Lesson 10
Different Options for Solving One Equation
10.1: Algebra Talk: Solve Each Equation (5 minutes)
Warmup
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 \((x3)\) as a chunk of the equation. For each equation, the equation is true if \((x3)\) 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.
Launch
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 wholeclass discussion.
Supports accessibility for: Memory; Organization
Student Facing
\(\begin {align} 100(x3) &= 1,\!000\end{align}\)
\(\begin {align} 500(x3) &= 5,\!000\end{align}\)
\(\begin {align} 0.03(x3) &= 0.3 \end{align}\)
\(\begin {align} 0.72(x+2) &= 7.2 \\ \end{align}\)
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 \(100(x3)=1,\!000\) but cover the \((x3)\) with your hand or with an eraser. “100 times something is 1,000. What is the something?”
Design Principle(s): Optimize output (for explanation)
10.2: Analyzing Solution Methods (15 minutes)
Activity
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 \(2x18\) by the distributive property, which doesn’t result in \(2x\).
Launch
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 wholeclass 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.
Supports accessibility for: Language; Socialemotional skills
Student Facing
Three students each attempted to solve the equation \(2(x9)=10\), but got different solutions. Here are their methods. Do you agree with any of their methods, and why?
Noah’s method:
\(\begin{align} 2(x9)&=10 \\ 2(x9)+9 &= 10+9 & \text{add 9 to each side} \\ 2x &= 19 \\ 2x \div 2 &= 19 \div 2 & \text{divide each side by 2} \\ x &= \frac{19}{2} \\ \end{align}\)
Elena’s method:
\(\begin{align} 2(x9) &= 10 \\ 2x18 &= 10 & \text{apply the distributive property} \\ 2x1818 &= 1018 & \text{subtract 18 from each side} \\ 2x &= \text8 \\ 2x \div 2 &= \text8 \div 2 & \text{divide each side by 2} \\ x &= \text4 \\ \end{align} \)
Andre’s method:
\(\begin{align} 2(x9) &= 10 \\ 2x18 &= 10 & \text{apply the distributive property} \\ 2x18+18 &= 10+18 & \text{add 18 to each side} \\ 2x &= 28 \\ 2x \div 2 &= 28 \div 2 & \text{divide each side by 2} \\ x &= 14 \\ \end{align}\)
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 \(2(x9)=10\).
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.
Design Principle(s): Optimize output (for explanation)
10.3: Solution Pathways (15 minutes)
Activity
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.
Launch
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 wholeclass discussion.
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.

\(2 ,\!000(x0.03)=6 ,\!000\)

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

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

\(\text10(x  1.7) = \text3\)

\(5.4 = 0.3(x + 8)\)
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.)
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: Cooldown  Solve Two Equations (5 minutes)
CoolDown
Cooldowns 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.
\(\begin {align} 100(x+0.06) &= 21 \\ 100x+6 &= 21 \\ 100x &= 15 \\ x &= \tfrac{15}{100} \\ \end {align}\)