6=3 x. Divide each side by 3, leaving 2=x.

Lesson Synthesis (5 minutes)Display the equation 4 x+6=9.2. Ask students to work with their partner to draw a corresponding hanger diagram. Then, one partner solves by reasoning about the equation, the other solves by reasoning about the diagram. Ask students to compare the two strategies and discuss how they are alike and how they are different.

7.4: Solve the Equation (Cool-down, 5 minutes)

Setup:None.


Solve the equation. If you get stuck, try using a diagram.

5 x+ 14=614

Possible responses

x=3

Tuesday 1/15:

Unit 6, Lesson 8: Reasoning about Solving Equations (Part 2)

Lesson Goals• Make connections between solving an equation of the form p(x+q)=r and finding an

unknown weight on a balanced hanger diagram.• Understand why a balanced hanger with multiple copies of a group of weights can be

represented by equations of both forms px+q=r and p(x+q)=r

Required Materials

8.1: Equivalent to 2(x+3) (5 minutes)

Setup:Groups of 2. 3 minutes of quiet work time, followed by sharing with a partner and whole-class discussion.


Select all the expressions equivalent to 2(x+3).

1. 2 ⋅(x+3)2. (x+3)23. 2 ⋅ x+2 ⋅3 4. 2 ⋅ x+35. (2 ⋅ x)+3 6. (2+x )3

Possible responses

1, 2, 3

8.2: Either Or (15 minutes)

Setup:Keep students in the same groups. 5–10 minutes of quiet work time, then sharing with a partner followed by a whole-class discussion.

Student task statement1. Explain why either of these equations could represent this hanger:

14=2(x+3) or 14=2 x+6

2. Find the weight of one circle. Be prepared to explain your reasoning.

Possible responses1. Answers vary. The diagram shows 14 balanced with 2 groups of x+3, and this corresponds

to 14=2(x+3). The diagram also shows 14 balanced with 2 x’s and another 6 units of weight, which corresponds to 14=2 x+6.

2. 4 units. Reasoning varies.

8.3: Use Hangers to Understand Equation Solving, Again (15 minutes)

Setup:Keep students in the same groups. Give 5–10 minutes of quiet work time and time to share their responses with a partner, followed by a whole-class discussion.


Here are some balanced hangers. Each piece is labeled with its weight.

For each diagram:

1. Assign one of these equations to each hanger:

2(x+5)=16

3( y+200)=3,000

20.8=4 (z+1.1)

203

=2(w+ 23 )

2. Explain how to figure out the weight of a piece labeled with a letter by reasoning about the diagram.

3. Explain how to figure out the weight of a piece labeled with a letter by reasoning about the equation.

Possible responses1. Each equation corresponds to the diagram with the variable that matches.2. See lesson plan.3. See lesson plan.

Lesson Synthesis (5 minutes)Display the equation 4 (x+7)=40. Ask one partner to solve by dividing first and the other to solve by distributing first. Then, check that they got the same solution and that it makes the equation true. If they get stuck, encourage them to draw a diagram to represent the equation.

8.4: Solve Another Equation (Cool-down, 5 minutes)

Setup:None.


Solve the equation 3(x+4.5)=36. If you get stuck, use the diagram.

Possible responses

x=7.5

Wednesday 1/16:

Unit 6, Lesson 9: Dealing with Negative Numbers

Lesson Goals• Understand that the strategy of doing the same to each side can be applied to equations

with negative numbers.

• Solve equations of the form px+q=r and p(x+q)=r where p, q, and r are rational numbers

Required Materials

9.1: Which One Doesn’t Belong: Rational Number Arithmetic (10 minutes)

Setup:Students in groups of 2–4. Display the image for all to see. 2 minutes of quiet think time, followed by small-group discussions.


Which equation doesn’t belong?

