Lesson 10

Different Options for Solving One Equation

10.1: Algebra Talk: Solve Each Equation (5 minutes)


The purpose of this Algebra Talk is to promote seeing structure in equations of the form \(p(x+q)=r\). The goal is for students to see \((x-3)\) as a chunk of the equation. For each equation, the equation is true if \((x-3)\) is 10. These understandings help students develop fluency. While four problems are given, it may not be possible to share every strategy. Consider gathering only two or three different strategies per problem, saving most of the time for the final question.


Display one equation at a time. Give students 30 seconds of 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

\(\begin {align} 100(x-3) &= 1,\!000\end{align}\)

\(\begin {align} 500(x-3) &= 5,\!000\end{align}\)

\(\begin {align} 0.03(x-3) &= 0.3 \end{align}\)

\(\begin {align} 0.72(x+2) &= 7.2 \\ \end{align}\)

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 ___’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 _____’s strategy?”
  • “Do you agree or disagree? Why?”

It may help to display an equation like \(100(x-3)=1,\!000\) but cover the \((x-3)\) with your hand or with an eraser. “100 times something is 1,000. What is the something?”

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)

10.2: Analyzing Solution Methods (15 minutes)


From previous work in the unit, students should already understand that distributing first is a valid solution method, though this activity reinforces that understanding. The purpose of this activity is to make explicit a common pitfall (in Noah’s method). Monitor for students with different, valid reasons for agreeing or disagreeing. For example, disagree with Noah because . . .

  • his answer \(\frac{19}{2}\) doesn’t make the original equation true.
  • a tape diagram shows that adding 9 to each side does not result in a diagram that can be represented with \(2x=19\).
  • when you add 9 to the left side, you are adding it to \(2x-18\) by the distributive property, which doesn’t result in \(2x\).


Arrange students in groups of 2. Give 5–10 minutes quiet work time and time to share their reasoning with their partner, followed by a whole-class discussion.

Explain to students that their job is to analyze three solution methods for errors. They should share with their partner whether they agree or disagree with each method, and explain why.

Engagement: Develop Effort and Persistence. Encourage and support opportunities for peer interactions. Invite students to talk about their ideas with a partner before writing them down. Display sentence frames to support students when they explain their strategy. For example, “I agree/disagree with _____ because…” or “Instead of _____, he/she should have...”
Supports accessibility for: Language; Social-emotional skills

Student Facing

Three students each attempted to solve the equation \(2(x-9)=10\), but got different solutions. Here are their methods. Do you agree with any of their methods, and why?

Noah’s method:

\(\begin{align} 2(x-9)&=10 \\ 2(x-9)+9 &= 10+9 & \text{add 9 to each side} \\ 2x &= 19 \\ 2x \div 2 &= 19 \div 2 & \text{divide each side by 2} \\ x &= \frac{19}{2} \\ \end{align}\)

Elena’s method:

\(\begin{align} 2(x-9) &= 10 \\ 2x-18 &= 10 & \text{apply the distributive property} \\ 2x-18-18 &= 10-18 & \text{subtract 18 from each side} \\ 2x &= \text-8 \\ 2x \div 2 &= \text-8 \div 2 & \text{divide each side by 2} \\ x &= \text-4 \\ \end{align} \)

Andre’s method:

\(\begin{align} 2(x-9) &= 10 \\ 2x-18 &= 10 & \text{apply the distributive property} \\ 2x-18+18 &= 10+18 & \text{add 18 to each side} \\ 2x &= 28 \\ 2x \div 2 &= 28 \div 2 & \text{divide each side by 2} \\ x &= 14 \\ \end{align}\)

Student Response

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

Anticipated Misconceptions

If students aren’t sure how to begin analyzing Noah’s method, ask them to explain what it means for a number to be a solution of an equation. Alternatively, suggest that they draw a tape diagram of \(2(x-9)=10\).

Activity Synthesis

Invite students to share as many unique reasons they agree or disagree with each method as time allows. See the Activity Narrative for anticipated approaches. Pay particular attention to Noah’s method, since this represents a common error. 

