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

# Matching Equations To Problems ## Overview

Students match equations such as 3x − 50 = 90 and 3(x − 50) = 90 to real-world and mathematical situations. They identify the steps needed to solve these equations.

# Key Concepts

Students solve equations such as 3x − 50 = 90 by using first the addition property and then the multiplication property of equality.

Students also solve equations such as 3(x − 50) = 90. Equations with parentheses were introduced in the Challenge Problem of Lesson 6. Now, in this lesson, students use two methods to solve the equation. First method: use the multiplication property of equality and then the addition property of equality; second method: use the distributive property to eliminate the parentheses, then use the addition property of equality, and then the multiplication property of equality.

# Goals and Learning Objectives

• Match equations to problems.
• Solve two-step equations.

# Lesson Guide

Have students work in pairs. Have students read the problem and then look at each equation, one by one, to see if it matches the problem situation. For each equation, students enter their response, Yes or No. Conduct a class poll to see student responses. As they see the class responses, students may decide to reconsider their response. Explain that pairs should be ready to defend their responses. When all pairs have completed the activity, discuss the responses.

ELL: Word problems are highly language dependent and may be challenging for ELLs. Word problems require the ability to think abstractly and manipulate concepts through language. Provide ELLs with extra time to process information. Allow them to use the language available to them, either English or their language of origin.

# Mathematics

Help students see how an equation for the problem situation can be built.

• The cost of a pencil is x − 50.
• The pen costs 3 times as much as a pencil.
• Since the pen costs 90 cents, 3(x − 50) = 90. Equation B shows this relationship.
• Equation D is an equivalent equation; 3(x − 50) and 3x − 150 are equivalent because of the distributive property.
• The expression 3x − 50 is not equivalent to 3(x − 50).
• x − 50 ⋅ 3 is not equivalent to 3(x − 50).

Therefore, Equations A and C do not represent the problem situation.

# Interventions

Student does not know how to identify what x stands for.

• Look at the problem. What are you trying to find?

Student writes an equation for the story but the equation does not match any of the choices.

• Try using the properties to write equations equivalent to the one you wrote.

Student writes an equation with parentheses and uses the distributive property incorrectly.

• Did you multiply each term inside the parentheses by the factor outside the parentheses?

# Pencil, Pen, and Notebook

A pencil costs 50 cents less than a notebook.

A pen costs 3 times as much as a pencil.

A pen costs 90 cents.

• Let x = the cost of the notebook. What does the notebook cost?
• Review each equation in the table. Does it represent the situation described above? Select "Yes" or "No" for each.

HANDOUT: Determine the Cost of the Notebook

# Lesson Guide

Discuss the Math Mission. Students will represent situations using equations, and construct each step of the solution process.

## Opening

Represent situations using equations, and construct each step of the solution process.

# Lesson Guide

Have students  work in pairs or groups of three on the card sorting activity. They match each of the six situations to one of the six equations. Point out to students that they may need to write some expressions before they can find the equation that matches the problem. Remind students that it is important that they can explain and justify their choices. Have students present and explain their matches.

Students continue to work in their groups to prepare a presentation of a complete solution for one of the equations from the card sort. Point out that they need to justify each step of the solution. Make sure that every card sort is used by at least one student group. Point out that for the equations with parentheses, there are two methods for solving the equation.

# Mathematical Practices

Mathematical Practice 2: Reason abstractly and quantitatively.

Students make sense of the quantities in the situations when they match the situations to the equations. When they work through the solution to the equation, they are reasoning abstractly. When they have completed a solution process, they can go back to the situation and check that the solution to the equation makes sense in the context of the situation.

# Interventions

Students are matching situations and equations but not explaining their reasoning.

• What value does x represent?
• Ask each group member to explain the choice.

Student is unsure about an alternative solution for an equation with parentheses.

• What can you do to eliminate the parentheses from the equation?
• How do you solve this new, equivalent equation?