15=−5⋅−3

2+−5=−3

4−−2=6

−3 ⋅−4=−12

Possible responses1. 15=−5⋅−3 doesn’t belong because it’s the only one with a number on the left and an

operation on the right.2. 2+−5=−3 doesn’t belong because it’s the only one with addition.3. 4−−2=6 doesn’t belong because it’s the only one with subtraction.4. −3 ⋅−4=−12 doesn’t belong because it’s the only one that is not true.

9.2: Old and New Ways to Solve (15 minutes)

Setup:5–10 minutes of quiet work time followed by a whole-class discussion.


Solve each equation. Be prepared to explain your reasoning.

1. x+6=42. x−−4=−63. 2(x−1)=−2004. 2 x+−3=−23

Possible responses1. x=−22. x=−103. x=−994. x=−10

Anticipated misconceptions

Some students may need some additional support remembering and applying strategies for performing operations on signed numbers. Draw their attention to any anchor charts or notes that are available from the previous unit.

9.3: Keeping It True (15 minutes)

Setup:Groups of 2. 2 minutes of quiet work time on the first question, pause for a discussion, and then time to complete the task with their partner.


Here are some equations that all have the same solution.

x ¿−6x−3 ¿−9−9 ¿ x−3900 ¿−100(x−3)900 ¿(x−3) ⋅(−100)900 ¿−100x+300

1. Explain how you know that each equation has the same solution as the previous equation. Pause for discussion before moving to the next question.

2. Keep your work secret from your partner. Start with the equation −5=x. Do the same thing to each side at least three times to create an equation that has the same solution as the starting equation. Write the equation you ended up with on a slip of paper, and trade equations with your partner.

3. See if you can figure out what steps they used to transform −5=x into their equation. When you think you know, check with them to see if you are right.

Possible responses1. Answers vary. Sample responses:

– subtract 3 from each side– swap the two sides of the equation– multiply each side by -100– swap the factors -100 and x−3 (the commutative property of multiplication)

– apply the distributive property2. Answers vary.3. Answers vary.


Some students may need some additional support remembering and applying strategies for performing operations on signed numbers. Draw their attention to any anchor charts or notes that are available from the previous unit.

Lesson Synthesis (5 minutes)Ask students to think of one or two important things they learned in this lesson, and share them with a partner.

9.4: Solve Two More Equations (Cool-down, 5 minutes)

Setup:None.


Solve each equation. Show your work, or explain your reasoning.

1. −3 x−5=162. −4( y−2)=12

Possible responses1. x=−72. y=−1

Thursday 1/17:

Unit 6, Lesson 10: Different Options for Solving One Equation

Lesson Goals• Identify and correct errors in solution pathways for equations of the form p(x+q)=r• Understand that there are two approaches to solving equations of the form p(x+q)=r:

first dividing each side by p, or first expanding with the distributive property, and that one might be more efficient depending on the numbers involved

Required Materials

10.1: Algebra Talk: Solve Each Equation (5 minutes)

Setup:Display one equation at a time. Allow 30 seconds of quiet think time, followed by a whole-class discussion.


100(x−3) ¿1,000500(x−3) ¿5,0000.03(x−3) ¿0.30.72(x+2) ¿7.217(x+2) ¿

107

Possible responses

13, 13, 13, 8, 8

10.2: Analyzing Solution Methods (15 minutes)

Setup:Groups of 2. 5–10 minutes quiet work time and time to share their reasoning with their partner, followed by a whole-class discussion.


Three students each attempted to solve the equation 2(x−9)=10, but got different solutions. Here are their methods. Do you agree with any of their methods, and why?

Noah’s method:

2(x−9) ¿10 ¿2(x−9)+9 ¿10+9 ¿each side ¿2 x¿=19¿¿2 x ÷2¿=19÷2¿divide eachside by 2¿ x ¿=192

¿¿

Elena’s method:

2(x−9) ¿10 ¿2x−18 ¿10 apply the distributive property2 x−18−18 ¿10−18 ¿ ¿ ¿

=−8 ¿¿2x ÷2¿=−8÷2¿ divide eachside by 2 ¿x ¿=−4¿¿

Andre’s method:

2(x−9) ¿10 ¿2 x−18 ¿10 apply the distributive property2 x−18+18 ¿10+18 ¿ ¿ ¿

=28¿¿2x ÷2¿=28÷2¿divide eachside by 2¿ x ¿=14¿¿

Possible responses

Disagree, disagree, agree.


