# Lesson 20

Writing and Solving Inequalities in One Variable

## 20.1: Dinner for Drama Club (5 minutes)

### Warm-up

In this warm-up, students practice writing an inequality to represent a constraint, reasoning about its solutions, and interpreting the solutions. The work here engages students in aspects of mathematical modeling (MP4).

To write an inequality, students need to attend carefully to verbal clues so they can appropriately model the situation. The word "budget," for instance, implies that the exact amount given or any amount less than it meets a certain constraint, without explicitly stating this. When thinking about how many pounds of food Kiran can buy, students should also recognize that the answer involves a range, rather than a single value.

Kiran is getting dinner for his drama club on the evening of their final rehearsal. The budget for dinner is $60. Kiran plans to buy some prepared dishes from a supermarket. The prepared dishes are sold by the pound, at$5.29 a pound. He also plans to buy two large bottles of sparkling water at $2.49 each. 1. Represent the constraints in the situation mathematically. If you use variables, specify what each one means. 2. How many pounds of prepared dishes can Kiran buy? Explain or show your reasoning. ### Student Response Teachers with a valid work email address can click here to register or sign in for free access to Student Response. ### Activity Synthesis Make sure students see one or more inequalities that appropriately model the situation. Then, focus the discussion on the solution set. Discuss questions such as: • "What strategy did you use to find the number of pounds of dishes Kiran could buy?" • "Does Kiran have to buy exactly 10.4 pounds of dishes?" (No.) "Can he buy less? Why or why not?" (Yes. He can buy any amount as long as the cost of the food doesn't exceed$55.02. This means he can buy up to 10.4 pounds.)
• "What is the minimum amount he could buy?" (He could buy 0 pounds of food, but it wouldn't make sense if his goal is to feed the club members.)

## 20.2: Gasoline in the Tank (10 minutes)

### Activity

This activity enables students to use inequalities to solve a problem about a situation whose constraints might be unfamiliar and in which multiple quantities are unknown.

To reason about the problem, students need to interpret the descriptions carefully and consider their assumptions about the situation. To make sense of the situation, some students may define additional variables or use diagrams, tables, or other representations. Along the way, they engage in aspects of modeling (MP4).

Monitor for these approaches students may use to find the possible number of hours of mowing:

• Start with an empty tank and think about the hours of mowing if there is 0, 1, 2, 3,... gallons of gasoline in the tank. (If the tank is empty, Han could mow 0 hours. If it has 1 gallon, he could mow $$\frac{1}{0.4}$$ or 2.5 hours. And so on, up to 5 gallons.)
• Write a series of inequalities and equations (for instance, $$c \leq 5$$, where $$c$$ is the capacity of the tank in gallons, $$c = 0.4x$$, and $$0.4x \leq 5$$) and then solve $$0.4x \leq 5$$.
• Start with a full tank and the maximum hours of mowing and work down to an empty tank. (If the tank has 5 gallons, Han could mow $$\frac {5}{0.4}$$ hours. If it has 0 gallons, then Han could mow 0 hours.)

Many students are likely to say that $$x \leq 12.5$$ represents all possible hours of mowing. Look for students who also specify that $$x$$ must be positive or who also write $$x\geq 0$$ (or $$0\leq x \leq 12.5$$) for the solution set. Ask them to share their thinking with the class during discussion.

### Launch

Speaking, Reading: MLR5 Co-Craft Questions. Use this routine to increase awareness of language used to talk about solutions of inequalities. Display only the context of this problem without revealing the questions that follow. Give students 1–2 minutes to write their own mathematical questions about the situation, then invite them to share their questions with a partner. Listen for and amplify any questions that involve possible constraints. Once students compare their questions, reveal the remainder of the task.
Design Principle(s): Maximize meta-awareness; Support sense-making
Representation: Internalize Comprehension. Represent the same information through different modalities by using a diagram. If students are unsure where to begin, suggest that they draw a diagram of a 5-gallon tank and an unknown quantity of gas. Encourage students to label the diagram with their own interpretations of the significant elements. Look for phrases that indicate students are fully internalizing comprehension of the the diagram and exploring how to apply the information mathematically. For example, phrases that lend themselves to translation into mathematical symbols such as “there is at most 5 gallons.”
Supports accessibility for: Conceptual processing; Visual-spatial processing

### Student Facing

Han is about to mow some lawns in his neighborhood. His lawn mower has a 5-gallon fuel tank, but Han is not sure how much gasoline is in the tank.

