# 19: Equations and Inequalities

In this chapter, you will explore linear equations, develop a strategy for solving them, and relate them to real-world situations.

## US racial inequality may be as deadly as COVID-19

The COVID-19 pandemic is causing a catastrophic increase in US mortality. How does the scale of this pandemic compare to another US catastrophe: racial inequality? Using demographic models, I estimate how many excess White deaths would raise US White mortality to the best-ever (lowest) US Black level under alternative, plausible assumptions about the age patterning of excess mortality in 2020. I find that 400,000 excess White deaths would be needed to equal the best mortality ever recorded among Blacks. For White mortality in 2020 to reach levels that Blacks experience outside of pandemics, current COVID-19 mortality levels would need to increase by a factor of nearly 6. Moreover, White life expectancy in 2020 will remain higher than Black life expectancy has ever been unless nearly 700,000 excess White deaths occur. Even amid COVID-19, US White mortality is likely to be less than what US Blacks have experienced every year. I argue that, if Black disadvantage operates every year on the scale of Whites’ experience of COVID-19, then so too should the tools we deploy to fight it. Our imagination should not be limited by how accustomed the United States is to profound racial inequality.

The COVID-19 pandemic is likely to kill people in the United States on a scale not seen in a century, since the 1918 flu. That catastrophic flu killed tens of millions worldwide and over half a million in the United States. Yet mortality levels that, for White Americans, were unprecedented nevertheless were lower than mortality for US Blacks in any given (nonpandemic) year. Past research has shown that the infectious mortality experienced by urban Whites in 1918 was lower than the infectious mortality of urban nonwhites in every documented year through 1920 (1). Fig. 1 shows that a similar pattern pertains to Whites’ and Blacks’ age-adjusted total mortality and life expectancy. Whites’ life expectancy in 1918 was far lower than in other twentieth century years—yet higher than Black life expectancy in all years but one between 1900 and 1918. Similarly, Whites’ age-adjusted mortality in 1918 was lower than Black mortality in all years but one from 1900 to 1931. By all major mortality measures—infectious mortality, total mortality, and life expectancy—in the early decades of the twentieth century, Blacks in the United States experienced a scale of death comparable to Whites’ experience of the 1918 flu every year.

US Black and White (A) logged deaths per 100,000 and (B) life expectancy, 1900–2017.

A century later, stark inequalities in survival persist. Will graphs of the early twenty-first century look like graphs of the early twentieth century, with a deadly pandemic causing a spike in mortality for Whites that nevertheless remains lower than the mortality Blacks experience routinely, outside of any pandemic? This question cannot be answered definitively until the final toll of COVID-19 is known. As a framework for answering it, I estimate how many White deaths from COVID-19 would be required for White mortality in 2020 to reach the levels of Black mortality in its best recorded year.

The results provide context for understanding the scale of racial inequality in mortality in the United States. Despite recent scholarly focus on rising White mortality (2), that racial inequality remains extreme. As Fig. 1 underscores, best-ever Black age-adjusted mortality and life expectancy are equivalent to White rates from, respectively, nearly 20 or 30 y earlier. For COVID-19 to raise mortality as much as racial inequality does, it would need to erase two to three decades of mortality progress for Whites.

## Lesson 19

A teacher is choosing between two options for a class field trip to an orchard.

• At Orchard A, admission costs $9 per person and 3 chaperones are required. • At Orchard B, the cost is$ 10 per person, but only 1 chaperone is required.
• At each orchard, the same price applies to both chaperones and students.

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Attribution: Apple tree, by pxhere. Public Domain. pxhere. Source.

Which orchard would be cheaper to visit if the class has:

To help her compare the cost of her two options, the teacher first writes the equation (9(n + 3) = 10(n + 1)) , and then she writes the inequality (9(n + 3) < 10(n + 1)) .

1. What does (n) represent in each statement?
2. In this situation, what does the equation (9(n + 3) = 10(n + 1)) mean?
3. What does the solution to the inequality (9(n + 3) < 10(n + 1)) tell us?

Graph the solution to the inequality on the number line. Be prepared to show or explain your reasoning.

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### 19.3: Part-Time Work

To help pay for his tuition, a college student plans to work in the evenings and on weekends. He has been offered two part-time jobs: working in the guest-services department at a hotel and waiting tables at a popular restaurant.

• The job at the hotel pays $18 an hour and offers$ 33 in transportation allowance per month.
• The job at the restaurant pays $7.50 an hour plus tips. The entire waitstaff typically collects about$ 50 in tips each hour. Tips are divided equally among the 4 waitstaff members who share a shift.

