Lesson 8

Which Variable to Solve for? (Part 1)

8.1: Which Equations? (5 minutes)


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.

Student Facing

  1. The table shows the relationship between the base length, \(b\), and the area, \(A\), of some parallelograms. All the parallelograms have the same height. Base length is measured in inches, and area is measured in square inches. Complete the table.
    \(b\) (inches) \(A\) (square inches)
    1 3
    2 6
    3 9
  2. Decide whether each equation could represent the relationship between \(b\) and \(A\). Be prepared to explain your reasoning.

    1. \(b = 3A\)
    2. \(b = \frac {A}{3}\)
    3. \(A = \frac {b}{3}\)
    4. \(A = 3b\)

Student Response

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Anticipated Misconceptions

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.

Activity Synthesis

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.)

8.2: Post-Parade Clean-up (15 minutes)


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.

Student Facing

After a parade, a group of volunteers is helping to pick up the trash along a 2-mile stretch of a road.

The group decides to divide the length of the road so that each volunteer is responsible for cleaning up equal-length sections. 

Mummers' Parade in Philly.
  1. Find the length of a road section for each volunteer if there are the following numbers of volunteers. Be prepared to explain or show your reasoning.

    1. 8 volunteers
    2. 10 volunteers
    3. 25 volunteers
    4. 36 volunteers
  2. Write an equation that would make it easy to find \(\ell\), the length of a road section in miles for each volunteer, if there are \(n\) volunteers.
  3. Find the number of volunteers in the group if each volunteer cleans up a section of the following lengths. Be prepared to explain or show your reasoning.

    1. 0.4 mile
    2. \(\frac {2}{7}\) mile
    3. 0.125 mile
    4. \(\frac {6}{45}\) mile
  4. Write an equation that would make it easy to find the number of volunteers, \(n\), if each volunteer cleans up a section that is \(\ell\) miles. 

Student Response

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Student Facing

Are you ready for more?

Let's think about the graph of the equation \(y = \frac{2}{x}\).

  1. Make a table of \((x,y)\) pairs that will help you graph the equation. Make sure to include some negative numbers for \(x\) and some numbers that are not integers.
  2. Plot the graph on the coordinate axes. You may need to find a few more points to plot to make the graph look smooth.

    Blank coordinate plane, x, negative 3 to 3 by 1, y, negative 3 to 3 by 1.


  3. The coordinate plane provided is too small to show the whole graph. What do you think the graph looks like when \(x\) is between 0 and \(\frac12\)? Try some values of \(x\) to test your idea.
  4. What is the largest value that \(y\) can ever be?

Student Response

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Activity Synthesis

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 \(n\).

Representation: Internalize Comprehension. Use color coding and annotations to highlight connections between representations. As students share their reasoning, scribe their thinking on a visible display. Invite students to describe connections they notice between approaches that use diagrams, proportional reasoning, or multiplication, with those who used an equation. For example, ask “Where does \(l\) appear in this (diagram)?”
Supports accessibility for: Visual-spatial processing

8.3: Filling and Emptying Tanks (15 minutes)


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.

Engagement: Provide Access by Recruiting Interest. Leverage choice around perceived challenge. Invite students to choose between working on questions about Tank A or the questions about Tank B. Chunking this task into more manageable parts may also support students who benefit from additional processing time.
Supports accessibility for: Organization; Attention; Social-emotional skills

Student Facing

  1. Tank A initially contained 124 liters of water. It is then filled with more water, at a constant rate of 9 liters per minute. How many liters of water are in Tank A after the following amounts of time have passed?

    1. 4 minutes
    2. 80 seconds
    3. \(m\) minutes
  2. How many minutes have passed, \(m\), when Tank A contains the following amounts of water?

    1. 151 liters
    2. 191.5 liters
    3. 270.25 liters
    4. \(p\) liters
  3. Tank B, which initially contained 80 liters of water, is being drained at a rate of 2.5 liters per minute. How many liters of water remain in the tank after the following amounts of time?

