Lesson 19

This warm-up activates what students know about the solutions to an inequality and ways to find the solutions. For the first time, students refer to the solutions as the solution set of the inequality. Throughout their work with one-variable inequalities, students will use the terms “solutions” and “solution set” interchangeably.

By now, students are likely to have internalized that a solution to an equation in one variable is a value that makes the equation true. The work here makes it explicit that we can extend this understanding of solution to inequalities in one variable.

Some students may have been taught to solve inequalities the same way we solve equations, with the added rule along the lines of "flip the symbol when dividing or multiplying both sides of an inequality by a negative number." They may or may not have understood why the rule is the way it is. (If a student shares this method, emphasize that they will look at different strategies in the lesson and adopt whichever ways that they can explain or justify.)

If students approach the last question (finding a solution to \(7(3-x)>14\)) by peforming operations directly on the inequality but neglect to reverse the inequality symbol, they would find solutions that result in false statements. Use this opportunity to point out that we may run into problems with this method. It is not essential to discuss why or to suggest better approaches at this point. There will be other opportunities in this lesson to reason about the solutions and to witness the same issue (in the second non-optional activity—Equality and Inequality).

Display a number line for all to see that looks like this:

Tell students that, if needed, they could use the number line to help them in reasoning about the inequalities.

Invite students to share some solutions to the first inequality and explain what it means for a value to be a solution to \(y\le9.2\). Be sure to mention some negative numbers that are solutions. If necessary, show the solution set on a number line. Then, focus the discussion on the second inequality.

Ask each student to mark the one solution they have for \(7(3-x)>14\) on the class number line (from the launch). (Students could draw a point or put a dot sticker, if available, on the number line.) Discuss with students:

• "How did you know that the value you chose is a solution?" (When substituted for \(x\) in the inequality, the value makes a true statement.)
• "What do you notice about all the points that are on the line?" (They are all to the left side of 1.)
• "On the number line, we can see that the solutions are values that are less than 1. All these values form the solution set to the inequality. Is there a way to write the solution set concisely, without using the number line and without writing out all the numbers less than 1?" (We can write \(x<1\)).
• "Does the solution set have anything to do with the solution to the equation \(7(3-x)=14\)?" (The solution to the equation \(x=1\).)
• "Why does the solution set to the inequality \(7(3-x)>14\) involve numbers less than 1?" (The inequality can be taken to say "7 times \(3-x\) is greater than 14." For the inequality to be true, \(3-x\) must be greater than 2. For \(3-x\) to be greater than 2, \(x\) must be less than 1.)

Highlight that we can use a number line to concisely show the solution set to an inequality, but we can also write another inequality that shows the same information.

This activity encourages students to interpret an inequality and its solution set in terms of a situation. A context can help students intuit why the solutions to an inequality form a ray on the number line.

The activity also prompts students to think about the solutions to an inequality in terms of a related equation. Here the situation involves choosing between two options. An equation can be written to represent the two options being equal. The solution to that equation can be seen as a tipping point, on either side of which one option would be better.

Students may recall this way of solving inequalities from middle school, but they may also solve by testing different possible values, or by reasoning about the relationship between quantities in other ways. The work here encourages students to reason quantitatively and abstractly (MP2), and to make sense of problems and persevere in solving them (MP1).

Monitor the strategies students use to find the solutions to \(9(n + 3) < 10(n + 1)\), and identify students using different approaches. Students may:

• Try different values of \(n\) until the inequality is no longer true.
• Try different numbers higher than 12 (based on their work on the first question) and find that, up to 17 students, the cost to go to Orchard B is lower. Beyond 17 students, the cost for Orchard A is lower.
• Solve the equation \(9(n + 3) = 10(n + 1)\) to find the number of students at which the costs for both options would be equal. That number is 17. Then, try a higher or lower number to see which side of the equation has a smaller value.
• Reason about the difference in the cost per student and cost for chaperones. The cost per student at Orchard A ($9) is $1 lower than at Orchard B ($10). But because 3 chaperones are required at Orchard A ($27 for 3 chaperones) and only 1 at Orchard B ($10 for 1 chaperone), the cost for chaperones is $17 higher at Orchard A than at Orchard B. So if 17 students go on the trip, the cost would be the same at both places. If more than 17 students go, Orchard A would be cheaper.

