Lesson 15
Solving Equations with Rational Numbers
15.1: Number Talk: Opposites and Reciprocals (5 minutes)
Warmup
The purpose of this number talk is to:
 Remind students that the sum of a number and a number of the same magnitude with the opposite sign is zero.
 Remind students that the product of a number and its reciprocal is one.
 Establish common vocabulary for referring to these numerical relationships.
There may not be time for students to share every possible strategy. Consider gathering only one strategy for the equations with one variable, and a few strategies for the equations with two variables, since these have many possible answers, and they serve to generalize the relationship and provide an opportunity to introduce or reintroduce relevant vocabulary.
Launch
Display one equation at a time. (When you get to \(c \boldcdot d = 1\), ensure students understand that \(c\) and \(d\) represent different numbers.) Give students 30 seconds of quiet think time per problem and ask them to give a signal when they have an answer and a strategy. Follow with a wholeclass discussion.
Supports accessibility for: Memory; Organization
Student Facing
The variables \(a\) through \(h\) all represent different numbers. Mentally find numbers that make each equation true.
\(\frac35 \boldcdot \frac53 = a\)
\(7 \boldcdot b = 1\)
\(c \boldcdot d = 1\)
\(\text6 + 6 = e\)
\(11 + f = 0\)
\(g + h = 0\)
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Anticipated Misconceptions
Some students may use the strategy of guess and check for equations with two variables and use inverse operations to solve equations with one variable. Ask these students what they notice about the numbers in the equation and encourage them to find the value of the variable without using inverse operations.
Activity Synthesis
Ask students to share their reasoning for each problem. Record and display the responses for all to see.
If the following ideas do not arise as students share their reasoning, make these ideas explicit:
 The sum of a number and its opposite is 0.
 The product of a number and its reciprocal is 1.
 If you want to find a number that you can add to something and get 0 as a sum, use its opposite.
 If you want to find a number that you can multiply something by and get 1 as a product, use its reciprocal.
To involve more students in the conversation, ask some of the following questions:
 “Who can restate ___’s reasoning 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)
15.2: Match Solutions (10 minutes)
Activity
Students solved equations of the form \(x+p = q\) and \(px=q\) in grade 6, but the equations only involved positive values. This activity bridges their understanding of a solution to an equation as a value that makes the equation true with their understanding of operations involving negative numbers from this unit. This activity builds on the work students have done in this lesson evaluating expressions at different values.
Monitor for students who:
 take an arithmetic approach by substituting in values and evaluating. (Does \(\text2 \boldcdot (\text4.5) = \text9\)?)
 take an algebraic approach by writing an equivalent equation using an inverse operation (If \(\text2 \boldcdot x=\text9\), then \(x=\text9 \div \text2\).)
Launch
Allow students 5 minutes quiet work time followed by whole class discussion.
The digital version has an applet that allows students to see how many correct answers they have at any time.
Supports accessibility for: Memory; Organization
Student Facing
Match each equation to a value that makes it true by dragging the answer to the corresponding equation. Be prepared to explain your reasoning.
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Launch
Allow students 5 minutes quiet work time followed by whole class discussion.
The digital version has an applet that allows students to see how many correct answers they have at any time.
Supports accessibility for: Memory; Organization
Student Facing

Match each equation to its solution.

\(\frac12 x=\text5\)

\(\text2x=\text9\)

\(\text\frac12 x=\frac14\)

\(\text2x=7\)

\(x+\text2 = \text6.5\)

\(\text2+x=\frac12\)


\(x = \text4.5\)

\(x = \text\frac12\)

\(x = \text10\)

\(x = 4.5\)

\(x = 2 \frac12\)