He knows, however, that the lawn mower uses 0.4 gallon of gasoline per hour of mowing.

What are all the possible values for $$x$$, the number of hours Han can mow without refilling the lawn mower?

Write one or more inequalities to represent your response. Be prepared to explain or show your reasoning.

### Anticipated Misconceptions

For students struggling to express the value of $$x$$ as an inequality, suggest they first try reasoning about the question and finding some possible hours of mowing if Han had, say, 1 gallon or 2 gallons of gasoline.

One likely incorrect answer is $$0 \leq x \leq 2$$, the result of multiplying gallons by 0.4 gallons per hour. If students make this mistake, ask, “About how long can the mower mow with one gallon of gas?” Then, ask if it is reasonable that the mower mows for 2 hours with 5 gallons of gas.

### Activity Synthesis

Select previously identified students to share their responses, in the order as listed in the Activity Narrative. Also invite students who explicitly stated that $$x$$ must be positive to explain why they included that lower boundary in their response.

Emphasize that although it is probably understood from the context that the amount of gasoline in the tank cannot be negative and that the lowest possible number of hours of mowing is 0, it is not apparent from a mathematical statement like $$x\leq12.5$$. Writing $$0\leq x \leq 12.5$$ (or $$x\geq0$$ and $$x\leq 12.5$$) makes this assumption explicit and leaves less room for misinterpretation.

## 20.3: Different Ways of Solving (20 minutes)

### Activity

This activity highlights some ways to decide whether the solution set to an inequality is greater than or less than a particular boundary value (identified by solving a related equation).

Students are presented with two approaches of solving an inequality. Both characters (Priya and Andre) took similar steps to solve the related equation, which gave them the same boundary value. But they took different paths to decide the direction of the inequality symbol.

One solution path involves testing a value on both sides of the boundary value, and another involves analyzing the structure of the inequality. The former is more familiar given previous work. The latter is less familiar and thus encourages students to make sense of problems and persevere in solving them (MP1). It is also a great opportunity for students to practice looking for and making use of structure (MP7).

Expect students to need some support in reasoning structurally, as it is something that takes time and practice to develop.

### Launch

Arrange students in groups of 2. Give students a few minutes of quiet time to make sense of what Andre and Priya have done, and then time to discuss their thinking with their partner. Pause for a class discussion before students move to the second set of questions and try to solve inequalities using Andre and Priya's methods.

Select students to share their analyses of Andre and Priya's work. Make sure students can follow Priya's reasoning and understand how Priya decided on $$x<3$$ by focusing her comparison on $$4x + 3$$ and $$18-x$$.

Representation: Develop Language and Symbols. Display or provide copies of Priya’s work and Andre’s work on separate displays. Write Priya and Andre’s descriptive phrases (as given on the student-facing materials) in contrasting colors/texts, so students can more easily sort and process information. Some students may benefit from time to read and interpret each strategy one at a time.
Supports accessibility for: Conceptual processing; Memory

### Student Facing

Andre and Priya used different strategies to solve the following inequality but reached the same solution.

$$2(2x + 1.5) < 18 - x$$

1. Make sense of each strategy until you can explain what each student has done.

Andre

\displaystyle \begin {align} 2 (2x + 1.5) &=18 - x\\ 4x + 3 &= 18 - x\\ 4x -15 &= \text-x\\ \text -15 &= \text-5x\\ 3& = x \end{align}

Testing to see if $$x = 4$$ is a solution:

\displaystyle \begin {align} 2(2 \cdot 4 + 1.5) &< 18 - 4\\ 2(9.5) &<14\\ 19 &<14 \end{align}

The inequality is false, so 4 is not a solution. If a number greater than 3 is not a solution, the solution must be less than 3, or $$3 > x$$.

Priya

\displaystyle \begin {align}2 (2x + 1.5) &= 18 - x\\4x + 3 &= 18 - x\\ 5x + 3 &= 18\\ 5x &= 15\\ x &= 3 \end{align}

In $$4x + 3 = 18 - x$$, there is $$4x$$ on the left and $$\text-x$$ on the right.