Speaking: MLR8 Discussion Supports. To help students produce statements that explain why they agree (or disagree) with the different solution methods, provide sentence frames such as “I notice ___ so I …,” “That could be true because…” and “This method works because…” This will provide students with a structure to communicate their reasoning while promoting access to both content and language.
Design Principle(s): Optimize output (for explanation)

10.3: Solution Pathways (15 minutes)


The purpose of this activity is to practice solving equations of the form \(p(x+q)=r\), recognizing that there are two valid approaches, and making judgments about which one is more sensible for a given equation.


Display this equation and a hanger diagram to match: \(3(x+2)=21\). Tell students, “Any time you want to solve an equation in this form, you have a choice to make about how to proceed. You can either divide each side by 3 or you can distribute the 3.” Demonstrate each solution method side by side, while appealing to reasoning about the hanger diagram.

Keep students in the same groups. 5–10 minutes of quiet or partner work time followed by a whole-class discussion.

Representation: Internalize Comprehension. Activate or supply background knowledge. Allow students to use calculators to ensure inclusive participation in the activity.
Supports accessibility for: Memory; Conceptual processing

Student Facing

For each equation, try to solve the equation using each method (dividing each side first, or applying the distributive property first). Some equations are easier to solve by one method than the other. When that is the case, stop doing the harder method and write down the reason you stopped.

  1. \(2 ,\!000(x-0.03)=6 ,\!000\)

  2. \(2(x+1.25)=3.5\)

  3. \(\frac14 (4 + x) = \frac43\)

  4. \(\text-10(x - 1.7) = \text-3\)

  5. \(5.4 = 0.3(x + 8)\)

Student Response

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

Activity Synthesis

Reveal the solution to each equation and give students a few minutes to resolve any discrepancies with their partner.

Display the list of equations in the task, and ask students to help you label them with which solution method was easier, either “divide first” or “distribute first.” Discuss any disagreements and the reasons one method is easier than the other. (There is really no right or wrong answer here. Some people might prefer moves that eliminate fractions and decimals as early as possible. Some might want to minimize the number of computations.)

Speaking: MLR8 Discussion Supports. Use this routine to help students produce mathematical language to communicate about which method is more efficient. Give groups of students 3–4 minutes to discuss which method would be easiest for each problem. Next, select groups to share how their solution methods minimize the number of computations needed, or address eliminating fractions and decimals. Consider providing sentence frames such as: “Dividing first was easier because ________, ” or “Distributing first was easier because_____.” Call on students to restate and/or revoice their peers' descriptions using mathematical language.
Design Principle(s): Optimize output (for justification)

Lesson Synthesis

Lesson Synthesis

Possible questions for discussion:

  • “What are the two main ways we can approach solving equations like the ones we saw today?” (divide first or distribute first)
  • “What kinds of things do we look for to decide which approach is better?” (powers of ten, operations that result in whole numbers, moves that will eliminate fractions or decimals)
  • “How can we check if our answer is a solution to the original equation?” (Substitute our answer for the variable and see if it makes the equation true.)

10.4: Cool-down - Solve Two Equations (5 minutes)


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

Student Lesson Summary

Student Facing

Equations can be solved in many ways. In this lesson, we focused on equations with a specific structure, and two specific ways to solve them. 

Suppose we are trying to solve the equation \(\frac45(x+27)=16\). Two useful approaches are:

  • divide each side by \(\frac45\)
  • apply the distributive property

In order to decide which approach is better, we can look at the numbers and think about which would be easier to compute. We notice that \(\frac45 \boldcdot 27\) will be hard, because 27 isn't divisible by 5. But \(16 \div \frac45\) gives us \(16 \boldcdot \frac54\), and 16 is divisible by 4. Dividing each side by \(\frac45\) gives:

\(\begin{align} \tfrac45 (x+27) &= 16 \\ \tfrac54 \boldcdot \tfrac45 (x+27) &= 16 \boldcdot \tfrac54 \\ x+27 &= 20 \\ x &= \text- 7 \\ \end{align} \)

Sometimes the calculations are simpler if we first use the distributive property. Let's look at the equation \(100(x+0.06)=21\). If we first divide each side by 100, we get \(\frac{21}{100}\)  or 0.21 on the right side of the equation. But if we use the distributive property first, we get an equation that only contains whole numbers. 

\(\begin {align} 100(x+0.06) &= 21 \\ 100x+6 &= 21 \\ 100x &= 15 \\ x &= \tfrac{15}{100} \\ \end {align}\)