Some students may find the solutions to \(9(n + 3) < 10(n + 1)\) by manipulating the inequality to isolate \(n\). Depending on the operations performed, they may once again end up with an incorrect solution set if they forget to reverse the inequality symbol. (For example, in the final step of solving, they may go from \(\text-1n < \text-17\) to \(n <17\).) If this happens, bring the issue to students' attention during activity synthesis.

Read the first part of the task statement with the class and make sure students understand the given information. 

Arrange students in groups of 2. For the first set of questions, ask one partner to find the cost of going to Orchard A and the other partner to find the cost of going to Orchard B, and then compare the costs. Before students move on to the second set of questions, pause to hear which option works best for 8, 12, and 30 students.

If students struggle to interpret the meaning of the equation \(9(n + 3) = 10(n + 1)\) and of the inequality \(9(n + 3) < 10(n + 1)\), ask them to think about what each side of the equal sign or the inequality symbol represents.

Make sure students understand the meaning of the inequality in context and recognize that there are various ways to find the solutions.

Select previously identified students to share how they found the solution set, in the sequence shown in the Activity Narrative (starting with guessing and checking, and ending with reasoning more structurally). It is not necessary to discuss all the listed strategies, but if the idea of solving a related equation doesn't come up, point it out.

Explain that one way to think about the solutions to the inequality is by thinking about the solution to a related equation. In this context, the solution to \(9(n + 3) = 10(n + 1)\) gives us the number of students at which it would cost the same to go to either orchard. This is a boundary value for \(n\). On one side of the boundary, the cost of Option A would be higher. On the other, it would be lower. We can test a value of that is higher and one that is lower than this boundary value to see which one makes the inequality \(9(n + 3) < 10(n + 1)\) true.

If a student brings up “flipping the symbol when multiplying or dividing by a negative number” as a strategy, invite them to explain why it works. Emphasize that in general it is more helpful and reliable to use reasoning strategies that we understand and can explain. If we use a rule without some idea of how it came about or why it works, we might end up misapplying it (for example, flipping the inequality symbol anytime we see a negative sign, even if we’re simply adding or subtracting). If we forget or misremember the rule, we would be stuck or make errors.

This optional activity gives students another opportunity to make sense of inequalities and their solutions in context. The context helps to reinforce why it makes sense for the solutions to an inequality to be represented by a set of points on one side of a particular value on the number line. It also reiterates the idea that solving a related equation can help us find that boundary value. In the context of part-time jobs presented here, that boundary value is a break-even point at which the two options are equal. On each side of that point, one job is more lucrative than the other.

Students are given an expression that represents the monthly pay for \(h\) hours of work at each job. The expression for the pay at the restaurant is \(7.50h + \frac {50}{4}h\)\(\). Some students may notice that this expression could be rewritten as \(20h\) and use this simpler equivalent expression throughout the activity. Invite students who do this to share their rationale during class discussion. 

Display the situation in the task statement and the equation \(7.50h + \frac {50}{4}h = 18h + 33\). To familiarize students with the quantities in the context, discuss:

• "What does the expression \(7.50h + \frac {50}{4}h\) represent?" (the pay at the restaurant for \(h\) hours of work) "What about \(18h + 33\)?" (the pay at the hotel for \(h\) hours of work)
• "What does the equation tell us about the situation?" (the student earning the same amount from both jobs for \(h\) hours of work)
• "What does the solution mean in this situation?" (the number of hours of work that would enable the student to earn the same amount from either job) 

Consider keeping students in groups of 2. Give them a few minutes of quiet work time, followed by time to discuss their responses with their partner.

Some students might be stymied by the fact that one of the expressions has a fractional coefficient for one of its terms. Remind these students that \(\frac{50}{4}h\) is just a number times \(h\). Rewriting \(\frac{50}{4}\) as a decimal may help students to see this concretely.