If students aren’t sure how to begin analyzing Noah’s method, ask them to explain what it means for a number to be a solution of an equation. Alternatively, suggest that they draw a tape diagram of 2(x−9)=10.

10.3: Solution Pathways (15 minutes)

Setup:Keep students in the same groups. 5–10 minutes of quiet or partner work time followed by a whole-class discussion.


For each equation, try to solve the equation using each method (dividing each side first, or applying the distributive property first). Some equations are easier to solve by one method than the other. When that is the case, stop doing the harder method and write down the reason you stopped.

1. 2,000(x−0.03)=6,000

2. 2(x+1.25)=3.5

3.14(4+x )=4

3

4. −10(x−1.7)=−3

5. 5.4=0.3(x+8)

Possible responses1. 3.032. 0.5

3.43 or equivalent

4. 25. 10

Lesson Synthesis (5 minutes)Discussion questions: What are the two main ways we can approach solving equations like the ones we saw today? What kinds of things do we look for to decide which approach is better? How can we check if our answer is a solution to the original equation?

10.4: Solve Two Equations (Cool-down, 5 minutes)

Setup:None.


Solve each equation. Show or explain your method.

1. 8.88=4.44 (x−7)

2. 5( y+ 25 )=−13

Possible responses1. x=92. y=−3

Friday 1/18:

Unit 6, Lesson 11: Using Equations to Solve Problems

Lesson Goals• Solve word problems leading to equations of the form px+q=r or p(x+q)=r

Required Materials• sticky notes• tools for creating a visual display

11.1: Remember Tape Diagrams (5 minutes)

Setup:Groups of 2. 5 minutes of quiet think time and time to share their work with a partner followed by a whole-class discussion.


1. Write a story that could be represented by this tape diagram.2. Write an equation that could be represented by this tape diagram.

Possible responses1. Answers vary. Sample story: A baker put a cookies in each of three boxes. Then, he put 2

more cookies in each box, and there were 24 total cookies in the 3 boxes.2. 24=3(a+2) or 24=3a+6 or equivalent

11.2: At the Fair (15 minutes)

Setup:Students in groups of 2. 5–10 minutes quiet work time and partner discussions followed by a whole-class discussion.

Student task statement1. Tyler is making invitations to the fair. He has already made some of the invitations, and he

wants to finish the rest of them within a week. He is trying to spread out the remaining work, to make the same number of invitations each day. Tyler draws a diagram to represent the situation.

1. Explain how each part of the situation is represented in Tyler’s diagram:

How many total invitations Tyler is trying to make.

How many invitations he has made already.

How many days he has to finish the invitations.

2. How many invitations should Tyler make each day to finish his goal within a week? Explain or show your reasoning.

3. Use Tyler’s diagram to write an equation that represents the situation. Explain how each part of the situation is represented in your equation.

4. Show how to solve your equation.

2. Noah and his sister are making prize bags for a game at the fair. Noah is putting 7 pencil erasers in each bag. His sister is putting in some number of stickers. After filling 3 of the bags, they have used a total of 57 items.

1. Explain how the diagram represents the situation.2. Noah writes the equation 3(x+7)=57 to represent the situation. Do you agree

with him? Explain your reasoning.3. How many stickers is Noah's sister putting in each prize bag? Explain or show your

reasoning.3. A family of 6 is going to the fair. They have a coupon for $1.50 off each ticket. If they pay

$46.50 for all their tickets, how much does a ticket cost without the coupon? Explain or show your reasoning. If you get stuck, consider drawing a diagram or writing an equation.

