9.1: Which Would You Choose? (5 minutes)
The purpose of this warm-up is for students to reason about two situations that can be represented with linear equations. Since the number of babysitting hours determines which situation would be most profitable, there is no one correct answer to the question. Students are asked to explain their reasoning.
Give students 2 minutes of quiet work time followed by a whole-class discussion.
If you were babysitting, would you rather
- Charge \$5 for the first hour and \$8 for each additional hour?
- Charge \$15 for the first hour and \$6 for each additional hour?
Explain your reasoning.
Poll the class on which situation they would choose. Invite students from each side to explain their reasoning. Record and display these ideas for all to see. If no one reasoned about babysitting for less than 5 hours and therefore chose the second option, mention this idea to students.
Students may not use linear equations or graphs to decide which situation they would choose. If there is time, ask students for the equation and graph we could use to model each scenario.
9.2: Water Tanks (10 minutes)
The goal of this activity, and the two that follow, is for students to solve an equation in a real-world context while previewing some future work solving systems of equations. Here, students first make sense of the situation using a table of values describing the water heights of two tanks and then use the table to estimate when the water heights are equal. A key point in this activity is the next step: taking two expressions representing the water heights in two different tanks for a given time and recognizing that the equation created by setting the two expressions equal to one another has a solution that is the value for time, \(t\), when the water heights are equal.
Give students 2–3 minutes to read the context and answer the first problem. Select students to share their answer with the class, choosing students with different representations of the situation if possible. Give 3–4 minutes for the remaining problems followed by a whole-class discussion.
Supports accessibility for: Conceptual processing; Visual-spatial processing
Design Principle(s); Maximize meta-awareness
The amount of water in two tanks every 5 minutes is shown in the table.
|time (minutes)||tank 1 (liters)||tank 2 (liters)|
- Describe what is happening in each tank. Either draw a picture, say it verbally, or write a few sentences.
- Use the table to estimate when the tanks will have the same amount of water.
- The amount of water (in liters) in tank 1 after \(t\) minutes is \(30t + 25\). The amount of water (in liters) in tank 2 after \(t\) minutes is \( \text-20t + 1000\). Find the time when the amount of water will be equal.
The purpose of this discussion is to elicit student thinking about why setting the two expressions in the task statement equal to one another is both possible and a way to solve the final problem.
Consider asking the following questions:
- “What does \(t\) represent in the first expression? The second?” (In each expression, \(t\) is the time in minutes since the tank’s water level started being recorded.)
- “After we substitute a time in for \(t\) and simplify one of the expressions to be a single number, what does that number represent? What units does it have?” (The number represents the amount of liters in the water tank.)
- “How accurate was your estimate about the water heights using the table?” (My estimate was within a few minutes of the actual answer.)
- “If you didn’t know which expression in the last problem belonged to which tank, how could you figure it out?” (One of the expressions is increasing as \(t\) increased, which means it must be Tank 1. The other is decreasing as \(t\) increases, so it must be Tank 2.)
- “How did you find the time the two water heights were equal using the expressions?” (Since each expression gives the height for a specific time, \(t\), and we want to know when the heights are equal, I set the two expressions equal to each other and then solved for the \(t\)-value that made the new equation true.)
9.3: Elevators (15 minutes)
In this activity, students work with two expressions that represent the travel time of an elevator to a specific height. As with the previous activity, the goal is for students to work within a real-world context to understand taking two separate expressions and setting them equal to one another as a way to determine more information about the context.
Give students 1 minute to read the context and the problems. You may wish to share with the class that programming elevators in buildings to best meet the demands of the people in the building can be a complicated task depending on the number of floors in a building, the number of people, and the number of elevators. For example, many large buildings in cities have elevators programmed to stay near the ground floor in the morning when employees are arriving and then stay on higher floors in the afternoon when employees leave work.
Arrange students in groups of 2. Give 2–3 minutes of quiet work time for the first two question and then ask students to pause and discuss their solutions with their partner. Give 3–4 minutes for partners to work on the remaining questions followed by a whole-class discussion.
Design Principle(s): Support sense-making
A building has two elevators that both go above and below ground.