Invite students to share how they determined which job pays better for different hours of work. Display the number lines that show the solution set to each inequality.

If time permits, solicit some ideas from students as to why one job—the restaurant job—continues to be the more lucrative option after 16.5 hours.

One way to make sense of this is to compare the hourly rates at the two jobs.

• The pay at the restaurant, \(7.5h + \frac {50}{4}h\) could be rewritten as \(20h\), which tells us the rate of change is $20 per hour. 
• At the hotel, the rate is $18 per hour. The $33 transportation allowance makes the total pay for this job higher up to a point, but then the higher rate of the restaurant job starts to overtake this one. 

If we graph the monthly pay for each job (\(y=20h\) and \(y=18h+33\)), we can see that the two graphs intersect when \(h\) is 16.5. The graph of \(y=20h\) is has a lower \(y\)-value to the left of that point but higher value to the right of it.

Students could explore this way of solving inequalities in one variable (by graphing) in an optional activity later in the lesson.

The purpose of this activity is to further develop the idea that we can solve an inequality by first solving a related equation. Previously, students reasoned about the solutions to an inequality in context. Here, they transition to solving an inequality without a context.

This activity offers another opportunity to point out the trouble with isolating the variable directly on the inequality statement, or with assuming that the solution set can be expressed using the same symbol as in the original inequality. (For example, if in the final step of solving students go from \(\text-24x \leq \text-48\) to \(x \leq 2\), they would end up with the wrong solution set.)

Display a blank number line for all to see. Ask students to share some values on the number line that are and are not solutions. Use different colors or different symbols to mark on the number line the solutions and non-solutions. 

Emphasize that if we solve an inequality by using a related equation, it is important to make sure that the solution to the equation is correct because that solution gives us a boundary from which we could check the solutions to the inequality. If the boundary value is off, we may not be able to correctly find the solution set to the inequality.

This optional activity allows students to visualize an inequality in one variable in another way—by graphing the expressions on each side and comparing the values of the expressions at different values of the variable. Doing so allows them to see the values at which the inequality is satisfied.

This activity works best when each student has access to devices that can run the Desmos applet because students will benefit from seeing the relationship in a dynamic way. If students don't have individual access, projecting the applet would be helpful during the synthesis. (Students can still graph the equations in the activity using the graphing technology available in the classroom.)

Some students may need help parsing the phrase “for what values of \(x\) is the \(y\)-value...” Ask them: “When \(x\) is 0, what is the \(y\)-value in \(y=\text{-} \frac12 x + 6\)? What about in \(y= 4x−3\)? Which \(y\)-value is greater?” Compare the \(y\)-values with another value of \(x\). Then, ask students, “Is there a value of \(x\) that would make the two \(y\)-values equal? What is that \(x\)-value?”

Focus the discussion on how this way of solving an inequality in one variable is like and unlike the strategy of solving a related equation, which students used in an earlier activity. Discuss questions such as:

• “Previously, we saw that we could solve an inequality like this by first solving a related equation: \(\text{-} \frac12 x + 6 = 4x−3\). Is the method of graphing similar to that process in any way?” (It is similar in that we can find the value of \(x\) that makes the two expressions equal.)
• “How is the graphing method different?” (Instead of comparing the values of the expressions by calculation, we can graph \(y=\text{[expression on the left]}\) and \(y=\text{[expression on the right]}\) and see where the two graphs intersect. The intersection tells us the \(x\)-value that produces the same \(y\)-value.)
• “Previously, to find the solutions to \(\text{-} \frac12 x + 6 < 4x−3\), we would test \(x\)-values that are greater and less than the solution to \(\text{-} \frac12 x + 6 = 4x−3\) and see which one would make the inequality true. How is the graphing method similar and how is it different?” (It is similar in that we are still comparing the values of the two expressions. It is different in that the graphs allow us to compare visually and see which graph has a greater \(y\)-value on either side of their intersection.)