Lesson 22
Solving Rational Equations
22.1: Notice and Wonder: Thoughtful Multiplication (5 minutes)
Warm-up
This warm-up begins where students left off at the end of the previous lesson, considering how to multiply strategically to “clear the denominators” in a rational equation. The purpose of this warm-up is to elicit the idea that when multiplying to clear denominators, we only need to multiply by the least common denominator shared by the different terms in the expression. This idea will be used in the following activities as students solve rational equations and investigate extra solutions that sometimes arise in the solving process.
While students may notice and wonder many things about these images, the multiplication by \(x(x+2)\) and not \(x(x+2)^2\) (the result of multiplying together all the denominators) is the important discussion point.
Launch
Display the partially solved equation for all to see. Ask students to think of at least one thing they notice and at least one thing they wonder. Give students 1 minute of quiet think time, and then 1 minute to discuss the things they notice and wonder with their partner, followed by a whole-class discussion.
Student Facing
What do you notice? What do you wonder?
\(\displaystyle \begin{align} \frac{3}{x(x-2)} &= \frac{2x+1}{x-2} \\ \frac{3}{x(x-2)} \boldcdot x(x-2) &= \frac{2x+1}{x-2} \boldcdot x(x-2) \\ 3 &= 2x^2 + x \\ 0 &= 2x^2+x-3 \\ \end{align}\)
Student Response
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Activity Synthesis
Ask students to share the things they noticed and wondered. Record and display their responses for all to see. If possible, record the relevant reasoning on or near the image. After all responses have been recorded without commentary or editing, ask students, “Is there anything on this list that you are wondering about now?” Encourage students to respectfully disagree, ask for clarification, or point out contradicting information. If students do not recall or have not heard the phrase “clearing the denominators” for this type of multiplying, tell them it is a useful description for the first few steps shown.
If time allows, have students finish solving for \(x\) by either using the quadratic formula or factoring (-1.5, 1).
22.2: Rational Solving (15 minutes)
Activity
The purpose of this activity is for students to understand how steps used to solve a rational equation sometimes lead to nonequivalent equations, giving rise to so-called extraneous solutions. For example, multiplying each side of a rational equation by \(x+1\) creates a new equation that is true when \(x= \text-1\), since \(0=0\), even if the original equation was not true at \(x= \text-1\).
Monitor for students who notice that neither 0 nor -1 can be substituted for \(x\) in the original expressions due to division by 0, and students who notice that multiplying by a variable on each side of an equation is different than the type of steps they have used in the past to solve linear or quadratic equations (specifically, that multiplying by \(x+1\) is the same as multiplying by 0 when \(x= \text-1\)).
Launch
Design Principle(s): Optimize output (for explanation)
Student Facing
Jada is working to find values of \(x\) that make this equation true:
\(\displaystyle \frac{5x+5}{x+1} = \frac{5}{x}\)
She says, “If I multiply both sides by \(x(x+1)\), I find that the solutions are \(x=1\) and \(x= \text-1\), but when I substitute in \(x= \text-1\), the equation does not make any sense.”
- Is Jada’s work correct? Explain or show your reasoning.
- Why does Jada’s method produce an \(x\) value that does not solve the equation?
Student Response
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Student Facing
Are you ready for more?
- What are the solutions to \(x^2=1\)?
- What are the solutions to \(\frac{x^2}{x-1} = \frac{1}{x-1}\)?
- How can you solve \(\frac{x^2}{x-1} = \frac{1}{x-1}\) by inspection?
- How does the denominator influence the solution(s) to \(\frac{x^2}{x-1} = \frac{1}{x-1}\)?
Student Response
For access, consult one of our IM Certified Partners.
Activity Synthesis
The goal of this discussion is for students to share what they think is happening to make \(x= \text-1\) appear as a solution to the original equation even though it is not.
Invite previously identified students to share things they noticed about the original equation and what it means when we multiply each side of an equation by a variable. Consider these questions to help further the discussion:
- “Jada multiplied by \(x(x+1)\). What values of \(x\) make that expression equal to 0?” (0 and -1.)
- “What are values of \(x\) that we cannot substitute into the original equation?” (We cannot substitute 0 or -1 for \(x\) in the original equation, because both values create an invalid equation due to division by 0.)
- “What happens when you multiply each side of any equation by 0? What values of \(x\) make this new equation true but not necessarily the original equation?” (You get the equation \(0=0\), which is true for all values of \(x\).)
Help summarize the discussion by telling students that sometimes steps we do to solve an equation result in a new equation that is not equivalent to the original equation. Two equations are equivalent if they have the exact same solutions. Since \(x= \text-1\) is a solution to the new equation but not the original equation, some people call this an extraneous solution. The step that created a new equation that is not equivalent to the original equation was when each side was multiplied by an expression that can have the value 0. This can make two previously unequal (or undefined) sides equal for some value of the variable, so the original equation and the new equation are not equivalent. These inequivalent equations do not always arise from multiplying both sides of an equation by an expression, so we should always check the solution in the original equation to be sure. In later units, students will see other types of equation solving steps that can result in new equations that have solutions that do not satisfy the original equation.
22.3: More Rational Solving (15 minutes)
Activity
The purpose of this activity is for students to practice solving rational equations and identifying extraneous solutions, if they exist. Students are expected to use algebraic methods to solve the equations and should be discouraged from using graphing technology.
Launch
Arrange students in groups of 2. Tell students to complete the first problem individually and then check their work with their partner before completing the following problems together.
Design Principle(s): Support sense-making; Maximize meta-awareness
Supports accessibility for: Organization; Attention
Student Facing
- Here are a lot of equations. For each one, use what you know about division to identify values of \(x\) that cannot be solutions to the equation.