In the first card sorting activity, these are the correct pairs:

Situation 1 → 2x + 12 = 60

Situation 2 → 2(x + 3) = 60

Situation 3 → 6(x − 2) = 54

Situation 4 → 2x + 6 = 54

Situation 5 → 6x − 54 = 6

Situation 6 → 2(x + 6) = 54

6(x − 2) = 54

6(x − 2) = 54

6x − 12 = 54                     Distributive property, distribute 6 over x − 2.

6x − 12 + 12 = 54 + 12    Addition property of equality, add 12 to each side.

$6x\cdot \frac{1}{6}=66\cdot \frac{1}{6}$                 Multiplication property of equality, multiply each side by $\frac{1}{6}$.

$x=\frac{66}{6}=11$                     Multiply.

2x + 6 = 54

2x + 6 = 54

2x + 6 − 6 = 54 − 6        Addition property of equality, add −6 to each side (which is the
same as subtracting 6).

$2x\cdot \frac{1}{2}=48\cdot \frac{1}{2}$             Multiplication property of equality, multiply each side by $\frac{1}{2}$.

x = 24                            Multiply.

2(x + 6) = 54

2(x + 6) = 54

2x + 12 = 54                    Distributive property, distribute 2 over x + 6.

2x + 12 + (−12) = 54 + (−12)            Addition property of equality, add −12 to each side.

2x $\frac{1}{2}$ = 42 ⋅ $\frac{1}{2}$                 Multiplication property of equality, multiply each side by $\frac{1}{2}$.

x = 21                                Multiply.

6x − 54 = 6

6x − 54 = 6

6x − 54 + 54 = 6 + 54        Addition property of equality, add 54 to each side.

6x $\frac{1}{6}$ = 60 ⋅ $\frac{1}{6}$                  Multiplication property of equality, multiply each side by $\frac{1}{2}$.

x = 10                                 Multiply.

2x + 12 = 60

2x + 12 = 60

2x + 12 − 12 = 60 − 12      Addition property of equality, add −12 to each side
(which is the same as subtracting 12).

2x $\frac{1}{2}$ = 48 ⋅ $\frac{1}{2}$                 Multiplication property of equality, multiply each side by $\frac{1}{2}$.

x = 24                                Multiply.

2(x + 3) = 60

2(x + 3) = 60

2x + 6 = 60                        Distributive property, distribute 2 over x + 3.

2x + 6 − 6 = 60 − 6            Addition property of equality, add −6 to each side.

2x $\frac{1}{2}$ = 54 ⋅ $\frac{1}{2}$                  Multiplication property of equality, multiply each side by $\frac{1}{2}$.

x = 27                                 Multiply.

# Situations and Equations

Match the situations to the equations.

INTERACTIVE: Matching Situations and Equations

## Hint:

You may need to write equivalent expressions in order to find some of the matching equations.

# Preparing for Ways of Thinking

As students work on the card sort interactive, make notes of students' approaches and support student reasoning. The main challenge for students is to write the equation that matches the situation language and then find an equation that is equivalent to the one they wrote. Look for students who are using the distributive property correctly.

During the Ways of Thinking discussion, alternate between asking students who you noted gave a good explanation, and students who listened to a classmate who provided a good explanation.

As groups work on preparing the solutions for an equation, walk around, watch, and listen. Look for students who can explain each step of the solution process.

Look for students who attempt the Challenge Problem. In the Ways of Thinking discussion, include both correct and incorrect solutions.

ELL: When forming pairs, be aware of your ELLs and ensure that they have a productive learning environment. Different types of partnerships include:

• Pairing them up with English speakers so they can learn language skills.
• Pairing them up with students who are at the same level of language skills, so they can take a more active role and they can work things out together.
• Pairing them up with students whose proficiency level is lower, so they play the role of the “supporter.”

You can also pair them up based on their math proficiency.

SWD: During partner work, monitor student discussions and provide guiding questions from the Intervention section to help students see the mathematics and find their own way to a solution. If a student is struggling with a particular concept, hold an individual conference. If many students are struggling with a concept, pull a small group to discuss the concept.

