Lesson 8

In this warm-up, students look for a relationship between two quantities by interpreting a verbal description and analyzing pairs of values in a table. They then use the observed relationship to find unknown values of one quantity given the other, and to think about possible equations that could represent the relationship more generally (MP8).

The work here reinforces the idea that the relationship between two quantities can be expressed in more than one way, and that some forms might be more helpful than others, depending on what we want to know. In this context, for instance, if we know the area of the parallelogram and want to know its base length, the equation \(b = \frac{A}{3}\) is more helpful than \(A=3b\).

Give students access to four-function calculators, if requested.

Some students may think that the height must be known before they could find the missing area or base. Encourage them to look for a pattern in the table and to reason from there.

Invite students to share their responses and explanations. Then, focus the whole-class discussion on the second question. Discuss with students:

• "Are the two equations we chose equivalent? How do you know?" (Yes. There is an acceptable move that takes one to the other. If we divide each side of \(A=3b\) by 3, we have \(\frac{A}{3}=b\), which can also be written as \(b=\frac{A}{3}\). If we multiply each side of \(b=\frac{A}{3}\) by 3, we have \(3b=A\) or \(A = 3b\).)
• "If we know the base, which equation would make it easier to find the area? Why?" (\(A=3b\). The variable for area is already isolated. All we have to do is multiply the base by 3 to find the area.)
• "If we know the area, which equation would make it easier to find the base? Why?" (\(b=\frac{A}{3}\). The variable for the base is already isolated. We can just divide the area by 3 to find the base.)

In an earlier lesson, students encountered relationships where two quantities that could vary multiplied to a constant. The focus there was on studying the relationships and describing them using equations—in any form. If \(x\) and \(y\) are quantities that vary and \(C\) a constant, they might write \(xy=C\), \(y = \frac{C}{x}\), or \(x = \frac {C}{y}\). 

In this activity, students encounter a similar relationship. The goal here, however, is for students to recognize—by repeatedly calculating the value of one quantity and then the value of the other quantity—that a particular form of equation might be handy for finding one quantity but not so handy for finding the other. 

When answering the first couple of parts of the first question (finding the length of a section given the number of volunteers), students can use various strategies to efficiently reason about the answers. They might draw diagrams, use proportional reasoning, or think in terms of multiplication (asking, for example, "8 times what number equals 2?"). They might also write an equation such as \(n \boldcdot \ell = 2\), substitute the number of volunteers for \(n\), and then solve the equation.

As students progress through the parts, they will likely notice that some strategies become less practical for finding the value of interest. One strategy, however, will remain efficient: dividing 2 by the number of volunteers, or evaluating \(2 \div n\) or \(\frac2n\) at the given values of \(n\).

Likewise, when answering the third question (finding the number of volunteers given the length of a section), students could start out with a variety of strategies and fairly easily find the number of volunteers. Later, however, the number of volunteers becomes a bit more cumbersome to find except when using division (that is, dividing 2 by the given length, or evaluating \(2 \div \ell\) or \(\frac {2}{\ell}\) at the given values of \(\ell\)).

Identify students who use these or other approaches and select them to share their strategies during discussion. 

Arrange students in groups of 2 and provide access to calculators. Give students a few minutes of quiet work time and then time to share their responses with their partner. Follow with a whole-class discussion.

Select students to present their strategies for solving either set (or both sets) of questions. Start with students using the least straightforward approach and end with those who wrote \(\ell = \frac{2}{n}\) for the first set of questions (or \(n = \frac{2}{\ell}\) for the second set of questions).

Emphasize that isolating the variable that we're interested in—before we substitute any known values—can be an efficient way for solving problems. Once we pin down the variable of interest first and see what expression is equal to it, we can simply evaluate that expression and bypass some tedious steps.

Highlight that isolating a variable is called “solving for a variable.” In road clean-up context, if we want to know the length of a road section each volunteer would be responsible for, we can solve for \(\ell\). If we want to know how many volunteers would be needed, we can solve for \(n\).

