Lesson 2

The purpose of this warm-up is to give students an opportunity to ground their understanding of equality in the context of weight, which is a context that will be used throughout the lesson.

Tell students they will look at a picture, and their job is to think of at least one thing they notice and at least one thing they wonder about the picture. Display the problem for all to see and give 1 minute of quiet think time. Ask students to give a signal when they have noticed or wondered about something.

Ask students to share their ideas. Record and display the responses for all to see. In the interest of time, you can ask if anything students wondered was a “why” question, meaning the question begins with the word why. Refer to MLR 2 (Collect and Display).

If not brought up during the first part of the discussion, ask students why they think the left hanger is balanced while the right hanger is unbalanced. Students should understand that a hanger will only balance if the weight of the unknown objects in both socks is the same. If they are not the same, then the heavier side is lower than the lighter side. 

The purpose of this task is for students to understand and explain why they can add or subtract expressions from each side of an equation and still maintain the equality, even if the value of those expressions are not known. Both problems have shapes with unknown weight on each side to promote students thinking about unknown values in this way before the transition to equations.

While the focus of this activity is on the relationship between both sides of the hanger and not equations, some students may start the second problem by writing and solving an equation to find the weight of a square. While students are working, identify those using equations and those not using equations to answer the second problem during the whole-class discussion.

In the first question, students may think the hanger will stay in balance since removing the 5 triangles results in three shapes on each side.

Begin the discussion by asking if students think the hanger will stay in balance, tip to the left, or tip to the right. Select 2–3 students to explain their vote. Make sure the class understands that removing unequal amounts of weight from the two sides results in the hanger tipping before moving on. Use MLR 2 (Collect and Display) to capture student reasoning about it being okay to add or remove terms of the same “size” from both sides of an equation. 

For the second question, select previously identified students to explain their answers, with the students who used equations going last. Record and display the specific equations the selected students wrote for all to see, such as \(x+x+x+1+1=x+1+1+1+1+1\) or \(3x+2=x+5\), and use it to help the class visualize how that student solved for the weight of a square.

The outcome of this discussion should be that it is okay to add or remove terms of the same “size” from both sides of an equation, and the sides will still be equal. This can be thought of in terms of shapes hanging on hangers, where you can remove one square from both sides or add two triangles to both sides, and the hanger will stay in balance. Equations are a more abstract representation of this, but the same concept holds: you can remove one \(x\) from both sides or add two 3s to both sides and the equation is still true with the left side equal to the right side. Removing equal weights from both sides can leave the hanger with 2 squares on the left and 3 triangles (or just 3) on the right. In equation form, this is the same as \(2x = 3\). Finally, you can halve the amount of weight on both sides of the hanger and keep it in balance, which is the same as multiplying \(2x = 3\) by \(\frac{1}{2}\) (or dividing both sides by 2).

Building on the previous activity, students now solve two more hanger problems and write equations to represent each hanger. In the first problem, the solution is not an integer, which will challenge any student who has been using guess-and-check in the previous activities to look for a more efficient method. In the second problem, the solution is any weight, which is a preview of future lessons when students purposefully study equations with one solution, no solution, and infinite solutions. The goal of this activity is for students to transition their reasoning about solving hangers by maintaining the equality of each side to solving equations using the same logic. In future lessons, students will continue to develop this skill as equations grow more complex culminating in solving systems of equations at the end of this unit.

As students work, identify those using strategies to find the weight of one square/pentagon that do not involve an equation. For example, some students may cross out pairs of shapes that are on each side (such as one circle and one square from each side of hanger A) to reason about a simpler problem while others may replace triangles with 3s and circles with 6s first before focusing on the value of 1 square. This type of reasoning should be encouraged and built upon using the language of equations.

Triangles weigh 3 grams in this activity instead of 1 gram as in the previous activity.

Select previously identified students to share their strategies for finding the unknown weight without using an equation. Ask students to be clear how they are changing each side of the hanger equally as they share their solutions.

Next, record the equations written by students for each hanger and display for all to see in two lists. Assign half the class to the list for Hanger A and the other half to the list for Hangar B. Give students 1–2 minutes to examine the equations for their assigned hanger and be prepared to explain how different pairs of equations are related. The goal here is for student to use the language they developed with the hangers (e.g., “remove 6 from each side”) on equations.

For example, for Hanger A, you might contrast \(3 + 6 + 6 +6 = 4x+ 6\) with \(21 + x = 6 + 5x\). Possible student responses:

• Removing an \(x\) from each side of the second equation would result in the first equation.
• \(x = 3.75\) grams makes both equations true.
• You can subtract 6s from the sides of each equation and they are still both true.

For Hanger B, examining equations should illuminate why it is impossible to know the weight of the unknown shape. If we start with \(6+6+x+x = x+x+3+3+6\) and keep removing things of equal weight from each side, we might end up with an equation like \(2x=2x\). Any value of \(x\) will work to make this equation true. For example if \(x\) is 10, then the equation is \(20=20\). It is also possible to keep removing things of equal weight from each side and end up with an equation like \(6 = 6\), which is always true.