Lesson 4
Practice Solving Equations and Representing Situations with Equations
4.1: Number Talk: Subtracting From Five (5 minutes)
Warmup
The purpose of this number talk is to have students recall subtraction where regrouping needs to happen in preparation for the problems they will solve in the lesson.
Launch
Display one problem 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
Find the value of each expression mentally.
\(52\)
\(52.1\)
\(52.17\)
\(52\frac78\)
Student Response
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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?”
Design Principle(s): Optimize output (for explanation)
4.2: Row Game: Solving Equations Practice (15 minutes)
Activity
The purpose of this activity is for students to practice solving equations. Some students may use the “do the same to each side” strategy they developed in their work with balanced hangers. Others may use strategies like substituting values until they find a value that makes the equation true, or asking themselves questions like “2 times what is 18?” As students progress through the activity, the equations become more difficult to solve by strategies other than “do the same thing to each side.”
Launch
Display an equation like \(2x=12\) or similar. Ask students to think about the balanced hangers of the last lesson and to recall how that helped us solve equations by doing the same to each side. Tell students that, after obtaining a solution via algebraic means, we end up with a variable on one side of the equal sign and a number on the other, e.g. \(x=6\). We can easily read the solution—in this case 6—from an equation with a letter on one side and a number on the other, and we often write solutions in this way. Tell students that the act of finding an equation's solution is sometimes called solving the equation.
Arrange students in groups of 2, and ensure that everyone understands how the row game works before students start working. Allow students 10 minutes of partner work followed by a wholeclass discussion.
Supports accessibility for: Organization; Attention
Design Principle(s): Support sensemaking
Student Facing
Solve the equations in one column. Your partner will work on the other column.
Check in with your partner after you finish each row. Your answers in each row should be the same. If your answers aren’t the same, work together to find the error and correct it.
column A  column B 

\(18=2x\)  \(36=4x\) 
\(17=x+9\)  \(13=x+5\) 
\(8x=56\)  \(3x=21\) 
\(21=\frac14 x\)  \(28=\frac13 x\) 
\(6x=45\)  \(8x=60\) 
\(x+4\frac56=9\)  \(x+3\frac56=8\) 
\(\frac57x=55\)  \(\frac37x=33\) 
\(\frac15=6x\)  \(\frac13=10x\) 
\(2.17+x=5\)  \(6.17+x=9\) 
\(\frac{20}{3}=\frac{10}{9}x\)  \(\frac{14}{5}=\frac{7}{15}x\) 
\(14.88+x=17.05\)  \(3.91+x=6.08\) 
\(3\frac34x=1\frac14\)  \(\frac75x=\frac{7}{15}\) 
Student Response
For access, consult one of our IM Certified Partners.
Activity Synthesis
Draw students' attention to \(21=\frac14 x\), and ask selected students to explain how they thought about solving this equation. Some may share strategies like “onefourth of what number is 21?” Ideally, one student will say “divide each side by \(\frac14\)” and another will say “multiply each side by 4.” From their studies in earlier units, students should understand that multiplying by 4 has the same result as dividing by \(\frac14\). Next, turn students attention to \(\frac57x=55\) and ask them to describe the two ways to think about solving it. “Divide each side by \(\frac57\)” gets the same result as “Multiply each side by \(\frac75\).”
4.3: Choosing Equations to Match Situations (15 minutes)
Activity
The purpose of this activity is for students to practice matching equations to situations and then solving those equations using their new strategies. Monitor for students who draw diagrams (tape, hanger, or their own creations) that describe the relationships and those who solve the equations by doing the same to each side of one or more equations.
Launch
Allow students 10 minutes of quiet work time, followed by a wholeclass discussion.
Design Principle(s): Support sensemaking
Student Facing
Circle all of the equations that describe each situation. If you get stuck, consider drawing a diagram. Then find the solution for each situation.