- \(\dfrac{x^2+x-6}{x-2} = 5\)
- \(\dfrac{2x+1}{x} = \dfrac{1}{x-2}\)
- \(\dfrac{10}{x+8} = \dfrac{5}{x-8}\)
- \(\dfrac{x^2+x+1}{13} = \dfrac{2}{x-1}\)
- \(\dfrac{x+1}{4x} = \dfrac{x-1}{3x}\)
- \(\dfrac{1}{x} = \dfrac{1}{x(x+1)}\)
- \(\dfrac{x+2}{x} = \dfrac{3}{x-2}\)
- \(\dfrac{1}{x-3} = \dfrac{1}{x(x-3)}\)
- \(\dfrac{(x+1)(x+2)}{x+1} = \dfrac{x+2}{x+1}\)
- Without solving, identify three of the equations that you think would be least difficult to solve and three that you think would be most difficult to solve. Be prepared to explain your reasoning.
- Choose three equations to solve. At least one should be from your “least difficult” list and one should be from your “most difficult” list.
Student Response
For access, consult one of our IM Certified Partners.
Anticipated Misconceptions
Some students may forget how to solve a quadratic equation. Remind them of options such as factoring or using the quadratic formula when they have an expression equal to zero.
Activity Synthesis
The purpose of this discussion is for students to share strategies for solving different types of equations. Some students may have thought that an equation was in the “least difficult” category, while others thought that the same equation was in the “most difficult” category. Remind students that once you feel confident about the strategies for solving an equation, it may move into the “least difficult” category, and recognizing good strategies takes practice and time.
Informally poll the class for each equation as to whether they placed it in the “most difficult” category, “least difficult” category, or if it was somewhere in the middle. Record and display the results for all to see.
For questions with a split vote, have a group share something that made it seem difficult about it and another group share something that made it seem less difficult. If there are any questions that everyone thought would be more difficult or everyone thought would be less difficult, ask students why it seemed that way. Ask students, “Were there any equations that were more difficult to solve than you expected? Were there any that were less difficult to solve than you expected?” If not brought up, ask how students worked out the equation \(\frac{2x+1}{x} = \frac{1}{x-2}\), which has irrational solutions.
Lesson Synthesis
Lesson Synthesis
The purpose of this writing prompt is for students to reflect on what they have learned about how so-called extraneous solutions can arise when solving rational equations. Ask students to respond to the following prompt: “How can extra solutions arise in the process of solving an equation?” Encourage students to use equations from the lesson to help in their explanations if needed.
If time allows, ask students to share what they’ve written with a partner and then select 2–3 students to share something from either their paper or their partner’s with the class. Key understandings are that when we multiply each side of an equation by an expression that can have the value 0, we sometimes get an equation that is not equivalent to the initial equation, and that we should always check to see if a solution satisfies the original equation.
22.4: Cool-down - Find Rational Solutions (5 minutes)
Cool-Down
For access, consult one of our IM Certified Partners.
Student Lesson Summary
Student Facing
Consider the equation \(\frac{x+2}{x(x+1)} = \frac{2}{(x+1)(x-1)}\). We could solve this equation for \(x\) by multiplying each expression by \(x(x+1)(x-1)\) to get an equation with no variables in denominators, and then rearranging it into an expression that equals 0. Here is what that looks like:
\(\displaystyle \begin{align} \frac{x+2}{x(x+1)} \boldcdot x(x+1)(x-1) &= \frac{2}{(x+1)(x-1)} \boldcdot x(x+1)(x-1) \\ (x+2)(x-1) &= 2x \\ x^2 + x - 2 &= 2x \\ x^2 - x - 2 &= 0 \\ (x-2)(x+1) & = 0 \\ \end{align}\)
The last equation, \((x-2)(x+1) = 0\), leads us to believe that the original equation has two solutions: \(x=2\) and \(x=\text-1\). Substituting \(x=2\) into the original equation, we get \(\frac{2+2}{2(2+1)} = \frac{2}{(2+1)(2-1)}\), which is true since each side is equal to \(\frac23\). But, substituting \(x=\text-1\) into the original equation, we get \(\frac{\text-1+2}{\text-1(\text-1+1)} = \frac{2}{(\text-1+1)(\text-1-1)}\), which isn’t a valid equation since division by 0 is not allowed. This means \(x=\text-1\) isn’t a solution, so what happened to make us think that it was?
Let’s consider the simpler equation \(x - 5 = 0\). This equation has one solution, \(x=5\). But if we multiply each side by \((x-1)\) the result is a new equation, \((x-1)(x-5)=0\), which has solutions 5 and 1. The 1 is a solution to the new equation because when \(x=1\), \(x-1=0\). But if we substitute 1 for \(x\) into the original equation, we get \(1-5 = \text-4 = 0\), which is not a valid equation, so 1 is not a solution to the original equation. Because we multiplied each side of the original equation by an expression that has the value 0 when \(x=1\), the two sides \(x-5\) and 0 that were unequal at that specific \(x\)-value are now equal. For this example, \(x=1\) is sometimes called an extraneous solution.
In the original example, \(x=\text-1\) is the extraneous solution. While \(x=\text-1\) is a solution to the equation we wrote after we multiplied the original equation by \(x(x+1)(x-1)\) on each side, it is not a solution to the original equation since they are not equivalent. It should be noted that even though we multiplied by \(x\), \((x+1)\), and \((x-1)\), only one extraneous solution was added. This shows that multiplying by an expression that can equal zero does not always cause an extraneous solution. So how do we tell if a solution is extraneous or not? We substitute it into the original equation and make sure the result is a valid equation.