• Presentations will vary.

# Challenge Problem

• Lengths of the three sides: 5$\frac{1}{2}$ in., 5$\frac{1}{2}$ in., 1$\frac{1}{2}$ in.
• An equation that could be used to solve the problem:
xx + (x − 4 ) = 12$\frac{1}{2}$

3x − 4 = 12$\frac{1}{2}$

3x − 4 + 4 = 12$\frac{1}{2}$ + 4

3x = 16$\frac{1}{2}$

3x ÷ 3 = 16$\frac{1}{2}$ ÷ 3x = 5

the longer sides are each 5$\frac{1}{2}$  inches long

5$\frac{1}{2}$ − 4 = 1$\frac{1}{2}$
the shorter side is 1$\frac{1}{2}$  inches long

# Prepare a Presentation

Select one of the equations from the card sort and solve the equation.

• Justify each step of your solution. One equation has been done for you as an example.
• Prepare a presentation that shows and justifies each step of your solution and explains how each step relates to the situation. # Challenge Problem

An isosceles triangle has a perimeter of 12$\frac{1}{2}$ inches.

The length of the shortest side is 4 inches less than the length of one of the two equal sides.

• What are the lengths of the three sides of the triangle?
• Solve the problem by writing and solving an equation.

# Mathematics

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

• How are using equivalent expressions helpful in the Pen, Pencil, and Notebook activity?
• How are equivalent expressions helpful in the card sort interactive activity?
• How does each step of the equation represent each part of the situation in the card sort interactive?
• How did you figure out what x represents in each situation?
• Does the solution to the equation make sense in terms of the problem?
• What properties can be used to solve the equations?
• Is subtracting the same number from both sides of an equation making use of the addition property of equality? Why?
• Why is dividing both sides of an equation by the same number making use of the multiplication property of equality? [Dividing by a number is the same as multiplying by the reciprocal of the number.]
• How did you attempt to solve the Challenge Problem?

SWD: Students with disabilities may struggle to follow the discussion and concurrently take notes during the Ways of Thinking portion of the lesson. Create scaffolded notes/questions based on the responses for some students. Use as reference and support during the Ways of Thinking portion of the lesson.

# Ways of Thinking: Make Connections

Take notes about the steps and justifications your classmates used to solve the equation they chose.

## Hint:

• How does each step of the equation represent each part of the situation?
• How did you figure out which quantity in the situation x represents?
• Does the solution to the equation make sense in terms of the problem?

# Lesson Guide

Have students work in pairs to read and discuss the Summary of the Math. It may be helpful to have each student explain the solution to his or her partner. Additionally, you may want to give each pair an equation, such as 5(y + 7) = 65.

SWD: Some students may struggle with the task of writing a Summary of the Math from the lesson. Possible supports:

• Prior to writing the summary, have students discuss their ideas with a partner or adult and rehearse what they might write.
• Allow students to map out their ideas in outline form or in a concept web.

# Summary of the Math: Solve Equations

Follow the steps and justifications below.

• Identify variables in the starting equation:

y − 2x − 7 = x + 3

• Add 2x + 7 to both sides by the addition property of equality:

(y −2x − 7) + (2x + 7) = (x + 3) + (2x + 7)

• Rearrange terms using properties of operations:

y + (2x − 2x) + (7 − 7) = (x + 2x) + (3 + 7)

• Combine like terms using the distributive property:

y + x(2 − 2) + 0 = x(1 + 2) + 10

• Simplify both sides to get a new equation:

y = 3x + 10

In this example, 2x + 7 is added to both sides of the equation. But any number could have been added, and the resulting equation would still have been true, as long as the same number was added to both sides. Notice how parentheses are used to keep terms organized, and how the associative property and commutative property of addition are used to rearrange the terms.

## Hint:

Can you explain how to use properties of operations to solve equations?