Lesson 8

Reasoning about Solving Equations (Part 2)

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

Warm-up

Students worked with the distributive property with variables in grade 6 and with numbers in earlier grades. In order to understand the two ways of solving an equation of the form \(p(x+q)=r\) in the upcoming lessons, it is helpful to have some fluency with the distributive property. 

Launch

Arrange students in groups of 2. Give 3 minutes of quiet work time and then invite students to share their responses with their partner, followed by a whole-class discussion.

Student Facing

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

  1. $2 \boldcdot (x+3) $
  2. $(x + 3)2 $
  3. $2 \boldcdot x + 2 \boldcdot 3$ 
  4. $2 \boldcdot x + 3 $
  5. $(2 \boldcdot x) + 3$ 
  6. $(2 + x)3$

Student Response

Student responses to this activity are available at one of our IM Certified Partners

Activity Synthesis

Focus specifically on why 1 and 3 are equivalent to lead into the next activity. You may also recall the warm-ups from prior lessons and ask if \(2(x+3)\) is equal to 10, or other numbers, how much is \(x+3\)?

Ask students to write another expression that is equivalent to \(2(x+3)\) (look for \(2x+6\)).

8.2: Either Or (15 minutes)

Activity

This activity continues the work of using a balanced hanger to develop strategies for solving equations. Students are presented with a balanced hanger and are asked to explain why each of two different equations could represent it. They are then asked to find the unknown weight. Note that no particular solution method is prescribed. Give students a chance to come up with a reasonable approach, and then use the synthesis to draw connections between the diagram and each of the two equations. Students notice the structure of equations and diagrams and find correspondences between them and between solution strategies.

Launch

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.

Representation: Develop Language and Symbols. Use virtual or concrete manipulatives to connect symbols to concrete objects or values. For example, create a balanced hanger using concrete objects. Be sure to use individual pieces for each part of the diagram. Demonstrate moving pieces off of the hanger to create an equation. Invite students to show different ways to create the same equation.
Supports accessibility for: Visual-spatial processing; Conceptual processing
Conversing, Representing: MLR1 Stronger and Clearer Each Time. Use this routine to help students improve their written response to the first question, by providing them with multiple opportunities to clarify their explanations through conversation. Give students time to meet with 2–3 partners, to share and get feedback on their responses. Provide listeners with prompts for feedback that will help their partners strengthen their ideas and clarify their language. For example, students can ask their partner, "How is \(2(x+3)\) represented in the hanger?" or "Can you say more about..." After both students have shared and received feedback, provide students with 3-4 minutes to revise their initial draft, including ideas and language from their partner. This will help students communicate why the same hanger can be represented with equations in either form.
Design Principle(s): Optimize output (for explanation); Cultivate conversation

Student Facing

  1. Explain why either of these equations could represent this hanger:

    Balanced hanger diagram, left side, rectangle 14, right side, circle x, square 3, circle x, square 3.

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

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

Student Response

Student responses to this activity are available at one of our IM Certified Partners

Activity Synthesis

Have one student present who did \(7=x+3\) first, and another student present who subtracted 6 first. If no one mentions one of these approaches, demonstrate it. Show how the hanger supports either approach. The finished work might look like this for the first equation:

Balanced hanger diagram.

\(\begin {align} 14&=2(x+3) \\ 7&=x+3 \\ \end{align}\)

Balaned hanger diagram.

\(\begin {align} 7&=x+3 \\ 4&=x \\ \end{align}\)

For the second equation, rearrange the right side of the hanger, first, so that 2 \(x\)’s are on the top and 6 units of weight are on the bottom. Then, cross off 6 from each side and divide each side by 2. Show this side by side with “doing the same thing to each side” of the equation.

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

Activity

The first question is straightforward since each diagram uses a different letter, but it’s there to make sure students start with \(p(x+q)=r\). If some want to rewrite as \(px+pq=r\) first, that’s great. We want some to do that but others to divide both sides by \(p\) first. Monitor for students who take each approach.

Launch

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.

Action and Expression: Internalize Executive Functions. To support development of organizational skills in problem-solving, chunk this task into more manageable parts. For example, show only 2 hangers and 2 equations. If students finish early, assign the remaining hangers and equations.
Supports accessibility for: Memory; Organization

Student Facing

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

Four balanced hanger diagrams.

For each diagram:

  1. Assign one of these equations to each hanger: 

    \(2(x+5)=16\)

    \(3(y+200)=3\!,000\)

    \(20.8=4(z+1.1)\)

    \(\frac{20}{3}=2\left(w+\frac23\right)\)

  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.

Student Response

Student responses to this activity are available at one of our IM Certified Partners

Activity Synthesis

Select one hanger for which one student divided by \(p\) first and another student distributed \(p\) first. Display the two solution methods side by side, along with the hanger.

Representing, Conversing: MLR7 Compare and Connect. Use this routine to prepare students for the whole-class discussion. Give students quiet think time to consider what is the same and what is different about the two solution methods. Next, ask students to discuss what they noticed with a partner. Listen for and amplify mathematical language students use to describe how each solution method can be represented by the hanger.
Design Principle(s): Cultivate conversation

Lesson Synthesis

Lesson Synthesis

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

Cool-Down

Cool-downs for this lesson are available at one of our IM Certified Partners

Student Lesson Summary

Student Facing

The balanced hanger shows 3 equal, unknown weights and 3 2-unit weights on the left and an 18-unit weight on the right.

There are 3 unknown weights plus 6 units of weight on the left. We could represent this balanced hanger with an equation and solve the equation the same way we did before.

\(\begin {align} 3x+6&=18 \\ 3x&=12 \\ x&=4 \\ \end{align}\)

Balanced hanger. Left side, circle labeled x, square labeled 2, circle labeled x, square labeled 2, circle labeled x, square labeled 2. Right side, rectangle labeled 18.

Since there are 3 groups of \(x+2\) on the left, we could represent this hanger with a different equation: \(3(x+2)=18\).

Balanced hanger, three groups are indicated, each group contains 1 circle labeled x and 1 square labeled 2. Right side, rectangle labeled 18.  To the side, an equation 3 ( x + 2 ) = 18.

The two sides of the hanger balance with these weights: 3 groups of \(x+2\) on one side, and 18, or 3 groups of 6, on the other side.

Balanced hanger. to the side, an equation.

The two sides of the hanger will balance with \(\frac13\) of the weight on each side: \(\frac13 \boldcdot 3(x+2) = \frac13 \boldcdot 18\).

Balanced hanger, left side, 1 circle labeled x and 1 square labeled 2, right side, rectangle labeled 6.  To the side, an equation says x + 2 = 6.

We can remove 2 units of weight from each side, and the hanger will stay balanced. This is the same as subtracting 2 from each side of the equation.

Balanced hanger.

An equation for the new balanced hanger is \(x=4\). This gives the solution to the original equation.

Balanced hanger, left side, circle labeled x, right side, rectangle labeled 4. To the side, an equation x = 4.

Here is a concise way to write the steps above:

\(\begin{align} 3(x+2) &= 18 \\ x + 2 &= 6 & \text{after multiplying each side by } \tfrac13 \\ x &= 4 & \text{after subtracting 2 from each side} \\ \end{align} \)