Lesson 7
Explaining Steps for Rewriting Equations
7.1: Math Talk: Could It be Zero? (5 minutes)
Warmup
This Math Talk encourages students to to rely on the structure of equations, properties of operations, and what they know about solutions to equations to mentally solve problems. It also prompts students to recall that dividing a number by 0 leads to an undefined result, preparing them for the work later in the lesson. (In that activity, students will consider why dividing by a variable is not considered an acceptable move when writing equivalent equations or solving equations.)
To determine if 0 is a solution to the equations, students could substitute 0 into the expressions and evaluate them. For some equations, however, the answer can be efficiently found by making use of structure (MP7). In explaining their strategies, students need to be precise in their word choice and use of language (MP6).
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 wholeclass discussion.
Supports accessibility for: Memory; Organization
Student Facing
Is 0 a solution to each equation?
\(4(x + 2) = 10\)
\(12  8x = 3(x + 4)\)
\(5x = \frac12 x\)
\(\frac {6}{x} + 1 = 8\)
Student Response
For access, consult 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?”
Design Principle(s): Optimize output (for explanation)
7.2: Explaining Acceptable Moves (15 minutes)
Activity
Previously, students saw that certain moves can be made to an equation to create an equivalent equation. In this activity, they deepen that understanding by explaining why, if a given equation is true for a certain value of the variable, performing one of those moves leads to a second equation that is also true for the same variable value. This requires more than simply stating what the moves are, and offers students opportunities to construct logical arguments (MP3).
Clarify the distinction with an example. Ask students: "Consider these equations: \(3x  5 = 16\) and \(3x = 21\). The first equation is a true statement for a certain value of \(x\). Can you explain why the second equation must also be true for the same value of \(x\)?"
Explain that an answer such as “Adding 5 to each side of the first equation gives the second equation” is a description of the move rather than an explanation for why the second equation must be true.
An explanation may sound something like: "We know that \(3x5\) and 16 are equal when \(x\) has a particular value. If we add the same number, like 5, to both \(3x5\) and to 16, the results are still equal for the same variable value. This means the statement \(3x = 21\) must be true for the same value of \(x\)."
In this partner activity, students take turns giving explanations, and listening to and critiquing a partner's explanations (MP3). As students discuss their thinking, listen for explanations that are particularly clear so that they can be shared with the class later.
Launch
Tell students that they have performed a number of moves to write equivalent equations and solve equations. They will now practice explaining to a partner why those moves are legitimate. Demonstrate what it means to explain or defend the steps rather than simply describing them, as shown in the Activity Narrative.
Arrange students in groups of 2. Ask the partners in each group to choose different equations from each column. Give students a few minutes of quiet time to think about and write explanations about the equations in column A. Then, give students time to take turns sharing their explanations with their partner before moving on to column B.
Tell students that when one student explains, the partner’s job is to listen and make sure that they agree and that the explanation makes sense. If they don't agree, the partners discuss until they come to an agreement.
Repeat the process with the equations in column B.
Design Principle(s): Support sensemaking; Maximize metaawareness
Student Facing
Here are some pairs of equations. While one partner listens, the other partner should:
 Choose a pair of equations from column A. Explain why, if \(x\) is a number that makes the first equation true, then it also makes the second equation true.
 Choose a pair of equations from column B. Explain why the second equation is no longer true for a value of \(x\) that makes the first equation true.
Then, switch roles until you run out of time or you run out of pairs of equations.
A  B  

1. 
\(16=4(9x)\) \(16=364x\) 
\(9x = 5x + 4\) \(14x = 4\) 
2. 
\(5x = 24 + 2x\) \(3x = 24\) 
\(\frac12x  8 = 9\) \(x  8 = 18\) 
3. 
\(\text3(2x +9)=12\) \(2x+9=\text 4\) 
\(6x  6 = 3x\) \(x  1 = 3x\) 
4. 
\(5x=3x\) \(5x = \textx + 3\) 
\(\text11(x 2) = 8\) \(x2 = 8 + 11\) 
5. 
\(18 = 3x  6 + x\) \(18 = 4x  6\) 
\(4  5x = 24\) \(5x = 20\) 
Student Response
For access, consult one of our IM Certified Partners.
Anticipated Misconceptions
Students may have trouble seeing why some equations in column B are not equivalent (particularly the second item, which contains a common error). Encourage these students to choose one pair of equations, solve one equation, and then substitute the solution into the other equation to see what goes wrong.
Activity Synthesis
Invite previously identified students to share their explanations on at least a couple of pairs of equations from each column. If not already clear from students' explanations, emphasize that:
 If two expressions are equal, then performing the same addition (or subtraction) or multiplication (or division) to both expressions maintains the equality.
 Applying the distributive, commutative, and associative properties (of multiplication or addition) to an expression doesn't change its value, so doing so to one side of a true equation doesn't change the equality.
Explain to students that next they will look at some examples where the moves made to write equivalent equations appear to be acceptable but the resulting equations turn out to be false statements.
