Subject:
Algebra
Material Type:
Lesson Plan
Level:
Middle School
7
Provider:
Pearson
Tags:
7th Grade Mathematics, Equations, Problem Solving
Language:
English
Media Formats:
Text/HTML

# Using Arithmetic & Equations To Solve Problems ## Overview

Students extend what they learned about solving equations in Grade 6. They learn to solve equations that require them to use both the addition and the multiplication properties of equality. They use what they know about solving equations such as 2x = 6 and x + 3 = 7 to solve equations such as 2x + 3 = 8. They connect solving problems using arithmetic to solving problems using equations. They solve equations containing both positive and negative rational numbers.

# Key Concepts

• Addition property of equality: If a = b, then a + c = b + c.
• Multiplication property of equality: If a = b, then ac = bc.
• For any equation, add or subtract the same value from both sides of the equation and the equation will still be true.
• For any equation, multiply or divide both sides of the equation by the same value and the equation will still be true.
• In this lesson, students use both properties to solve equations. They then solve equations that contain both positive and negative rational numbers.

# Goals and Learning Objectives

• Solve equations using both the addition and multiplication properties of equality.
• Relate solving problems using arithmetic to solving problems using equations.
• Solve equations containing both positive and negative rational numbers.

# Lesson Guide

Have students read about the properties of equality. Ask questions such as the following to make sure students understand the properties:

• If x + 7 = 9, what would you do to solve for x? (Add –7 to both sides of the equation.) What property of equality are you using? (addition property of equality)
• If 5z = 20, what would you do to solve for z? (Multiply both sides by 15.) What property of equality are you using? (multiplication property of equality)

ELL: If students are struggling to get started, ask these questions to ELLs. Ensure that your pace is adequate and you are providing ample wait-time to allow for a thoughtful response. Present the questions in writing and allow students to use dictionaries before responding, if there are words in your questions that they don't understand.

# Properties of Equality

Review the properties of equality.

Addition property of equality: If a = b, then a + c = b + c.

Multiplication property of equality: If a = b, then ac = bc.

So, for any equation that is true, you can do the following:

• Add the same value to both sides or subtract the same value from both sides, and the equation will still be true.
• Multiply or divide both sides by the same value, and the equation will still be true.

# Lesson Guide

Have students work in pairs to read the information about the LaRonde mine. Have each pair predict how deep a miner could go before the temperature gets dangerously warm.

• What temperature in degrees Fahrenheit would be dangerously warm?
• Does your strategy make sense?

# It's Hot Down Here!

The LaRonde mine in Quebec, Canada, is the deepest mine in North America. It is currently about 2.25 km (or almost 7,400 ft) deep. When it is completed, it will be about 3 km (or almost 10,000 ft) deep.

As the mine gets deeper, it gets closer to the center of Earth; so, the temperature in the mine increases. For every 1,000 ft of depth, the temperature rises about 5ºF.

Without a cooling system, miners would be unable to go very far down into the mine without becoming overheated.

• Predict how deep you think a miner could go down into the mine before getting dangerously warm.

# Lesson Guide

Discuss the Math Mission. Students will use the addition property of equality and the multiplication property of equality to solve equations.

## Opening

Use the addition property of equality and the multiplication property of equality to solve equations.

# Lesson Guide

Have students work in pairs. Ask questions such as the following, to make sure that students understand the problems:

• As you go down in the mine, what happens to the temperature?
• How does the temperature at 3,000 m compare to the temperature at 100 m?
• Can you use the properties of equality to solve the equation?

# Interventions

Student does not know how to begin to answer the questions about temperature in the mine.

• What is the temperature at ground level?
• Will the temperature in the mine be greater or less than 15°C?
• If the temperature rises 0.010°C at a depth of 1 m, how much will it rise at a depth of 2 m? 3 m? 10 m? 100 m?

