Lesson 21

Rational Equations (Part 2)

21.1: Math Talk: Adding Rationals (5 minutes)

Warm-up

In this warm-up, students have an opportunity to notice and make use of structure (MP7), because the skills they use to solve equations involving fractions also work to solve equations with more complex rational expressions.

Launch

Display one problem at a time. Give students quiet think time for each problem and ask them to give a signal when they have an answer and a strategy. Keep all problems displayed throughout the talk. Follow with a whole-class discussion.

Representation: Internalize Comprehension. To support working memory, provide students with sticky notes or mini whiteboards.
Supports accessibility for: Memory; Organization

Student Facing

Solve each equation mentally:

\(\dfrac{x}{2} = \dfrac{3}{4}\)

\(\dfrac{3}{x} = \dfrac{1}{6}\)

\(\dfrac{1}{4} = \dfrac{1}{x^2}\)

\(\dfrac{2}{x} = \dfrac{x}{8}\)

Student Response

Student responses to this activity are available at one of our IM Certified Partners

Activity Synthesis

Ask students to share their strategies for each problem. Record and display their responses for all to see. To involve more students in the conversation, consider asking:

  • “Who can restate \(\underline{\hspace{.5in}}\)’s reasoning in a different way?”
  • “Did anyone have the same strategy but would explain it differently?”
  • “Did anyone solve the problem in a different way?”
  • “Does anyone want to add on to \(\underline{\hspace{.5in}}\)’s strategy?”
  • “Do you agree or disagree? Why?”
Speaking: MLR8 Discussion Supports. Display sentence frames to support students when they explain their strategy. For example, "First, I _____ because . . . ." or "I noticed _____ so I . . . ." Some students may benefit from the opportunity to rehearse what they will say with a partner before they share with the whole class.
Design Principle(s): Optimize output (for explanation)

21.2: A Rational River (15 minutes)

Activity

Continuing their work from the previous lesson, the purpose of this activity is for students to write a rational equation to model a situation and then use it to answer questions. The difference between this activity and previous ones in which students solved rational equations is that now students must solve a rational equation in which the denominators of the two expressions on each side are different expressions involving the variable \(r\).

Monitor for students who either graph each expression to find the point of intersection or set up an equation that they solve to share during the class discussion.

Making graphing technology available gives students opportunity to choose appropriate tools strategically (MP5).

Launch

Arrange students in groups of 2. Tell students to read the activity and answer the first problem. After quiet work time, ask students to compare their responses to their partner’s and reach agreement on the two expressions. Select 2–3 students to share their expressions with the class, recording them for all to see. Once students are in agreement on the expressions, allow them to continue to the next problem.

Conversing: MLR8 Discussion Supports. Use this routine to help students explain their expression to their partner. Students should take turns sharing their expressions and explaining their reasoning to their partner. Display the following sentence frames for all to see: “_____ represents _____”, “Where does _____ show . . . ?”, and “That could (or couldn’t) be true because . . . .” Encourage students to challenge each other when they disagree. This will help students clarify their reasoning about modeling situations with rational expressions.
Design Principle(s): Support sense-making; Maximize meta-awareness

Student Facing

Noah likes to go for boat rides along a river with his family. In still water, the boat travels about 8 kilometers per hour. In the river, it takes them the same amount of time \(t\) to go upstream 5 kilometers as it does to travel downstream 10 kilometers.

family boating on a river
  1. If the speed of the river is \(r\), write an expression for the time it takes to travel 5 kilometers upstream and an expression for the time it takes to travel 10 kilometers downstream.
  2. Use your expressions to calculate the speed of the river. Explain or show your reasoning.

Student Response

Student responses to this activity are available at one of our IM Certified Partners

Anticipated Misconceptions

Students may not know how to use the information that the boat goes 8 kph in still water. Tell them that if a boat is going downstream, then the river is pushing it forward, so it will travel at its speed plus the river’s speed. If the boat is going upstream, then the river will push against it and slow it down.

Activity Synthesis

The purpose of this discussion is for students to share how they solved a rational equation in which the denominators of the rational expressions are not the same. During the discussion, students should make connections to how they solved rational equations in the previous lesson in order to reason that they can strategically multiply by a common denominator in order to get an equation without variables in denominators.

Select previously identified students to share how they answered the last question, starting with any who used graphs. The main focus of the discussion should be on how students identified a solution to the equation \(\frac{5}{8-r}=\frac{10}{8+r}\). Here are some questions for discussion:

  • “What are some possible first steps to solving the equation?” (Multiplying by \(8-r\) or \(8+r\).)
  • “Is it better to multiply each side by \(8-r\) first or \(8+r\) first?” (No matter the order, multiplying by one then the other results in the linear equation \(5(8+r) = 10(8-r)\).)

If time allows, display a graph of \(y=\frac{5}{8-r}\) and ask, “What is the meaning of the vertical asymptote \(r=8\) in this situation?” (If the boat goes 8 kilometers per hour and the river goes 8 kilometers per hour, then the boat does not go anywhere.)