The equation (7.50h + frac <50><4>h = 18h + 33) represents a possible constraint about a situation.

1. Solve the equation and check your solution.
2. Here is a graph on a number line.

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Put a scale on the number line so that the point marked with a circle represents the solution to the equation.

Does one job pay better if:

1. The student works fewer hours than the solution you found earlier? If so, which job?
2. The student works more hours than the solution you found earlier? If so, which job?

Be prepared to explain or show how you know.

Here are two inequalities and two graphs that represent the solutions to the inequalities.

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1. Put the same scale on each number line so that the circle represents the number of hours that you found earlier.
2. Match each inequality with a graph that shows its solution. Be prepared to explain or show how you know.

### 19.4: Equality and Inequality

1. Solve this equation and check your solution: (displaystyle ext-frac<4(x+3)>5 = 4x-12) .
2. Consider the inequality: (displaystyle ext-frac<4(x+3)>5 le 4x-12) .
1. Choose a couple of values less than 2 for (x) . Are they solutions to the inequality?
2. Choose a couple of values greater than 2 for (x) . Are they solutions to the inequality?
3. Choose 2 for (x) . Is it a solution?

Graph the solution to the inequality on the number line.

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Here is a different type of inequality: (x^2 leq 4) .

1. Is 1 a solution to the inequality? Is 3 a solution? How about -3?
2. Describe all solutions to this inequality. (If you like, you can graph the solutions on a number line.)
3. Describe all solutions to the inequality (x^2 geq 9) . Test several numbers to make sure your answer is correct.

### 19.5: More or Less?

Consider the inequality ( ext <->frac12 x + 6 < 4x−3) . Let's look at another way to find its solutions.

Use the provided applet or another graphing technology to graph (y= ext <->frac12 x + 6 ) and (y=4x−3) on the same coordinate plane.

Use your graphs to answer the following questions. If using the applet, the slider might be helpful.

1. Find the values of ( ext-frac <1><2>x+6) and (4x-3) when (x) is 1.
2. What value of (x) makes ( ext-frac <1><2>x+6) and (4x-3) equal?
3. For what values of (x) is ( ext- frac12 x + 6) less than (4x−3) ?
4. For what values of (x) is ( ext- frac12 x + 6) greater than (4x−3) ?

### Summary

The equation (frac12 t = 10) is an equation in one variable. Its solution is any value of (t) that makes the equation true. Only (t=20) meets that requirement, so 20 is the only solution.

The inequality (frac12t >10) is an inequality in one variable. Any value of (t) that makes the inequality true is a solution. For instance, 30 and 48 are both solutions because substituting these values for (t) produces true inequalities. (frac12(30) >10) is true, as is (frac12(48) >10) . Because the inequality has a range of values that make it true, we sometimes refer to all the solutions as the solution set.

One way to find the solutions to an inequality is by reasoning. For example, to find the solution to (2p<8) , we can reason that if 2 times a value is less than 8, then that value must be less than 4. So a solution to (2p<8) is any value of (p) that is less than 4.

Another way to find the solutions to (2p<8) is to solve the related equation (2p=8) . In this case, dividing each side of the equation by 2 gives (p=4) . This point, where (p) is 4, is the boundary of the solution to the inequality.

To find out the range of values that make the inequality true, we can try values less than and greater than 4 in our inequality and see which ones make a true statement.

Let's try some values less than 4:

• If (p=3) , the inequality is (2(3) <8) or (6 < 8) , which is true.
• If (p= ext-1) , the inequality is (2( ext-1) < 8) or ( ext-2 <8 ) , which is also true.

Let's try values greater than 4:

• If (p=5) , the inequality is (2(5)<8) or (10<8) , which is false.
• If (p=12) , the inequality is (2(12) <8) or (24<8) , which is also false.

In general, the inequality is false when (p) is greater than or equal to 4 and true when (p) is less than 4.

We can represent the solution set to an inequality by writing an inequality, (p<4) , or by graphing on a number line. The ray pointing to the left represents all values less than 4.

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## 19.2: Off to an Orchard (20 minutes)

### Activity

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 ( ext-1n < ext-17) to (n <17) .) If this happens, bring the issue to students' attention during activity synthesis.

### Launch

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.

A teacher is choosing between two options for a class field trip to an orchard.

• At Orchard A, admission costs $9 per person and 3 chaperones are required. • At Orchard B, the cost is$ 10 per person, but only 1 chaperone is required.
• At each orchard, the same price applies to both chaperones and students.