    1. 30 seconds
    2. 7 minutes
    3. \(t\) minutes
  4. For how many minutes, \(t\), has the water been draining when Tank B contains the following amounts of water?

    1. 75 liters
    2. 32.5 liters
    3. 18 liters
    4. \(v\) liters

Student Response

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Anticipated Misconceptions

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.

Activity Synthesis

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.)

If no students mentioned that they found the expression for the last part of the second and last questions by writing an equation and solving it for the variable of interest, demonstrate it for all to see.

For example, in Tank A, we know the relationship between the liters of water in the tank, \(p\), after \(m\) minutes is: \(124+9m=p\). To find the minutes at which the tank reaches \(p\) liters, we can isolate \(m\):

\(\begin{align}124+9m&=p\\9m &=p-124\\ m&=\dfrac{p-124}{9}\end{align}\)

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

Lesson Synthesis

Lesson Synthesis

To help students consolidate and reflect on their work in the lesson, present the following scenario.

Suppose a classmate who is absent today wanted to know:

  • What does it mean to "solve for a variable"?
  • Why should we solve for a variable?
  • How do we solve for a variable?

Ask students: "How would you respond to these questions and help your classmate catch up with what was missed?"

Consider arranging students in groups of 2 and asking partners to do a role play, taking turns being the absent classmate and the explainer. Display the equations from the lesson for all to see. Encourage students to use 1–2 of them to demonstrate how to solve for a variable.


\(n \boldcdot \ell = 2\)

\(124 + 9m = p\)


8.4: Cool-down - A Rectangular Relationship (5 minutes)


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Student Lesson Summary

Student Facing

A relationship between quantities can be described in more than one way. Some ways are more helpful than others, depending on what we want to find out. Let’s look at the angles of an isosceles triangle, for example.

Triangle with angles labeled a, b, and a.

The two angles near the horizontal side have equal measurement in degrees, \(a\)

The sum of angles in a triangle is \(180^{\circ}\), so the relationship between the angles can be expressed as:

 \(a + a+ b=180\)

Suppose we want to find \(a\) when \(b\) is \(20^{\circ}\).  

Let's substitute 20 for \(b\) and solve the equation.

\(\begin {align}a + a + b &=180\\ 2a + 20 &=180\\ 2a &=180 - 20\\ 2a &=160\\ a&=80 \end{align}\)

What is the value of \(a\) if \(b\) is \(45^{\circ}\)

\(\begin {align}a + a + b &=180\\ 2a + 45 &=180\\ 2a &=180 - 45\\ 2a &=135\\ a&=67.5 \end{align}\)

Now suppose the bottom two angles are \(34^\circ\) each. How many degrees is the top angle?

Let's substitute 34 for \(a\) and solve the equation.

 \(\begin {align}a + a + b &=180\\ 34 + 34 + b &=180\\ 68 + b &=180 \\ b &=112 \end{align}\)

What is the value of \(b\) if \(a\) is \(72.5^{\circ}\)?

 \(\begin {align}a + a + b &=180\\ 72.5 + 72.5 + b &=180\\ 145 + b &=180 \\ b &=35 \end{align}\)

Notice that when \(b\) is given, we did the same calculation repeatedly to find \(a\): we substituted \(b\) into the first equation, subtracted \(b\) from 180, and then divided the result by 2. 

Instead of taking these steps over and over whenever we know \(b\) and want to find \(a\), we can rearrange the equation to isolate \(a\):

\(\begin{align} a + a + b &= 180\\ 2a + b&=180\\2a &=180-b\\ a &=\dfrac{180-b}{2} \end{align}\)

This equation is equivalent to the first one. To find \(a\), we can now simply substitute any value of \(b\) into this equation and evaluate the expression on right side.

Likewise, we can write an equivalent equation to make it easier to find \(b\) when we know \(a\):

\(\begin{align} a + a + b &= 180\\ 2a + b&=180\\b &=180-2a \end{align}\)

Rearranging an equation to isolate one variable is called solving for a variable. In this example, we have solved for \(a\) and for \(b\). All three equations are equivalent. Depending on what information we have and what we are interested in, we can choose a particular equation to use.