\(x=\text3.5\)
Be prepared to explain your reasoning.
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Activity Synthesis
Select an equation for which there is a student who took an arithmetic approach and a student who took an algebraic approach. Ask students to share their reasoning for why a solution is correct. Sequence arithmetic approaches before algebraic approaches. An example of an arithmetic approach: “I know that 3.5 is the solution to \(\text2x=7\), because I know that \(\text2 \boldcdot (\text3.5)=7\). An example of an algebraic approach: “If \(\text2x=7\), then I know that \(x=7 \div \text2\).) Record their work, side by side, for all to see. If there are no equations for which students took both approaches, present both approaches anyway. Tell students that either approach is valid, but that in the next unit they will see some more complicated equations for which one approach might be simpler than the other.
Design Principle(s): Maximize metaawareness; Support sensemaking
15.3: Trip to the Mountains (10 minutes)
Activity
In this activity, students interpret equations that represent situations (MP2). The purpose is for students to see that equations of the form \(x + p = q\) can be solved by adding the opposite of \(p\) to the equation, regardless of whether \(p\) is positive or negative. Students also see that equations of the form \(px = q\) can be solved by multiplying the equation by the reciprocal of \(p\). Through this work, students see that the structure of equations can be used to reason about a path to a solution (MP7) even when negative values are included or when a variable can represent a negative number.
Launch
Give students 5–6 minutes of quiet work time followed by wholeclass discussion.
Supports accessibility for: Organization; Attention
Design Principle(s): Support sensemaking
Student Facing
The Hiking Club is on a trip to hike up a mountain.

The members increased their elevation 290 feet during their hike this morning. Now they are at an elevation of 450 feet.
 Explain how to find their elevation before the hike.
 Han says the equation \(e + 290 = 450\) describes the situation. What does the variable \(e\) represent?
 Han says that he can rewrite his equation as \(e=450 + \text290\) to solve for \(e\). Compare Han's strategy to your strategy for finding the beginning elevation.

The temperature fell 4 degrees in the last hour. Now it is 21 degrees. Write and solve an equation to find the temperature it was 1 hour ago.

There are 3 times as many students participating in the hiking trip this year than last year. There are 42 students on the trip this year.
 Explain how to find the number of students that came on the hiking trip last year.
 Mai says the equation \(3s=42\) describes the situation. What does the variable \(s\) represent?
 Mai says that she can rewrite her equation as \(s=\frac13 \boldcdot 42\) to solve for \(s\). Compare Mai's strategy to your strategy for finding the number of students on last year’s trip.