Clare has 8 fewer books than Mai. If Mai has 26 books, how many books does Clare have?
 \(26x=8\)
 \(x=26+8\)
 \(x+8=26\)

\(268=x\)
\(x = \text{______}\)

A coach formed teams of 8 from all the players in a soccer league. There are 14 teams. How many players are in the league?
 \(y=14\div8\)
 \(\frac{y}{8}=14\)
 \(\frac18y=14\)

\(y=14\boldcdot 8\)
\(y = \text{______}\)

Kiran scored 223 more points in a computer game than Tyler. If Kiran scored 409 points, how many points did Tyler score?
 \(223=409z\)
 \(409223=z\)
 \(409+223=z\)

\(409=223+z\)
\(z = \text{______}\)

Mai ran 27 miles last week, which was three times as far as Jada ran. How far did Jada run?
 \(3w=27\)
 \(w=\frac13\boldcdot 27\)
 \(w=27\div3\)

\(w=3\boldcdot 27\)
\(w = \text{______}\)
Student Response
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Student Facing
Are you ready for more?
Mai’s mother was 28 when Mai was born. Mai is now 12 years old. In how many years will Mai’s mother be twice Mai’s age? How old will they be then?
Student Response
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Anticipated Misconceptions
Students who continue to focus on key words misidentify the relationship in each situation. Encourage students to express the relationships in their own words and draw diagrams comparing the given quantities. For example, in the situation with Clare and Mai, they can draw a long rectangle representing Mai's books subdivided into two pieces. Filling in the information given in the story will help clear up the relationships; Clare's rectangle is labeled \(x\), and she has 8 fewer books than Mai, so Mai's rectangle is labeled \(x+8\) and also 26. Alternatively, they can show that the piece labeled \(x\) must equal \(268.\)
Activity Synthesis
Invite students to share their strategies for matching equations to the stories and for solving those equations. Include students who drew tape, hanger, or other types of diagrams to help them understand and reason about the relationships. Record the diagrams and strategies and have students compare them. Ask where they see information from the story in the parts of the diagrams and equations.
If no students bring it up, ask if any of the situations have a similar structure.
 The Clare/Mai and Kiran/Tyler situations share a similar structure where both the larger quantity and the difference between the smaller and larger quantities are known while the smaller quantity is unknown. Note that the first relationship is expressed with “fewer” and the second with “more.” This provides an opportunity for students to reason about the quantities, decontextualizing to see the similar structure and then contextualizing to understand the situations and answer questions.
 The soccer teams and Mai/Jada situations share similar structures in that equal parts add to a whole. The two problems differ in which quantities are known and unknown. In the soccer situation, the size of each group (8 players per team) and number of groups (14 teams) are known while the total is unknown. In the Mai/Jada multiplicative comparison situation, a total is known (27 miles) and the number of groups is known (3 times as many) but the size of each group is unknown.
Focusing on structure in this way helps students reason about the relationships between quantities in a situation, rather than focus on the words in the problem as hints to the operations needed in the equations.
For students who solved for the unknown by using the equations, ask which of the chosen equations they decided to solve and why.
Lesson Synthesis
Lesson Synthesis
The end of this lesson is a good place for students to take a moment and reflect on the learning of the past four lessons. Some questions to guide the discussion:
 “Describe some ways to understand how a situation can be represented mathematically.”
 “What have you learned about equations that surprised you?”
 “Share your thoughts about using diagrams to help understand relationships. Where have you seen diagrams used earlier this year? Where were they most helpful to you? Least helpful?”
 “Describe any connections you see between the types of diagrams used in the last four lessons.”
4.4: Cooldown  More Storytime (5 minutes)
CoolDown
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Student Lesson Summary
Student Facing
Writing and solving equations can help us answer questions about situations.
Suppose a scientist has 13.68 liters of acid and needs 16.05 liters for an experiment. How many more liters of acid does she need for the experiment?
 We can represent this situation with the equation:
\(\displaystyle 13.68 + x=16.05\)
 When working with hangers, we saw that the solution can be found by subtracting 13.68 from each side. This gives us some new equations that also represent the situation:
\(\displaystyle x=16.05  13.68\)
\(\displaystyle x=2.37\)
 Finding a solution in this way leads to a variable on one side of the equal sign and a number on the other. We can easily read the solution—in this case, 2.37—from an equation with a letter on one side and a number on the other. We often write solutions in this way.
Let’s say a food pantry takes a 54pound bag of rice and splits it into portions that each weigh \(\frac34\) of a pound. How many portions can they make from this bag?
 We can represent this situation with the equation:
\(\displaystyle \frac34 x = 54\)
 We can find the value of \(x\) by dividing each side by \(\frac34\). This gives us some new equations that represent the same situation:
\(\displaystyle x=54\div \frac34\)
\(\displaystyle x=72\)
 The solution is 72 portions.