• (100 ⋅ 0.010) + 15 = 16°C
• (3,000 ⋅ 0.010) + 15 = 45°C
• 1,100 m
Possible methods:
• Start by finding the change in temperature: 11°C. Then, since each degree of change represents 100 m of descent, multiply 11 by 100.
• 26 − 15 = 11
11 ÷ 0.010 = 1,100

# How Hot?

Suppose the temperature of a mine at ground level is 15°C, and for every 1 m of depth the temperature rises 0.010°C.

• What will the temperature be at a depth of 100 m?
• What will the temperature be at a depth of 3,000 m?
• If the temperature is 26°C, what is the depth in meters? ## Hint:

• What happens to the temperature for each 1 meter of increasing depth?
• Will the temperature at a depth of 3,000 meters be greater or less than the temperature at a depth of 100 meters?
• To find the depth based on the temperature, think about how much the temperature increased compared to the ground-level temperature.

# Lesson Guide

Have students work in pairs to write an equation and solve it.

# Mathematical Practices

Mathematical Practice 1: Make sense of problems and persevere in solving them.

Since students have the opportunity to solve the same problem using both arithmetic and algebra, they have the opportunity to compare the two methods and the two results and work to reconcile them. If pairs of students are struggling because their results do not match, allow them time to figure out why.

# Interventions

Student is not sure how to begin to solve the equation to determine the depth in the mine.

• Your goal is to isolate the variable on one side; that is, get d by itself on one side of the equation.
• Try adding or subtracting the same number on both sides. What number should you use?

Student does not know how to write the equation to find depth when the temperature is 40 °C.

• Is there an equation that relates depth and temperature?
• Which number represents temperature?
• What number should you replace with 40?

• To solve for d: $26=\frac{1}{100}d+15$

$26-15=\frac{1}{100}d+15-15$    Addition property of equality

$11=\frac{1}{100}d$
$11\cdot 100=\frac{1}{100}d\cdot 100$         Multiplication property of equality
$1,100=d$

Answer: 1,100 m • $40=\frac{1}{100}d+15$

To solve for d:

$40=\frac{1}{100}d+15$

$40-15=\frac{1}{100}d+15-15$     Addition property of equality

$25=\frac{1}{100}d$

$25\cdot 100=\frac{1}{100}d\cdot 100$             Multiplication property of equality

$2,500=d$

• Check: $\begin{array}{c}40=\frac{1}{100}\left(2500\right)+15\\ 40=25+15\\ 40=40\end{array}$

# How Deep?

You can use the following equation, in which d = depth in meters, to determine the depth in meters when the temperature is 26°C:

$26=\frac{1}{100}d+15$, where d = depth in meters

• Show how you can use the properties of equality to solve this equation.
• Write an equation that you can use to find the depth in meters when the temperature is 40°C.
• Solve the equation.
• Use substitution to check your solution. ## Hint:

• To solve an equation like x + 5 = 8, use the addition property of equality.
• To solve an equation like 1/ 2 x = 8, use the multiplication property of equality.
• Think about how this problem compares to the problem you just solved.

# Preparing for Ways of Thinking

Look for these types of responses to share during the Ways of Thinking discussion:

• Students who solve the problems in different ways (make note of students who use tables or diagrams to show the change in temperature in terms of the descent)
• Students who apply the properties of equality to solve the equations (notice the ways students apply the properties)
• Students who have difficulty working with fractions

SWD: Students with disabilities may struggle to determine salient information in lessons. Preview the goals with students to support saliency determination as they move through the instruction and tasks.

# Challenge Problem

• 59°F is equivalent to 15°C.

$\begin{array}{c}15=\frac{5}{9}\left(F-32\right)\\ \frac{9}{5}\left(15\right)=\frac{5}{9}\left(F-32\right)×\frac{9}{5}\\ 27=F-32\\ 27+32=F-32+32\\ 59=F\end{array}$

# Prepare a Presentation

Explain the situation at LaRonde mine from a mathematical viewpoint. Use your work to support your explanation.

# Challenge Problem

The relationship between temperature in degrees Celsius and degrees Fahrenheit is represented by this formula:

$C=\frac{5}{9}\left(F-32\right)$

where:
C = degrees Celsius
F = degrees Fahrenheit

• Use this formula to determine what temperature in degrees Fahrenheit is equivalent to 15°C.