If $$x$$ is a negative number, $$4x+3$$ could be positive or negative, but $$18-x$$ will always be positive.

For $$4x+3<18-x$$ to be true, $$x$$ must include negative numbers or $$x$$ must be less than 3.

2. Here are four inequalities.

1. $$\frac15 p > \text-10$$
2. $$4(x+7) \leq \text 4(2x + 8)$$
3. $$\text-9n <36$$
4. $$\frac {c}{3} < \text-2(c-7)$$

Work with a partner to decide on at least two inequalities to solve. Solve one inequality using Andre's strategy (by testing values on either side the given solution), while your partner uses Priya's strategy (by reasoning about the parts of the inequality). Switch strategies for the other inequality.

### Student Facing

#### Are you ready for more?

Using positive integers between 1 and 9 and each positive integer at most once, fill in values to get two constraints so that $$x=7$$ is the only integer that will satisfy both constraints at the same time.

$$\boxed{\phantom{3}}x+\boxed{\phantom{3}}<\boxed{\phantom{3}}x+\boxed{\phantom{3}}$$

$$\boxed{\phantom{3}}x+\boxed{\phantom{3}}>\boxed{\phantom{3}}x+\boxed{\phantom{3}}$$

### Anticipated Misconceptions

Students who perform procedural steps on the inequality may find incorrect answers. For instance, in the third inequality, they may divide each side by -9 and arrive at the incorrect solution $$x < \text-4$$. Encourage these students to check their work by substituting numbers into the original inequality.

### Activity Synthesis

Students should recognize that the solution set for each inequality should be the same regardless of the reasoning method used.

Select as many students as time permits to share how they used a strategy similar to Priya's to determine the solution set of each inequality. There may be more than one way to reason structurally about a solution set. Invite students who reason in different ways to share their thinking. Record and display their thinking for all to see.

Representing, Conversing: MLR7 Compare and Connect. Use this routine to prepare students for the whole-class discussion. At the appropriate time, invite students to create a visual display showing the two inequalities they selected to solve. Students should consider how to display their work so that another student can interpret what they see. Encourage students to add notes or details to their displays to help communicate their thinking. Begin the whole-class discussion by selecting and arranging 2–4 displays for all to see. Give students 2–3 minutes of quiet think time to interpret the displays before inviting the authors to present their work.
Design Principle(s): Optimize output; Cultivate conversation

## 20.4: Matching Inequalities and Solutions (15 minutes)

### Optional activity

This optional activity gives students additional practice in reasoning about the solutions to inequalities without a context. Students can match the inequalities and solutions in a variety of ways—by testing different values, by solving a related equation and then testing values on either side of that solution, by reasoning about the parts of an inequality or its structure, or by graphing each side of an inequality as a linear function.

For all of the inequalities, once students find the boundary value by solving a related equation, the range of the solutions can be determined by analyzing the structure of the inequality. Monitor for students who do so and ask them to share their reasoning during class discussion later.

As students work, also make note of any common challenges or errors so they could be addressed.

### Student Facing

Match each inequality to a graph that represents its solutions. Be prepared to explain or show your reasoning.

1. $$6x \leq 3x$$
2. $$\frac14 x > \text-\frac12$$
3. $$5x + 4 \geq 7x$$
4. $$8x - 2< -4(x-1)$$
5. $$\dfrac{4x-1}{3} > \text-1$$
6. $$\frac{12}{5}-\frac {x}{5} \leq x$$

### Activity Synthesis

Select students who used different strategies—especially those who made use of the structure of the inequalities—to share their thinking. If no students found solution sets by thinking about the features of the inequalities, demonstrate the reasoning process with one or two examples. For instance:

• $$6x \leq 3x$$: After finding $$x = 0$$ as the solution to $$6x=3x$$, we can reason that for $$6x$$ to be less than $$3x$$, $$x$$ must include negative numbers, so the solution must be $$x \leq 0$$.
• $$\dfrac{4x-1}{3}>\text-1$$: After finding $$x = \text- \frac12$$ as the solution to the related equation $$\dfrac{4x-1}{3}=\text-1$$, we can reason that as $$x$$ gets smaller, $$\dfrac{4x-1}{3}$$ is also going to get smaller. For that expression to be greater than -1, $$x$$ will have be to greater than $$\text-\frac12$$

If time permits, ask students to choose a different inequality and try reasoning this way about its solution.