In this activity, students solve problems involving two quantities in non-proportional linear relationships. As before, they are prompted to reason repeatedly about the value of one quantity given the other, and to generalize the process by writing an expression (MP8). They then connect the work here to the idea of writing an equation and isolating a variable of interest.

As students work, monitor for the different ways students reason about the unknown number of minutes in the second question (for Tank A) and the last question (for Tank B). Some students may find the missing value by reasoning about the quantities informally. Others may write an equation and reason about the equations, or solve the equation for a variable.

For example, to find the number of minutes that have passed when Tank B has 18 liters left, students may:

• Reason that the tank has lost 62 liters because \(80-62=18\) or \(80-18=62\). If each minute it loses 2.5 liters, then it would take \(\frac{62}{2.5}\) or 24.8 minutes for it to lose 62 liters.
• Write \(80 - 2.5t = 18\) and think: "I'm looking for a number that, when multiplied by 2.5 and subtracted from 80, gives 18," or "I'm looking for a number that, when multiplied by 2.5, is equal to \(80-18\)."

• Write \(80 - 2.5t = 75\) and solve for \(t\) this way:

  \(\begin {align} 80-2.5t &=18\\ \text-2.5t &= 18 - 80\\t &= \frac{18-80}{\text-2.5} \\t &= \frac{\text-62}{\text-2.5} \\ t&=\frac{62}{2.5}\\ t&=24.8 \end{align}\)

  Or this way:

  \(\begin {align} 80-2.5t &=18\\80-18-2.5t &= 0\\80-18 &= 2.5t \\\frac{80-18}{2.5}&= t \\ \frac{62}{2.5} &=t\\24.8 &=t \end{align}\)

Regardless of the approach students take, the important idea to spotlight (for Tank B) is that finding the time at which the water in the tank reaches a certain volume can be done by subtracting that volume (\(v\)) from the original volume (80 liters) and then dividing it by 2.5. That process can be summarized by the expression: \(\dfrac{80-v}{2.5}\).

If we start out with the equation \(80-2.5t = v\) and perform allowable moves to isolate \(t\), we will end up with \(t=\dfrac{80-v}{2.5}\).

Keep students in groups of 2 and provide continued access to calculators. 

If time is limited, consider asking one half of the class to answer the first two questions about Tank A and the other half to answer the last two questions about Tank B.

For students who struggle to write expressions for \(p\) liters and \(v\) liters, encourage them to revisit their previous three calculations. Some students may need to write out their work more carefully before noticing that they could perform the same steps using \(p\) or \(w\) in place of a number.

Invite students to share their expressions for the last part of each question. Then, focus the discussion on how the different strategies students used to answer the other parts of each question could be summed up by the expression, as shown in the example in the Activity Narrative.  

Discuss questions such as: 

• "How do you know that the expression gives us the liters of water in Tank A after \(m\) minutes?" (When finding the liters of water in the tank after 4 minutes and 80 seconds, we multiplied the minutes by 9 and then added it to 124. So for \(m\) minutes, we'd multiply \(m\) by 9 and add 124.)
• "How do you know that the expression gives us the minutes at which the tank reaches \(p\) liters?" (When finding the minutes at which the tank reaches 151 liters, 191.5 liters, and 270.25 liters, we subtracted 124 from each number then divided the difference by 9. So we can do the same for \(p\) liters.)

Representing, Conversing: MLR7 Compare and Connect. Prior to the whole-class discussion, invite students to create a visual display of their work about either Tank A or Tank B. Students should consider how to display their work so that other students will be able to interpret the connections between their calculations and the expressions they created. Students may wish to add notes or details to their displays to help communicate their thinking. Begin the whole-class discussion by selecting and arranging 2–4 displays for all to see. Give students 1–2 minutes of quiet think time to interpret the displays before inviting the authors to share their expressions as described in the lesson synthesis.
Design Principle(s): Cultivate conversation; Maximize meta-awareness