# Mathematics

Facilitate the discussion to help students understand the mathematics of the lesson informally. Ask questions such as the following:

• What happens to the temperature for each meter of depth?
• How did you determine the temperature at a depth?
• How did you use the properties of equality to solve the equation to determine the depth? Is there more than one way to solve the equation?
• Explain the equation that you used to find the depth when the temperature is 40°C.
• How do you use the formula to determine the temperature in Celsius given the temperature in Fahrenheit?
• Can you compare solving the problem by using arithmetic to solving the problem by solving an equation? What is true about the answer? Which do you prefer? Why?

# Ways of Thinking: Make Connections

Take notes about your classmates’ work with solving equations using properties of equality.

## Hint:

• As the mine gets deeper, why do you add the change in temperature to the ground-level temperature rather than subtract it?
• Why did you replace 26 with 40 in the equation?

# Lesson Guide

Have students work individually to apply what they have learned about using the properties of equality to solve equations. Students should show each step and indicate which properties they are using.

# Mathematics

A single property of equality is used to solve equations b and d; the other equations (a, c, e, f, g, and h) require both properties. Students now work with equations that contain both positive and negative rational numbers. As students complete these exercises, look for students who may be struggling with rational number operations. Pair those students with students who have been successful with these concepts.

1. x = 2
2. $x=5\frac{3}{5}$
3. x = −1
4. x = −9
5. $x=13\frac{1}{2}$
6. $x=-2\frac{2}{7}$
7. x = −2
8. $x=-1\frac{1}{6}$

# Use Properties of Equality

Solve for x using the properties of equality. Use substitution to check each solution.

1. 4x + 4 = 12
2. −5x = −28
3. 2x + 7 = 5
4. $\frac{2}{3}x=-6$
5. 2x + 8 = 35
6. −7x − 6 = 10
7. −4x + 6 = 14
8. $-2x+\frac{2}{3}=3$

# Lesson Guide

Have each student write a summary of the math in this lesson, then write a class summary. When done, if you think the summary is helpful, share it with the class.

# A Possible Summary

You can solve an equation like x + 5 = 13 by using the addition property of equality. You add –5 to both sides of the equation to get x + 5 + (−5) = 13 + (−5). Then you simplify both sides and get  x by itself on one side: x = 8. x is by itself on one side. The solution is 8.

You can solve an equation like −2x = 13 by using the multiplication property of equality. You multiply both sides of the equation by $-\frac{1}{2}$ to get $-\frac{1}{2}$(−2x)=$-\frac{1}{2}$ ⋅ 13. Then you simplify both sides and get x = −6$\frac{1}{2}$. The solution is −6$\frac{1}{2}$.

You can solve an equation like −2x + 5 = 13 by using both the addition and multiplication properties of equality. You add –5 to both sides of the equation to get −2x + 5 + (−5) = 13 + (−5). Then you simplify both sides to get −2x = 8. Next, you multiply both sides of the equation by −$\frac{1}{2}$ to get −$\frac{1}{2}$(−2x) = −$\frac{1}{2}$ ⋅ 8. You then simplify both sides and get x = −4. The solution is –4.

# Summary of the Math: Solve Equations Using Properties of Equality

Write a summary about using the properties of equality to solve equations.

## Hint:

• Do you define the addition property of equality and the multiplication property of equality?
• Do you explain how you can use these properties to solve equations, and provide an example? (You might try using the equation –2 x + 5 = 13 as an example.)

# Lesson Guide

Have each student write a brief reflection before the end of class. Review the reflections to determine the level of students' understanding of the properties of equality.

ELL: Asking students to reflect on their learning provides opportunities for ELLs to develop literacy in English and proficiency in mathematics. Make sure students use both academic and specialized mathematical language when reflecting on their learning at the end of each session.

# Reflection

Write a reflection about the ideas discussed in class today. Use the sentence starter below if you find it to be helpful.

One thing I know about the properties of equality is …