21.3: Rational Resistance (15 minutes)

Activity

In this activity, students use the formula for the total resistance of circuits in parallel to write a rational equation for an unknown resistance. The particular equation students write involves the sum of two rational expressions, which students have not seen before. They then solve the equation by graphing, which relates back to work earlier in the unit when students identified a solution to a 5th degree polynomial using a graph.

Monitor for students writing clear descriptions for how Clare could use graphs to determine the value of \(R\).

Launch

Begin the activity by asking students if they know what a circuit is and where it is used. If not suggested, tell students that one example of a circuit is a flashlight, in which the batteries, light, and switch together make a circuit. In a circuit, resistance is like friction—it makes it harder for electricity to flow. Often we want some resistance in a circuit so it can do work. For example, the filament in a light bulb glows because of its high resistance. In this activity, students are going to consider a law about circuits that are run in parallel, like the ones shown in this diagram.

Circuits in parallel.

For students who are still unsure about what a circuit is, it may be helpful for students to mentally picture \(R_1\), \(R_2\), and \(R_3\) in this picture as three light bulbs that are connected to the 12-volt battery on the far left. If students have not experienced subscript notation previously, let them know that when talking about the same type of thing, such as 3 light bulbs, the same letter with a different number written smaller and to the bottom right can be used to tell the difference between the objects.

Arrange students in groups of 2. Provide access to graphing technology. Display the task statement and first 2 questions for all to see. Give quiet work time for students to answer these questions, followed by sharing work with a partner. Select 2–3 students to share their equation with the class, recording student reasoning for all to see. 

Writing, Conversing: MLR1 Stronger and Clearer Each Time. Use this routine to help students improve their writing by providing them with multiple opportunities to clarify their explanations through conversation. Give students time to meet with 2–3 partners to share their response to the last question. Students should first check to see if they agree with each other about how Clare should use the graph to estimate the resistance of the first circuit \(R\) if \(R\)T is 85 ohms. Provide listeners with prompts for feedback that will help their partner add detail to strengthen and clarify their ideas. For example, students can ask their partner, “What did you do first?”, “How did you use the information in the graph to help you?” or “How did you find the point of intersection?” Next, provide students with 3–4 minutes to revise their initial draft based on feedback from their peers. This will help students produce written explanations of how they can use graphs to solve rational equations.
Design Principle(s): Optimize output (for explanation)
Representation: Internalize Comprehension. Differentiate the degree of difficulty or complexity by beginning with an example with more accessible values. Show that \(\frac12+\frac13=\frac56\), and that this is equivalent to \(\frac{1}{\tfrac65}\). Notice that while \(\frac56\) is more than \(\frac12\) and \(\frac13\)—which it should be, because we added two positive numbers—that \(\frac65\) is smaller than both 2 and 3, which is consistent with the idea that the bigger the denominator of a fraction, the smaller the number it represents. Create a display to show patterns in adding fractions with numerator 1, first with numbers (\(\frac12+\frac13=\frac56\)) and then algebraically (\(\tfrac1x + \frac1y\) has the common denominator \(xy\), and after rewriting both fractions to have this same denominator, we get \(\frac{y}{xy} + \tfrac{x}{xy} = \tfrac{x + y}{xy}\)) to use as a reference.
Supports accessibility for: Conceptual processing

Student Facing

Circuits in parallel follow this law: The inverse of the total resistance is the sum of the inverses of each individual resistance. We can write this as: \(\displaystyle \frac{1}{R_T}=\frac{1}{R_1} +\frac{1}{R_2}+ . . . + \frac{1}{R_n}\) where there are \(n\) parallel circuits and \(R_T\) is the total resistance. Resistance is measured in ohms.

  1. Two circuits are placed in parallel. The first circuit has a resistance of 40 ohms and the second circuit has a resistance of 60 ohms. What is the total resistance of the two circuits?
  2. Two circuits are placed in parallel. The second circuit has a resistance of 150 ohms more than the first. Write an equation for this situation showing the relationships between \(R_T\) and the resistance \(R\) of the first circuit.
  3. For this circuit, Clare wants to use graphs to estimate the resistance of the first circuit \(R\) if \(R_T\) is 85 ohms. Describe how she could use a graph to determine the value of \(R\) and then follow your instructions to find \(R\).

Student Response

Student responses to this activity are available at one of our IM Certified Partners

Student Facing

Are you ready for more?

Two circuits with resistances of 40 ohms and 60 ohms have a combined resistance of 24 ohms when connected in parallel. If we had used two circuits that each had a resistance of 48 ohms, they would have had that same combined resistance. 48 is called the harmonic mean of 40 and 60. A more familiar way to find the mean of two numbers is to add them up and divide by 2. This is the arithmetic mean. Here is how each kind of mean is calculated:

Harmonic mean of \(a\) and \(b\):

\(\dfrac{2ab}{a+b}\)

Arithmetic mean of \(a\) and \(b\):

\(\dfrac{a+b}{2}\)

The harmonic mean of 40 and 60 was 48, and their arithmetic mean is (40+60)/2=50. Experiment with other pairs of numbers. What can you conclude about the relationship between the harmonic mean and arithmetic mean?

