Lesson 13
More Balanced Moves
13.1: Different Equations? (5 minutes)
Warm-up
The purpose of this warm-up is to for students to use the structure of equations to recognize when they are the same without having to solve for the specific \(x\) value that makes the equations true.
Monitor for students who:
- solve each equation for \(x\), then compare the solutions
- solve Equation 1 for \(x\), then substitute it into Equations A–D
- manipulate Equations A–D to look like Equation 1 and vice versa
Launch
Give students 2–3 minutes quiet think time, then whole-class discussion.
Student Facing
Equation 1
\(x-3=2-4x\)
Which of these have the same solution as Equation 1? Be prepared to explain your reasoning.
Equation A
\(2x-6=4-8x\)
Equation B
\(x-5=\text-4x\)
Equation C
\(2(1-2x)=x-3\)
Equation D
\(\text-3=2-5x\)
Student Response
For access, consult one of our IM Certified Partners.
Activity Synthesis
Select students previously identified to share how they determined whether each equation had the same solution as Equation 1 in the sequence listed in the Activity Narrative. Point out that the question did not ask students what the solution was, only whether each equation had the same solution.
To help students make connections between the different methods their classmates used to solve the warm-up, ask:
- “Which method of answering the question was most efficient? After seeing all these ways to answer the question, which would you choose?”
- “What is an advantage of changing the equation to look like Equation 1? What is a disadvantage?” (An advantage is that I could see quickly whether it would be the same as Equation 1, and I didn't have to keep going to actually figure out the value of \(x\). A disadvantage would be that I never discovered what the value for \(x\) is that makes the equations true.)
- “How is this method (manipulating the equation to look like Equation 1) similar to what we did in previous lessons with the balance hangers?” (In order to keep the hangers balanced, I had to make sure to do the same thing to each side of the hanger. In order to have each equation still be true, I have to make sure to do the same thing to each side of an equation.)
By showing that two equations are related by a move (or series of moves), we know they must have the same solution.
If time allows, have students create another equation with the same solution as Equation 1 and trade with a partner. They should then explain the step(s) necessary to make it look like Equation 1 to each other.
13.2: Step by Step by Step by Step (15 minutes)
Activity
Before students work on solving complex equations on their own, in this activity they examine the work (both good and bad) of others. The purpose of this activity is to build student fluency solving equations by examining the solutions of others for both appropriate and inappropriate strategies (MP3).
Encourage students to use precise language when discussing the different steps made by the four students in the problem (MP6). For example, if a student says Clare distributed to move from \(12x + 3 = 3(5x + 9)\) to \(3(4x+1) = 3(5x + 9)\), ask them to be more specific about how Clare used the distributive property to help the whole class follow along. (Clare used the distributive property to re-write \(12x + 3\) as \(3(4x+1)\).)
Launch
Arrange students in groups of 2. Give 4–5 minutes of quiet work time and ask students to pause after the first two problems for a partner discussion. Give 2–3 minutes for partners to work together on the final problem followed by a whole-class discussion. Refer to MLR 3 (Clarify, Critique, Correct) to guide students in using language to describe the wrong steps.
Supports accessibility for: Organization; Attention
Student Facing
Here is an equation, and then all the steps Clare wrote to solve it:
\(\displaystyle \begin{align}14x - 2x + 3 &= 3(5x + 9)\\12x + 3& = 3(5x + 9)\\3(4x+1)& = 3(5x + 9)\\4x + 1 &= 5x + 9\\1 &= x + 9\\ \text{-}8 &= x \end{align}\)
Here is the same equation, and the steps Lin wrote to solve it:
\(\displaystyle \begin{align}14x - 2x + 3 &= 3(5x + 9)\\12x + 3 &= 3(5x + 9)\\12x + 3 &= 15x + 27\\12x &= 15x + 24\\ \text{-}3x &= 24\\x &= \text{-}8 \end{align}\)
- Are both of their solutions correct? Explain your reasoning.
- Describe some ways the steps they took are alike and different.
- Mai and Noah also solved the equation, but some of their steps have errors. Find the incorrect step in each solution and explain why it is incorrect.
