Lesson 6

The purpose of this warm-up is to help students recall what it means for two expressions to be equivalent. The given expressions are in forms that are unfamiliar to students but are not difficult to evaluate for integer values of the variable. This is by design—to pique students' curiosity while keeping the mathematics accessible.

Assign the first expression to one half of the class and the second expression to the other half. Give students a couple of minutes to evaluate their assigned expression at as many values of \(n\) as listed in the first question.

When evaluating their expression, some students may perform the operations in an incorrect order. For example, when finding the value of \(8-3\boldcdot 2\) in the second expression, they may find \(8-3\) and then multiply by 2. Ask them whether the subtraction or multiplication should be performed first. Remind them about the order of operations as needed.

Ask a few students from each group for their results. Then, ask students what they wonder about the results. Students are likely curious if the values of the two expressions will be the same for other values of \(n\). If they noticed that all the given values of \(n\) are odd numbers, they might wonder if even values of \(n\) would give the same result. If time permits, consider allowing students to try evaluating the expressions using a value of their choice.

Discuss questions such as:

• "Were you surprised that these expressions have the same result for different values of \(n\)?"
• "If (or when) you tried using other values of \(n\), what did you find?"
• "Do you think that the two expressions will have the same value no matter what value of \(n\) is used? How do you know?"

Tell students that it would be impossible to check every value of \(n\) to see if the expressions would give the same value. There are, however, ways to show that these expressions must have the same value for any value of \(n\). We call expressions that are equal no matter what value we use for the variable equivalent expressions.

Remind students that in middle school they had seen simpler equivalent expressions. For example, they know that \(3(x + 5)\) is equivalent to \(3x + 15\) by the distributive property (without trying different values of \(x\)).

Explain to students that they'll learn more about how to identify or write equivalent expressions and about equivalent equations in this unit.

In this activity, students examine simple relationships that can each be expressed with many equations. Writing multiple equations for the same relationship and doing so in context prepares students to later consider more rigorously what makes equations equivalent.

As students write equations for the first relationship, identify students who use all numbers and those who use both numbers and variables.

For the second question, students may explain Tyler's claim concretely (using values of \(m\) and \(y\)) or more abstractly (using the structure of the equations). For instance, they may reason that:

• When the middle child is 12, the youngest child is 7, and that substituting \(m=12\) and \(y=7\) to \(2m-2y\) gives \(2(12) - 2(7)\) or \(24 - 14\), which is 10. At all other ages of the two children, the expression \(2m-2y\) always has a value of 10.
• \(2m-2y=10\) is twice of \(m-y=5\). If the difference between \(m\) and \(y\) is 5, then twice the difference between \(m\) and \(y\) must be twice of 5, which is 10, so the two equations are still describing the same relationship.

Identify students who reason in these ways so they can share later.

Give students a couple of minutes to answer the first question and ask them to pause for a discussion before moving on.

Invite students to share their equations. Record and display the equations for all to see. Along the way, consider organizing them into categories (numerical equations, equations in one variable, and equations in two variables).

Solicit from students some thoughts on the following questions. (It is not necessary to resolve the questions at the moment.)

• "Consider the equations you wrote for each situation. Are they all equivalent? Why or why not?"
• "What do you think it means for two equations to be equivalent?"

Some students might say that each equation accurately represents the same relationship between the ages, so the equations must be equivalent. Those who wrote variable equations might say that the equations are equivalent because the same value for the variable makes each equation true. 

After students have had a chance to consider these questions, prompt them to complete the remaining questions.

Some students may say that \(2m-2y=10\) doesn’t describe the relationship between the ages because it doesn’t involve the number 5. Encourage students to think of some values for \(m\) and \(y\) that do make the equation true. After they find some pairs of \(m\) and \(y\), ask them what they notice about the pairs.

Select previously identified students to share their explanation of why Tyler's claim is true, in the order shown in the Activity Narrative. If no students mention the last approach, bring it up.