Student Response

Student responses to this activity are available at one of our IM Certified Partners

Anticipated Misconceptions

Students may be unsure how to use graphs to find the solution to \(\frac{1}{85} = \frac{1}{R}+\frac{1}{R+150}\). Ask them to consider a simpler problem, such as \(85=x(x+15)\), and how they could use a graph to identify the value(s) of \(x\) that make(s) the equation true.

Activity Synthesis

Earlier in this unit, students used graphs to identify what value of \(x\) made the equation \(1,\!580=300x^4+500x^3+250x^2+400x\) true. This activity asks students a similar question, but with a rational equation instead of just a polynomial. An important takeaway from this discussion is for students to recognize that even as equations become more complicated and include things like rational expressions added together, everything they have learned about identifying solutions to equations is still true. Specifically, the point of intersection for graphs of the expressions on the left and right side gives the \(R\) value that makes each side equal to \(\frac{1}{85}\).

Select previously identified students to share their graphing directions, and then select one set of directions to follow while displaying the graphs for all to see. If time allows, ask students to use graphing technology to identify the value of \(R\) needed for different values of \(R_T\).

Lesson Synthesis

Lesson Synthesis

Arrange students in groups of 2. Tell students that while they have solved this equation using a graph, now they are going to solve it using algebra. Display the equation for all to see and tell students that a first step, rewriting the equation to have a single fraction on each side, has been done for them.

\(\begin{array} \\ \dfrac{1}{85} = \dfrac{1}{R}+\dfrac{1}{R+150} \\ \dfrac{1}{85} = \dfrac{2R+150}{R(R+150)} \end{array}\)

Ask students to identify what they could multiply each side of the equation by to get a new equation with no variables in any denominators. After a brief quiet think time, ask students to compare their responses to their partner’s and decide if they are both correct, even if they are different. Select 2–3 students to share what they would multiply by, recording responses for all to see. (\(R\) and \(R+150\) or just \(R^2+150R\).)

Once students are in agreement, add

\(\begin{array} \\ \dfrac{1}{85} &= \dfrac{2R+150}{R(R+150)} \\ \dfrac{1}{85} \boldcdot R(R+150) &= 2R+150 \end{array}\)

At this point, tell students they have a choice. \(\frac{1}{85}\) is a number, so we can use the distributive property on the left expression, or we could multiply each side by 85. Ask students to use whichever method they choose to rewrite the equation in the form \(\text{expression}=0\). After a brief quiet think time, ask students to compare their responses to their partner’s and decide if they are both correct, even if they are different. Next, ask students what type of equation this is. Depending on the form students used, it may be more or less clear that they have a quadratic equation, which is a type of equation students learned to solve in a previous course using methods such as the quadratic formula. If time allows, ask students to solve this equation using a method of their choice. Students will have more opportunity to solve these types of equations in a future lesson.

Conclude this discussion by telling students that solving rational equations by multiplying strategically in order to rewrite the equation without variables in denominators is sometimes called “clearing the denominators”.

21.4: Cool-down - Solving Rational Equations (5 minutes)

Cool-Down

Cool-downs for this lesson are available at one of our IM Certified Partners

Student Lesson Summary

Student Facing

A boat travels about 6 kilometers per hour in still water. If the boat is on a river that flows at a constant speed of \(r\) kilometers per hour, it can travel at a speed of \(6+r\) kilometers per hour downstream and \(6-r\) kilometers per hour upstream. (And if the river current is the same speed as the boat, the boat wouldn’t be able to travel upstream at all!)

On one particular river, the boat can travel 4 kilometers upstream in the same amount of time it takes to travel 12 kilometers downstream. Since time is equal to distance divided by speed, we can express the travel time as either \(\frac{12}{6+r}\) hours or \(\frac{4}{6-r}\) hours. If we don’t know the travel time, we can make an equation using the fact that these two expressions are equal to one another, and figure out the speed of the river.

\(\displaystyle \begin{align} \frac{12}{6+r} &= \frac{4}{6-r} \\ \frac{12}{6+r} \boldcdot (6+r)(6-r) &= \frac{4}{6-r} \boldcdot (6+r)(6-r) \\ 12(6-r) &= 4(6+r) \\ 72-12r &= 24 + 4r \\ 48 &= 16r \\ 3 &= r \\ \end{align}\)

Substituting this value into the original expressions, we have \(\frac{12}{6+3}=\frac43\) and \(\frac{4}{6-3}=\frac43\), so these two expressions are equal when \(r=3\). This means that when the water flow in the river is about 3 kilometers per hour, it takes the boat 1 hour and 20 minutes to go 4 kilometers upstream and 1 hour and 20 minutes to go 12 kilometers downstream.

Even though we started out with a rational expression on each side of the equation, multiplying each side by the product of the denominators, \((6+r)(6-r)\), resulted in an equation similar to ones we have solved before. Multiplying to get an equation with no variables in denominators is sometimes called “clearing the denominators.”