Mai:
\(\displaystyle \begin{align}14x - 2x + 3 &= 3(5x + 9) \\ 12x + 3 &= 3(5x + 9) \\ 7x + 3 &= 3(9) \\7x + 3 &= 27 \\7x &= 24 \\ x &= \frac{24}{7} \end{align}\)Noah:
\(\displaystyle \begin{align}14x - 2x + 3 &= 3(5x + 9) \\ 12x + 3 &= 15x + 27 \\ 27x + 3 &= 27 \\ 27x& = 24 \\ x &= \frac{24}{27} \end{align}\)
Student Response
For access, consult one of our IM Certified Partners.
Activity Synthesis
Begin the discussion by asking, “How do you know when a solution to an equation is correct?” (One way to know it is correct is by substituting the value of \(x\) into the original equation and seeing if it makes the equation true.)
Display Clare and Lin’s solutions for all to see. Poll the class to see which solution they prefer. It is important to draw out that neither solution is better than the other, they are two ways of accomplishing the same task: solving for \(x\). Invite groups to share ways the steps Clare and Lin took are alike and different while annotating the two solutions with students’ observations. If none of the groups say it, point out that while the final steps may look different for Clare and Lin, their later steps worked to reduce the total number of terms until only an \(x\)-term and a number remained on either side of the equal sign.
Display Mai and Noah’s incorrect solutions for all to see. Invite groups to share an incorrect step they found and what advice they would give to Mai and Noah for checking their work in the future.
Design Principle(s): Optimize output (for comparison); Cultivate conversation
13.3: Make Your Own Steps (15 minutes)
Activity
The purpose of this lesson is to increase fluency in solving equations. Students will solve equations individually and then compare differing, though accurate, solution paths in order to compare their work with others. This will help students recognize that while the final solution will be the same, there is more than one path to the correct answer that uses principles of balancing equations learned in previous lessons.
Launch
Arrange students in groups of 3–4. Give students quiet think time to complete the activity and then tell groups to share how they solved the equations for \(x\) and discuss the similarities and differences in their solution paths.
Supports accessibility for: Memory; Conceptual processing
Design Principle(s); Maximize meta-awareness
Student Facing
Solve these equations for \(x\).
1. \(\frac{12+6x}{3}=\frac{5-9}{2}\)
2. \(x-4=\frac13(6x-54)\)
3. \(\text-(3x-12)=9x-4\)
Student Response
For access, consult one of our IM Certified Partners.
Student Facing
Are you ready for more?
I have 24 pencils and 3 cups. The second cup holds one more pencil than the first. The third holds one more than the second. How many pencils does each cup contain?
Student Response
For access, consult one of our IM Certified Partners.
Activity Synthesis
Students should take away from this activity the importance of using valid steps to solve an equation over following a specific solution path. Invite students to share what they discussed in their groups. Consider using some of the following prompts:
- “How many different ways did your group members solve each problem?”
- “When you compared solution paths, did you still come up with the same solution?” (Yes, even though we took different paths, we ended up with the same solutions.)
- “How can you make sure that the path you choose to solve an equation is a valid path?” (I can use the steps we discovered earlier when we were balancing: adding the same value to each side, multiplying (or dividing) by the same value to each side, distributing and collecting like terms whenever it is needed.)
- “What are some examples of steps that will not result in a valid solution?” (Performing an action to only one side of an equation and distributing incorrectly will give an incorrect solution.)
13.4: Trading Moves (20 minutes)
Activity
The goal of this activity is for students to build fluency solving equations with variables on each side. Students describe each step in their solution process to a partner and justify how each of their changes maintains the equality of the two expressions (MP3).
Look for groups solving problems in different, but efficient, ways. For example, one group may distribute the \(\frac12\) on the left side in problem 2 while another may multiply each side of the equation by 2 in order to re-write the equation with less factors on each side.
Launch
Arrange students in groups of 2. Instruct the class that they will receive 4 cards with problems on them and that their goal is to create a solution to the problems.
For the first two cards they draw, students will alternate solving by stating to their partner the step they plan to do to each side of the equation and why before writing down the step and passing the card. For the final two problem cards, each partner picks one and writes out its solution individually before trading to check each others’ work.