Expand Image

Attribution: Apple tree, by pxhere. Public Domain. pxhere. Source.

Which orchard would be cheaper to visit if the class has:

To help her compare the cost of her two options, the teacher first writes the equation (9(n + 3) = 10(n + 1)) , and then she writes the inequality (9(n + 3) < 10(n + 1)) .

1. What does (n) represent in each statement?
2. In this situation, what does the equation (9(n + 3) = 10(n + 1)) mean?
3. What does the solution to the inequality (9(n + 3) < 10(n + 1)) tell us?

Graph the solution to the inequality on the number line. Be prepared to show or explain your reasoning.

Expand Image

## Example 3: Variables on Both Sides and Dividing By a Negative Number

Let's take a look at how our new rule applies to solving equations. Pay special attention to the part where we divide by negative 1.

Our last example revisits how to solve equations and/or inequalities with fractions. I hope you remember the trick.

## 19: Equations and Inequalities

More than ever, students need to engage with mathematical concepts, think quantitatively and analytically, and communicate using mathematics. All these skills are central to a young person’s preparedness to tackle problems that arise at work and in life beyond the classroom. But the reality is that many students are not familiar with basic mathematics concepts and, at school, only practice routine tasks that do not improve their ability to think quantitatively and solve real-life, complex problems.

How can we break this pattern? This report, based on results from PISA 2012, shows that one way forward is to ensure that all students spend more “engaged” time learning core mathematics concepts and solving challenging mathematics tasks. The opportunity to learn mathematics content – the time students spend learning mathematics topics and practising maths tasks at school – can accurately predict mathematics literacy. Differences in students’ familiarity with mathematics concepts explain a substantial share of performance disparities in PISA between socio-economically advantaged and disadvantaged students. Widening access to mathematics content can raise average levels of achievement and, at the same time, reduce inequalities in education and in society at large.

In Unit 6, sixth graders move from expressions to equations and inequalities. They revisit familiar diagrams such as tape diagrams to model equations, and they discover new models such as balances and hanging mobiles. Students investigate what it means to be a solution to an equation or an inequality and how use equations and inequalities to model relationships between quantities. When using an equation or inequality to represent real-world situations, students must decontextualize the situation to represent it using variables and symbols and then recontextualize in order to interpret what their answer means in regard to the situation at hand (MP.2). In this unit, students bring concepts from three domains together: Ratios and Proportions, Number Sense, and Expressions and Equations. They re-visit percentages from Unit 2 and solve percent problems using equations. They study relationships between different quantities and draw on their ratio reasoning where relevant. A note on fluency: solving equations provides a good opportunity for students to continue development of and to demonstrate fluency with decimal operations and fraction division. Several problems involve computing with decimals and dividing by fractions include additional problems in practice for students as needed.

Several prior skills support students in this unit. In fifth grade, students analyzed patterns and relationships when they studied standard 5.OA.3. They also observed what happened when these relationships were plotted on the coordinate plane. In previous sixth-grade units, students studied algebraic and numerical expressions and collections of equivalent ratios. Students draw on all of these concepts and skills in this unit.

There are many future connections to the standards in this unit. In seventh grade, students will deeply investigate proportional relationships in the form $y=rx$ , understanding the value of $r$ as the constant of proportionality. They&rsquoll further investigate the graphs of these equations, and in eighth grade, students will compare across multiple representations of proportional relationships. Students will also become exposed to increasingly more complex equations and inequalities to solve.

Pacing: 17 instructional days (14 lessons, 2 flex days, 1 assessment day)

For guidance on adjusting the pacing for the 2020-2021 school year due to school closures, see our 6th Grade Scope and Sequence Recommended Adjustments.

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## 19: Equations and Inequalities

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## Example 1: Compound Inequalities

Solve and graph the following inequality: 10 < 2x + 8 < 16

Notice that this is a "conjunction" inequality. Conjunction inequalities usually include the word "and".

If we were to read this aloud, we would say, "10 is less than 2x + 8 AND 2x+8 is less than 16". Notice how the word "and" separates this into two inequalities?

There are actually two different ways to solve this inequality. This first method is to separate this into two different inequalities and solve each independently. The second method is to solve both parts at the same time. I will show you both ways and you can decide which is best for you.

### Method 1: Solve Two Inequalities Independently

For this method, we will use the word "and" to write two different inequalities.

Take note of how the expression in the middle is used twice in order to write two different inequalities.