Supports accessibility for: Visualspatial processing; Conceptual processing
7.3: It Doesn't Work! (15 minutes)
Activity
So far, students have seen only onevariable equations that have a solution. For these equations, performing acceptable moves always led to equivalent equations that have the same solution. In this activity, students encounter an example where the given equation has no solutions and performing the familiar moves leads to an untrue statement.
Prior to this point, students have added, subtracted, multiplied, and divided a number from both sides of an equation. They have also added a variable expression to (or subtracted a variable expression from) an equation. They recognize these moves as allowable for solving equations. Here, students also come across an equation that is divided by a variable expression and make sense of why it leads to a false statement.
Launch
Arrange students into groups of 3–4. Give groups 1–2 minutes of quiet time to analyze the first set of equations and then time to brainstorm why Noah's work results in a false statement. Follow with a class discussion.
Invite students to share their explanations as to why Noah ended up with \(6=1\). Students are likely to say that the moves Noah made were allowable and can be explained, but they may struggle to say why they end up with a false statement.
Draw students’ attention to the third line of Noah's work. Ask them to interpret the expression on each side of the equal sign: \(x + 6\) can be interpreted as 6 more than a number and \(x + 1\) as 1 more than that same number. Then, ask them to try to find a value of \(x\) that could make that equation true. (There is none!)
Highlight that it’s not possible for 6 more than some number, no matter what that number is, to be equal to 1 more than that number. If no value of \(x\) could make the third equation true, the same could be said about the original equation—it had no solution. So even though Noah performed acceptable moves, the final equation is a false equation. Because all of these equations are equivalent, this means the original equation is also a false equation. It has no solutions because no value of \(x\) could make the equation true.
Repeat the process to analyze the second set of equations. See Activity Synthesis for discussion questions.
Student Facing
Noah is having trouble solving two equations. In each case, he took steps that he thought were acceptable but ended up with statements that are clearly not true.
Analyze Noah’s work on each equation and the moves he made. Were they acceptable moves? Why do you think he ended up with a false equation?
Discuss your observations with your group and be prepared to share your conclusions. If you get stuck, consider solving each equation.

\(\begin {align} x + 6 &= 4x + 1  3x &\quad& \text{original equation}\\ x + 6 &= 4x  3x + 1 &\quad& \text{apply the commutative property}\\ x + 6 &= x + 1 &\quad& \text{combine like terms}\\ 6 &= 1 &\quad& \text{subtract }x \text{ from each side} \end {align}\) 
\(\begin {align} 2(5 + x) 1 &= 3x + 9 &\quad& \text {original equation}\\ 10 + 2x 1 &= 3x + 9 &\quad& \text{apply the distributive property}\\ 2x 1 &= 3x  1 &\quad& \text{subtract 10 from each side}\\ 2x &= 3x &\quad& \text{add 1 to each side}\\ 2 &= 3 &\quad& \text{divide each side by } x \end {align} \)
Student Response
For access, consult one of our IM Certified Partners.
Student Facing
Are you ready for more?
 We can’t divide the number 100 by zero because dividing by zero is undefined.
 Instead, try dividing 100 by 10, then 1, then 0.1, then 0.01. What happens as you divide by smaller numbers?
 Now try dividing the number 100 by 10, by 1, by 0.1, 0.01. What is the same and what is different?
 In middle school, you used tape diagrams to represent division. This tape diagram shows that \(6 \div 2 = 3\)
 Draw a tape diagram that shows why \(6 \div \frac12 = 12\).
 Try to draw a tape diagram that represents \(6 \div 0\). Explain why this is so difficult.
Student Response
For access, consult one of our IM Certified Partners.
Anticipated Misconceptions
Some students may point to a step that is valid and mistakenly identify it as an error. For instance, in the first set of steps, they may object to replacing \(+1  3x\) with \(3x+1\), thinking that it should be rearranged to \(+3x1\). Push their reasoning with a simpler example. Ask, for instance, if \(+7  4\) is equivalent to \(+4  7\). Remind students that we can think of \(+7 4\) as \(+7 + (\text4)\) and then apply the commutative property of addition to get \(\text4 + 7\).
If students hypothesize about two equations being equivalent but are not sure how to check if it's actually the case, suggest that a good way to check is by finding the solution to one equation, then checking whether that value is also a solution to the second equation.
Activity Synthesis
Invite students to share what they thought was the problem with Noah’s work. They are likely to say that Noah seems to have performed allowable moves and did them correctly. Then, draw students' attention to the second to last step: \(2x = 3x\). Ask students:
 “Earlier, when looking at \(x + 6 = x + 1\), we reasoned that there’s no value of \(x\) that could make this equation true. Is there a value of \(x\) that could make \(2x = 3x\) true?” (Yes, 0.)
 “If there is a value that can make \(2x=3x\) true, what do we know about the original equation? Is it also true?” (Yes, they are all equivalent equations.) "What is its solution?” (0)
 “So the equations are true up through the third step. The last step is where we have a false statement, after Noah shows division by \(x\). If \(x\) is 0, what might be problem with dividing by \(x\)?” (If \(x\) is 0, then we are dividing both sides by 0, which gives an undefined result.)