To help students understand how they are expected to solve the first two problems, demonstrate the trading process with a student volunteer and a sample equation. Emphasize that the “why” justification should include how their step maintains the equality of the equation. Use MLR 3 (Clarify, Critique, Correct) by reminding students to push each other to explain how their step guarantees that the equation is still balanced as they are working. For example, a student might say they are combining two terms on one side of the equation, which maintains the equality as the value of that side does not change, only the appearance.
Distribute 4 slips from the blackline master to each group. Give time for groups to complete the problems, leaving at least 5 minutes for a whole-class discussion. If any groups finish early, make sure they have checked their solutions and then challenge them to try and find a new solution to one of the problems that uses less steps than their first solution. Conclude with a whole-class discussion.
If time is a concern, give each group 2 cards rather than all 4 and have them only doing the trading steps portion of the activity, but make sure that all 4 cards are distributed throughout the class. Give 6–7 minutes for groups to complete their problems. Make sure each problem is discussed in a final whole-group discussion. Alternatively, extend the activity by selecting more problems for students to solve with their partners.
Student Facing
Your teacher will give you 4 cards, each with an equation.
- With your partner, select a card and choose who will take the first turn.
- During your turn, decide what the next move to solve the equation should be, explain your choice to your partner, and then write it down once you both agree. Switch roles for the next move. This continues until the equation is solved.
- Choose a second equation to solve in the same way, trading the card back and forth after each move.
- For the last two equations, choose one each to solve and then trade with your partner when you finish to check one another’s work.
Student Response
For access, consult one of our IM Certified Partners.
Activity Synthesis
Depending on your observations as students worked, you may wish to begin the discussion with a few common errors and ask students to explain why they are errors. For example, write out a solution to problem 2 where the second line has \(3.5x-6\) instead of \(3.5x-3\) and then ask students to find the error.
The goal of this discussion is for the class to see different, successful ways of solving the same equation. Record and display the student thinking that emerges during the discussion to help the class follow what is being said. To highlight some of the differences in solution paths, ask:
- “Did your partner ever make a move different than the one you expected them to? Describe it.”
- “For problem 4, could you start by halving the value of each side? Why might you want to do this?”
- “What’s an arithmetic error you made but then caught when you checked your work?”
Supports accessibility for: Conceptual processing
Design Principle(s): Cultivate conversation; Maximize meta-awareness
Lesson Synthesis
Lesson Synthesis
Display the following prompts one at a time and after each ask students if the move described maintains the equality of an equation:
- subtract a number from each side (maintains)
- subtract \(4x\) from each side (maintains)
- dividing each side of the equation by 7 (maintains)
- adding \(5x\) to one side and 10 to the other (maintains equality only if \(x=2\))
- add 4 to one side and add 5 to the other (does not maintain equality)
Ask students to write an equation and a solution to the equation that contains an error. Then, tell students to swap with a partner and try to find the error in their partner’s solution.
13.5: Cool-down - Mis-Steps (5 minutes)
Cool-Down
For access, consult one of our IM Certified Partners.
Student Lesson Summary
Student Facing
How do we make sure the solution we find for an equation is correct? Accidentally adding when we meant to subtract, missing a negative when we distribute, forgetting to write an \(x\) from one line to the next–there are many possible mistakes to watch out for!
Fortunately, each step we take solving an equation results in a new equation with the same solution as the original. This means we can check our work by substituting the value of the solution into the original equation. For example, say we solve the following equation:
\(\begin{align} 2x&=\text-3(x+5)\\ 2x&=\text-3x+15\\ 5x&=15\\ x&=3 \end{align}\)
Substituting 3 in place of \(x\) into the original equation,
\(\begin{align} 2(3) &= \text-3(3+5)\\ 6&= \text-3(8)\\ 6&=\text-24 \end{align}\)
we get a statement that isn't true! This tells us we must have made a mistake somewhere. Checking our original steps carefully, we made a mistake when distributing -3. Fixing it, we now have
\(\begin{align} 2x&=\text-3(x+5)\\ 2x&=\text-3x-15\\ 5x&=\text-15\\ x&=\text-3 \end{align}\)
Substituting -3 in place of \(x\) into the original equation to make sure we didn't make another mistake:
\(\begin{align} 2(\text-3) &= \text-3(\text-3+5)\\ \text-6&= \text-3(2)\\ \text-6&=\text-6 \end{align}\)
This equation is true, so \(x=\text-3\) is the solution.