At a certain time of day, the travel time it takes elevator A to reach height \(h\) in meters is \(0.8 h + 16\) seconds.
The travel time it takes elevator B to reach height \(h\) in meters is \(\text-0.8 h + 12\) seconds.
- What is the height of each elevator at this time?
- How long would it take each elevator to reach ground level at this time?
- If the two elevators travel toward one another, at what height do they pass each other? How long would it take?
- If you are on an underground parking level 14 meters below ground, which elevator would reach you first?
Are you ready for more?
- In a two-digit number, the ones digit is twice the tens digit. If the digits are reversed, the new number is 36 more than the original number. Find the number.
- The sum of the digits of a two-digit number is 11. If the digits are reversed, the new number is 45 less than the original number. Find the number.
- The sum of the digits in a two-digit number is 8. The value of the number is 4 less than 5 times the ones digit. Find the number.
Students may mix up height and time while working with these expressions. For example, they may think that at \(h = 0\), the height of the elevators is 16 meters and 12 meters, respectively, instead of the correct interpretation that the elevators reach a height of 0 meters at 16 seconds and 12 seconds, respectively.
This discussion should focus on the act of setting the two expressions equal and what that means in the context of the situation.
Consider asking the following questions:
- “If someone thought that the height of Elevator A before we started timing was 16 meters because they substituted 0 for the variable of the expression \(0.8h+16\) and got 16, how would you help them correct their answer?” (I would remind them that \(h\) is height and \(0.8h+16\) is the time, so when we start timing at 0 that means \(0.8h+16=0\), not that \(h=0\).)
- “Which of the elevators in the image is A and which is B? How do you know?” (A is the elevator on the right since at time 0 the height is negative, while B is the elevator on the left since at time 0 the height is positive.)
- “How did you find the height when the travel times are equal?” (Since each expression gives the time to travel to a height \(h\), I solved the equation \(0.8h+16=\text- 0.8h+12\), which gives the value of \(h\) when the two expressions are equal.)
Supports accessibility for: Visual-spatial processing; Conceptual processing
The work in this lesson is a prelude to a simple form of a system of equations, where each equation can be written in the form \(y=\) some expression (though students do not need to know the term "system of equations" at this point).
Arrange students in groups of 2. Ask partners to think of another situations where two quantities are changing and they want to know when the quantities are equal. Give groups time to to discuss and write down a few sentences explaining their situation. Invite groups to share their situation with the class. (For example, in a race where participants walk at steady rates but the slower person has a head start, when will they meet?) Consider allowing groups to share their situation by making a picture, a graph, in words, or by acting it out.
9.4: Cool-down - Printers and Ink (5 minutes)
Cool-downs for this lesson are available at one of our IM Certified Partners
Student Lesson Summary
Imagine a full 1,500 liter water tank that springs a leak, losing 2 liters per minute. We could represent the number of liters left in the tank with the expression \(\text-2x+1,\!500\), where \(x\) represents the number of minutes the tank has been leaking.
Now imagine at the same time, a second tank has 300 liters and is being filled at a rate of 6 liters per minute. We could represent the amount of water in liters in this second tank with the expression \(6x+300\), where \(x\) represents the number of minutes that have passed.
Since one tank is losing water and the other is gaining water, at some point they will have the same amount of water—but when? Asking when the two tanks have the same number of liters is the same as asking when \(\text-2x+1,\!500\) (the number of liters in the first tank after \(x\) minutes) is equal to \(6x+300\) (the number of liters in the second tank after \(x\) minutes),
Solving for \(x\) gives us \(x=150\) minutes. So after 150 minutes, the number of liters of the first tank is equal to the number of liters of the second tank. But how much water is actually in each tank at that time? Since both tanks have the same number of liters after 150 minutes, we could substitute \(x=150\) minutes into either expression.
Using the expression for the first tank, we get \(\text-2(150)+1,\!500\) which is equal to \(\text-300+1,\!500\), or 1,200 liters.
If we use the expression for the second tank, we get \(6(150)+300\), or just \(900+300\), which is also 1,200 liters. That means that after 150 minutes, each tank has 1,200 liters.