Explain that dividing by the variable in the equation is not done because if the solution happens to be 0, it could lead us to thinking that there is no solution while in fact there is (the solution is the number 0).
Revisit the lists of acceptable and unacceptable moves compiled in earlier activities. Add ”dividing by the variable” and ”dividing by 0” to the list of unacceptable moves.
Design Principle(s): Support sensemaking; Cultivate conversation
Lesson Synthesis
Lesson Synthesis
Display the following sets of equations and questions for all to see. Tell students that each set represents an original equation and the first step taken to solve it.
A
\(\begin{align} 5(x3)&=5\\x3&=1 \end{align}\)
B
\(\begin{align} 5x3&=5\\5x&=2 \end{align}\)
C
\(\begin{align} 5(x3)&=5x\\x3&=x \end{align}\)
D
\(\begin{align} (53)x&=5x\\ 53&=5 \end{align}\)
For each set of equations, ask students:
 "What move was made to the original equation to obtain the second equation?"
 "Is the solution to the second equation the same as the solution to the original equation? Why does it stay the same or why does it change?"
Make sure students see that:
 A: The move made is dividing each side by 5. The solution is the same (\(x=4\)) because dividing both sides by the same number keeps the two sides equal.
 B: The move made is adding 3 to the left side and 3 to the right. The solution of the first equation (\(x=\frac85\)) no longer makes the second equation true because the two sides of the equation are no longer equal.
 C: The move made is dividing each side by 5. It is a valid move, but because the original equation does not have a solution (5 times a number, \(5x\), cannot be equal to 5 times a smaller number, \(5(x3)\)), the second equation also does not have a solution (3 less than a number, \(x3\), cannot be equal to that number, \(x\)).
 D: The move made is dividing each side by \(x\). The second equation doesn't have a solution even though the original solution does. The move is problematic because it would eliminate the variable and make us miss the solution, namely \(x=0\).
7.4: Cooldown  If This, Then That (5 minutes)
CoolDown
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Student Lesson Summary
Student Facing
When solving an equation, sometimes we end up with a false equation instead of a solution. Let’s look at two examples.
Example 1: \(\qquad \qquad 4(x+1) = 4x\)
Here are two attempts to solve it.
\(\begin {align} 4(x+1) &= 4x &\hspace{0.1 cm} &\text{original equation}\\ x + 1 &= x &\hspace{0.1 cm} &\text{divide each side by 4} \\ 1 &=0 &\hspace{0.1 cm} &\text{subtract }x \text{ from each side} \end{align}\)
\(\begin {align} 4(x+1) &= 4x &\hspace{0.1 cm} &\text{original equation}\\ 4x + 4 &= 4x &\hspace{0.1 cm} &\text{apply the distributive prop}\\ 4 &= 0 &\hspace{0.1 cm} &\text{subtract }4x \text{ from each side}\end {align}\)
Each attempt shows acceptable moves, but the final equation is a false statement. Why is that?
When solving an equation, we usually start by assuming that there is at least one value that makes the equation true. The equation \(4(x +1) = 4x\) can be interpreted as: 4 groups of \((x+1)\) are equal to 4 groups of \(x\). There are no values of \(x\) that can make this true.
For instance, if \(x=10\), then \(x+1=11\). It's not possible that 4 times 11 is equal to 4 times 10. Likewise, 1.5 is 1 more than 0.5, but 4 groups of 1.5 cannot be equal to 4 groups of 0.5.
Because of this, the moves made to solve the equation would not lead to a solution. The equation \(4(x+1) = 4\) has no solutions.
Example 2: \(\qquad \qquad \qquad 2x  5 = \dfrac {x20}{4}\)
\(\begin{align} 2x  5 &= \dfrac{x20}{4} &\quad &\text{original equation}\\ 8x  20 &= x  20 &\quad &\text{multiply each side by 4}\\ 8x &= x &\quad &\text{add 20 to each side}\\ 8 &= 1 &\quad &\text{divide each side by }x \end {align}\)
Each step in the process seems acceptable, but the last equation is a false statement.
It is not easy to tell from the original equation whether it has a solution, but if we look at the equivalent equation \(8x = x\), we can see that 0 could be a solution. When \(x\) is 0, the equation is \(0 = 0\), which is a true statement. What is going on here?
The last move in the solving process was division by \(x\). Because 0 could be the value of \(x\) and dividing by 0 gives an undefined number, we don't usually divide by the variable we're solving for. Doing this might make us miss a solution, namely \(x=0\).
Here are two ways to solve the equation once we get to \(8x = x\):
\(\begin{align} 8x &= x &\hspace{0.15cm} &\text{}\\ 7x &= 0 &\hspace{0.15cm} &\text{subtract }x \text{ from each side}\\ x&=0 &\hspace{0.15cm} &\text{divide each side by 7} \end {align}\)
\(\begin{align} 8x &= x &\hspace{0.15cm} &\text{}\\ 0 &= \text7x &\hspace{0.15cm} &\text{subtract }8x \text{ from each side}\\ 0 &=x &\hspace{0.15cm} &\text{divide each side by 7} \end {align}\)