## Lesson Synthesis

### Lesson Synthesis

Display the following inequalities. Remind students that they are examples of inequalities written and solved in the lesson.

$$5.29p+4.98 \leq 60$$

$$0.4x \leq 5$$ and $$x \geq0$$

$$2(2x + 1.5) < 18-x$$

$$\frac15 x > \text-10$$

Next, tell students that they will see some statements about writing and solving inequalities. Their job is to decide whether they agree or disagree wtih each statement and be prepared to defend their response. (One way to collect their responses is by asking them to give a discrete hand signal.)

Display the following statements—one at a time—for all to see. After students indicate their agreement or disagreement, select a student from each camp to explain their reasoning. Then invite others who are not convinced by the reasoning to offer a counterexample or an alternative view.

• We can usually represent the constraints in a situation with a single inequality. (Disagree. Sometimes multiple inequalities are needed, depending on what's happening in the situation.)
• The only way to check the solutions to an inequality is to see if they make sense in a situation. (Disagree. We can also check by substituting some values in the solution set back into the inequality to see if they make the inequality a true statement.)
• The only way to find the solutions to an inequality in one variable is by testing different values for the variable and seeing which ones work. (Disagree. We can also reason about the solutions in context, solve a related equation and test a higher and lower value, or use the structure of the inequality.)
• To express the solutions to an inequality in one variable, we always use the same inequality symbol as in the original inequality. (Disagree. There are different ways to express a solution set. The symbol we use would depend on how we reason about the solutions. For example, $$x < 3$$ or $$3 > x$$ both represent the solution set to $$2(2x + 1.5) < 18-x$$.)

Make sure students see that there are reasons for disagreeing with each of these statements and can articulate some of the reasons.

## Student Lesson Summary

### Student Facing

Writing and solving inequalities can help us make sense of the constraints in a situation and solve problems. Let's look at an example.

Clare would like to buy a video game that costs \$130. She has saved \$48 so far and plans on saving \$5 of her allowance each week. How many weeks, $$w$$, will it be until she has enough money to buy the game? To represent the constraints, we can write $$48 + 5w \geq 130$$. Let’s reason about the solutions: • Because Clare has \$48 already and needs to have at least \$130 to afford the game, she needs to save at least \$82 more.
• If she saves \$5 each week, it will take at least $$\frac{82}{5}$$ weeks to reach \$82.
• $$\frac{82}{5}$$ is 16.4. Any time shorter than 16.4 weeks won't allow her to save enough.
• Assuming she saves \\$5 at the end of each week (instead of saving smaller amounts throughout a week), it will be at least 17 weeks before she can afford the game.

We can also solve by writing and solving a related equation to find the boundary value for $$w$$, and then determine whether the solutions are less than or greater than that value.

\begin {align} 48 + 5w &= 130\\ 5w & = 82\\ w &=\frac{82}{5} \\w&=16.4 \end{align}

• Substituting 16.4 for $$w$$ in the original inequality gives a true statement. (When $$w=16.4$$, we get $$130 \geq 130$$.)
• Substituting a value greater than 16.4 for $$w$$ also gives a true statement. (When $$w = 17$$, we get $$133\geq130$$.)
• Substituting a value less than 16.4 for $$w$$ gives a false statement. (When $$w=16$$, we get $$128\geq130$$.)
• The solution set is therefore $$w \geq 16.4$$

Sometimes the structure of an inequality can help us see whether the solutions are less than or greater than a boundary value. For example, to find the solutions to $$3x > 8x$$, we can solve the equation $$3x = 8x$$, which gives us $$x = 0$$. Then, instead of testing values on either side of 0, we could reason as follows about the inequality:

• If $$x$$ is a positive value, then $$3x$$ would be less than $$8x$$.
• For $$3x$$ to be greater than $$8x$$, $$x$$ must include negative values.
• For the solutions to include negative values, they must be less than 0, so the solution set would be $$x < 0$$.