Next, emphasize two main points:

• Equivalent equations have exactly the same solutions, or exactly the same values that make each of the equations true. All of the equations in the last question are equivalent because they have 3 as the solution and they all have no other solutions.
• Suppose we start with a true equation, where the two sides are equal. If we perform the same operation to both sides of the equation and get a new equation where the two sides are also equal, we can say that the two equations are equivalent. For instance:
  • Subtracting both sides of \(3a + 6 = 15\) by 6 gives \(3a=9\). If \(3a+6\) and 15 are equal, then the expressions or numbers we get by subtracting 6 from each one are also equal. We can conclude that \(3a+6=15\) and \(3a=9\) are equivalent.
  • Dividing both sides of \(3a + 6 = 15\) by 3 gives \(a + 2 = 5\). If \(3a+6\) is equal to 15, the result of dividing \(3a+6\) by 3 is equal to dividing 15 by 3. We can conclude that \(3a+6=15\) and \(a + 2 = 5\) are equivalent.

Ask students:

• "How can we show that \(\frac13a = 1\) is equivalent to \(3a = 9\)?" (Multiplying both sides of \(\frac13a = 1\) by 9 gives \(3a=9\).)
• "How can we show that \(a + 2 = 5\) is equivalent to \(\frac13a = 1\)?" (Subtracting 2 from both sides of \(a + 2 = 5\) and then multiplying both sides of the resulting equation, \(a=3\), by \(\frac13\) gives \(\frac13a = 1\).)

If no students notice that we have made these moves when solving equations, bring it to their attention. Highlight that solving an equation essentially involves writing a series of equivalent equations that eventually isolates the variable on one side.

This activity further develops the idea of equivalent equations and does so in the context of a situation. Students pay attention to the moves that create equations with the same solution and those that don't. They also make sense of both the moves and the solutions in terms of the given situation. The work here gives students opportunities to reason concretely and abtractly (MP2). In an upcoming lesson, they will think about why these moves lead to equations with the same solution.

Arrange students in groups of 2 and provide access to calculators. Ask students to read the opening paragraph and display the equation \(x - 0.1x + 2.70 = 56.70\) for all to see. Ask students to explain how the equation represents Noah's purchase.

Then, give students a couple of minutes to discuss the first question with their partner and ask them to pause for a whole-class discussion. Once students understand that the solution to the equation is the original price of a pair of jeans and that substituting 60 for \(x\) makes the equation true, move on to the rest of the activity.

Tell students that they will see a series of equations that are related to the original equation and that can be interpreted in terms of the situation. For each equation, their job is to determine either what move was made to the original equation, or what the equation means in context. Next, they should check if the equation has the same solution as the original. (To help students understand the task and if needed, consider working on Equation A as a class.)

If time is limited, ask one partner in each group to take Equations A, C, E and the other partner to take B, D, F.

Display the original equation and Equations A–F for all to see. Invite students to share what was done to the original equation to get each of those equations and whether they have the same solution. Along the way, compile a list of moves that lead to equations with the same solution and those that lead to different solutions. 

Ask students to observe the list and see what kinds of moves produce the Equations A–C and D–F. Record the moves that create equations with the same solutions, such as:

• adding or multiplying both sides of the equal sign by the same number
• applying properties of operations (commutative, associative, or distributive)
• combining like terms

Also record the moves that create equations with different solutions, such as:

• adding or multiplying a different number to the two sides, or performing an operation to only one side
• adding different expressions to the two sides
• performing different operations on each side

(The lists don't need to be comprehensive, as students will examine these moves more closely later.)

Next, give students a couple of examples of how the equations and their solutions could be interpreted in context. For example:

• Equation A shows the cost calculation in cents, so the original price of a pair of jeans is unchanged.
• Equation D shows the discounted price, excluding tax, to be $56.70, but in the original equation, the price $56.70 included tax. This means that, in Equation D, the original price of one pair of jeans is different than in the first equation. 

Prompt students to interpret 1–2 other equations and to explain why the solution in each equation (or the price of one pair of jeans) is equal or unequal to that in the initial equation. Here are possible interpretations:

• Equation B shows the discounted price before sales tax to be $54, which is $2.70 less than $56.70. This is also the case in the initial equation, so the original price of the jeans is still the same. 
• Equation C shows the discounted price as 90% of the original price, which is the same as 10% less than the original price (100%). The original price is unchanged.
• Equation E shows the discounted price, before tax, to be $59.40. This is more than the pre-tax price in the initial equation, so the original price of the jeans here must be higher.
• Equation F shows the price of 2 pairs of jeans, including the discount and tax, to be $56.70. This means the price of a pair of jeans must be much less than in the initial equation.