Possible responses1. 122 total invitations. 66 have been made already. 7 days to finish because there are

7 boxes with an equal, unknown amount x.2. 8. Sample reasoning: subtract 66 from 122 and divide the result, 56, by 7.3. 7 x+66=122. He makes the same amount x each of 7 days, so 7 x represents the

number made in 7 days. He already made 66, so add those on, for a total of 122.4. 7 x+66=122, 7 x+66+(−66)=122+(−66), 7 x=56, 7 x÷7=56÷7, x=8.1. Answers vary. Sample response: There are 3 groups of x stickers and 3 groups of 7

pencils. All together there are 57 items.2. Answers vary. Sample response: Yes, Noah’s equation says that 3 groups of x+7

gives a total of 57 items.3. 12

1. $9.25.

11.3: Running Around (15 minutes)

Setup:Keep students in the same groups and, optionally, provide tools for creating a visual display. 5–6 minutes quiet work time and a partner discussion followed by a whole-class discussion or gallery walk.


Priya, Han, and Elena, are members of the running club at school.

1. Priya was busy studying this week and ran 7 fewer miles than last week. She ran 9 times as far as Elena ran this week. Elena only had time to run 4 miles this week.

1. How many miles did Priya run last week?

2. Elena wrote the equation 19(x−7)=4 to describe the situation. She solved the

equation by multiplying each side by 9 and then adding 7 to each side. How does her solution compare to the way you found Priya's miles?

2. One day last week, 6 teachers joined 57 of the members of the running club in an after-school run.

Priya counted a total of 31 people running that day. How many members does the running club have?

3. Priya and Han plan a fundraiser for the running club. They begin with a balance of −80 because of expenses. In the first hour of the fundraiser they collect equal donations from 9 parents, which brings their balance to −44. How much did each parent give?

4. The running club uses the money they raised to pay for a trip to a canyon. At one point during a run in the canyon, the students are at an elevation of 128 feet. After descending at a rate of 50 feet per minute, they reach an elevation of −472 feet. How long did the descent take?

Possible responses1. 43 miles2. Answers vary. Sample response: Since Priya ran 9 times as far as Elena, I multiplied

9 by 4 to get 36 miles. This is the number of miles Priya ran this week, so 36 is 7 less than what she ran last week. I added 36 and 7 to get 43 miles. Elena solved an equation and I worked backwards with the information in the problem (or used a diagram). But we both took the same steps, multiply by 9 and add 7.

1. 35 members (one way is to solve the equation 57x+6=31

2. $4 (one way is to solve −80+9x=−44)3. 12 minutes (one way is to solve 128−50x=−472)


The phrases “9 times as far” and “9 times as many” may lead students to think about

multiplying by 9 instead of dividing (or multiplying by 19 ). Encourage students to act out the

situations or draw diagrams to help reason about the relationship between the quantities. Remind them to pay careful attention to what or who a comparison refers to.

Are you ready for more?

A musician performed at three local fairs. At the first he doubled his money and spent $30. At the second he tripled his money and spent $54. At the third, he quadrupled his money and spent $72. In the end he had $48 left. How much did he have before performing at the fairs?

Possible Responses

$29.

Arithmetic solution: Work backwards: 48+72=120, 120÷4=30, 30+54=84, 84÷3=28, 28+30=58, 58÷2=29.

Algebraic solution: Let m represent original amount of money. After first fair: 2 x−30. After second fair: 3(2 x−30)−54=6 x−144 . After third fair, 4 (6 x−144 )−72 = 24 x−648. Write equation 24 x−648=48 to show how much money is left. Solution is 696÷24=29.

Lesson Synthesis (5 minutes)Ask students to reflect on the work done in this unit so far. What strategies have they learned? What kinds of problems can they solve that they weren't able to, previously? Ask them to write down or share with a partner one new thing they have learned and one thing they still have questions or confusion about.

11.4: The Basketball Game (Cool-down, 5 minutes)

Setup:None.


Diego scored 9 points less than Andre in the basketball game. Noah scored twice as many points as Diego. If Noah scored 10 points, how many points did Andre score?

Possible responses

14 points.

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