The cost of the hiking trip this year is \(\frac23\) of the cost of last year's trip. This year's trip cost \$32. Write and solve an equation to find the cost of last year's trip.
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Student Facing
Are you ready for more?
A number line is shown below. The numbers 0 and 1 are marked on the line, as are two other rational numbers \(a\) and \(b\) .
Decide which of the following numbers are positive and which are negative.
\(a1\)
\(a2\)
\(\textb\)
\(a+b\)
\(ab\)
\(ab+1\)
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Anticipated Misconceptions
Students may be misled by words to add or multiply (or subtract or divide) by the wrong numbers. For example, the word "increased" in question 1 may lead students to simply add the numbers they see, while the words "three times as many" in question 4 may lead students to multiply the numbers in the problem. Encourage students to make sense of the situations by acting them out or using visual diagrams, which will help them understand the actions and relationships in the stories.
Activity Synthesis
Tell students: “We learned four things about the hiking trip in this activity: the students were climbing, the temperature was falling, there were more students this year than last, and the cost of the trip was less this year than last.” Then ask them:
 “Think about how you knew what operation described the rise in elevation, fall in temperature, rise in number of students, and fall in the cost. How did you know whether the situation used adding or multiplying?” (This conversation can highlight the problem with relying on “key words” . For example, when students see “times as many” , they might want to multiply the numbers they see in the problem. Encourage students to make sense of the situations by acting them out or drawing diagrams.)
 “How did you decide how to solve for the unknown quantity?”
 “What are some ways to know that a situation involves negative values?”
15.4: Card Sort: Matching Inverses (10 minutes)
Optional activity
The blackline master is a set of matching cards with fractions and integers. The students first recall the work from the previous section about additive inverses by matching them. They then match multiplicative inverses.
When matching multiplicative inverses, students should now use the fact that division follows the same structure as multiplication to identify that negative numbers require a negative inverse and positive numbers require a positive inverse. Monitor for students who use this step to make an initial sort.
Launch
Remind students of the meaning of additive inverse and multiplicative inverse. If \(x + y = 0\), then \(x\) and \(y\) are additive inverses. If \(x \boldcdot y = 1\), then \(x\) and \(y\) are multiplicative inverses. Display these definitions for all to see and use as a reference while they work through this activity. Ask students for some examples of additive inverse, and add these to the display (for example, 7 and 7, because \(7+\text7=0\)). Ask students for some examples of multiplicative inverses, and add these to the display (for example, \(\frac17\) and 7, because \(\frac17 \boldcdot 7 = 1\)).
If necessary, demonstrate productive ways for partners to communicate during a matching activity. For example, partners take turns identifying a match and explaining why they think it is a match. The other partner either accepts their explanation, or explains why they don't think it's a match. Then they change roles for the next match.
Arrange students in groups of 2. Distribute one set of paper slips per group. Instruct students to first match the numbers that are additive inverses. Then, they will resort the same cards into pairs of multiplicative inverses. Consider asking students to pause their work after matching the additive inverses for discussion before proceeding to match multiplicative inverses.
Design Principle(s): Support sensemaking; Cultivate conversation
Student Facing
Your teacher will give you a set of cards with numbers on them.
 Match numbers with their additive inverses.
 Next, match numbers with their multiplicative inverses.
 What do you notice about the numbers and their inverses?
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Anticipated Misconceptions
Students might need a reminder of the difference between the additive inverse and multiplicative inverse.
Activity Synthesis
The most important thing to recognize is that multiplicative inverses require that the numbers have the same sign in order for the product to be positive, and so negative numbers require a negative multiplicative inverse and positive numbers require a positive inverse. Contrast this with additive inverses, which must have opposite signs in order for their sum to be 0. Select students that used that strategy, as well as some who used calculation, to share their thinking and draw out this conclusion.
Lesson Synthesis
Lesson Synthesis
In this lesson students represented situations with equations and used inverses as a strategy to solve them. Bring the activities together by asking students:
 How can we solve an equation like \(x + (\text 9.2) = 7.5\)? (We can add the opposite of 9.2 to 7.5.)
 How can we solve an equation like \(x \boldcdot (\text 9.2) = 7.5\)? (We can multiply 7.5 by the reciprocal of 9.2.)
 Suppose we know that 60 is \(\frac45\) of a number. What is the difference between writing the equation \(\frac45x=60\) and writing the equation \(x=60\boldcdot \frac54\)? (The first equation describes the situation while the second shows a way to rewrite the equation to solve for the unknown.)
15.5: Cooldown  Hiking Trip (5 minutes)
CoolDown
Cooldowns for this lesson are available at one of our IM Certified Partners
Student Lesson Summary
Student Facing
To solve the equation \(x + 8 = \text5\), we can add the opposite of 8, or 8, to each side:
Because adding the opposite of a number is the same as subtracting that number, we can also think of it as subtracting 8 from each side.
\(\begin{align} x + 8 &= \text5\\ (x+ 8) + \text8&=(\text5)+ \text8\\ x&=\text13 \end{align}\)
We can use the same approach for this equation:
\(\begin{align} \text12 & = t +\text \frac29\\ (\text12)+ \frac29&=\left( t+\text\frac29\right) + \frac29\\\text11\frac79& = t\end{align}\)
To solve the equation \(8x = \text5\), we can multiply each side by the reciprocal of 8, or \(\frac18\):
Because multiplying by the reciprocal of a number is the same as dividing by that number, we can also think of it as dividing by 8.
\(\begin{align} 8x & = \text5\\ \frac18 ( 8x )&= \frac18 (\text5)\\ x&=\text\frac58 \end{align}\)
We can use the same approach for this equation:
\(\begin{align} \text12& =\text\frac29 t\\ \text\frac92\left( \text12\right)&= \text\frac92 \left(\text\frac29t\right) \\ 54& = t\end{align}\)