Skills and Strategies Lesson 14 CCSS Solutions of Linear Equationsodonovanacademy.org/ourpages/auto/2013/8/16/53929464/1... · 2013. 8. 16. · Part 2: Modeled Instruction Lesson

©Curriculum Associates, LLC Copying is not permitted.L14: Solutions of Linear Equations124

Solutions of Linear EquationsLesson 14 Part 1: Introduction

You’ve learned how to solve linear equations and how to check your solution. In this lesson, you’ll learn that not every linear equation has just one solution. Take a look at this problem.

Jason and his friend Amy are arguing. Jason says that a linear equation always has just one solution. Amy says that some linear equations have more than one solution. Who’s right? Amy asked Jason to explore solutions to the following equation.

2x 1 1 1 x 5 3(x 2 2) 1 7

Explore It

Use the math you already know to solve this problem.

Remember that a solution to an equation is a number that makes the equation true. To check to see if a number is a solution to this equation, replace x with its value.

Is 6 a solution to the equation? Show your work.



Can an equation have more than one solution? Explain. Who is right—Jason or Amy?

Develop Skills and Strategies

CCSS8.EE.C.7a

©Curriculum Associates, LLC Copying is not permitted.125L14: Solutions of Linear Equations

Lesson 14Part 1: Introduction

Find Out More

Look at how you could solve Amy’s equation.

2x 1 1 1 x 5 3(x 2 2) 1 7

First, simplify each side:

Use the distributive property. 2x 1 1 1 x 5 3x 2 6 1 7

Combine like terms. 3x 1 1 5 3x 1 1

Once both sides of an equation are in simplest form, you can say a lot about the solution without actually solving the equation. You can just look at the structure.

Think about how you might solve this equation with pictures. Look at the pan balance below. The left pan represents 3x 1 1. So does the right pan.

x x x x x x

If you take away 3x from both sides, you end up with 1 5 1, a true statement. If you take away 1 from both sides, you end up with 3x 5 3x, a true statement. You can replace x with any number and you will always get a true statement. The pan will remain balanced. This equation has infinitely many solutions.

You’ve seen that a linear equation can have one solution or, in a case like this, infinitely many solutions. You will also see that a linear equation can have no solution.

Reflect

1 Once both sides of an equation are in simplest form, how can you tell if it has infinitely many solutions?

Part 2: Modeled Instruction Lesson 14


Read the problem below. Then explore how to identify when an equation has one solution, infinitely many solutions, or no solution.

Yari and her friends, Alyssa and David, were each given an equation to solve.

Yari: 2(x 1 10) 2 17 5 5 1 2x 2 2

Alyssa: 5x 1 3 2 3x 5 2(x 1 3) 2 5

David: 2(x 2 3) 1 9 5 5 1 x 2 1

Whose equation has one solution? Infinitely many solutions? No solution?

Model It

You can use properties of operations to simplify each side of Yari’s equation.2(x 1 10) 2 17 5 5 1 2x 2 2

2x 1 20 2 17 5 3 1 2x

2x 1 3 5 3 1 2x

Model It

You can use properties of operations to simplify each side of Alyssa’s equation.5x 1 3 2 3x 5 2(x 1 3) 2 5

2x 1 3 5 2x 1 6 2 5

2x 1 3 5 2x 1 1

Model It

You can use properties of operations to simplify each side of David’s equation.2(x 2 3) 1 9 5 5 1 x 2 1

2x 2 6 1 9 5 4 1 x

2x 1 3 5 4 1 x

x x x x

x x x x

x x x

The variable terms are the same on both sides of the equation but the constants are different. There is no value for x that will make the equation true.

The variable terms are different. There is only one value for x that will make the equation true.

The variable terms and the constants are the same on both sides of the equation. No matter what value you choose for x, the equation will always be true.

Part 2: Guided Instruction Lesson 14


Connect It

Now you will use the models to solve this problem.

2 Look at Model It for Yari’s equation. What do you notice about both sides of the equation? What equation do you get if you subtract 2x from both sides of the equation?

3 Look at Model It for Alyssa’s equation. How is it different than Yari’s equation? How is it similar?

4 Look at the pan balance for Alyssa’s equation. Is there any way to balance the pan? Explain. What equation do you get if you subtract 2x from both sides of the equation?

5 Look at the Model It for David’s equation. Are the variable terms on each side of the equation the same or different? Solve David’s equation.

6 Explain how you know when an equation has one solution, no solution, or infinitely many solutions.

Try It

Use what you just learned about equations with one solution, no solution, or infinitely many solutions. Show your work on a separate sheet of paper.

Replace c and d in the equation cx 1 d 5 8x 1 12 with the given values. Explain why the equation has one solution, no solution, or infinitely many solutions.

7 c 5 6 and d 5 34 ___________________ c 5 8 and d 5 6 ____________________8

Part 3: Guided Practice Lesson 14


Student Model

Study the student model below. Then solve problems 9–11.

Michelle looks at the equation 26x 2 30 5 6x 2 30 and says there is no solution. Is she correct? Explain.

Look at how you can show your work.

Solution:

9 Draw lines to match each linear equation to its correct number of solutions.

Show your work.

5(4 2 x) 5 25x 1 20 no solution

25(4 2 x) 5 25x 1 20 infinitely many solutions

5(5 2 x) 5 25x 1 20 one solution

How could you convince Michelle that there is a solution to this equation?

Pair/Share

When the variable terms on both sides of an equation are the same, what does that tell you about the solution(s) to the equation?

Pair/Share

The student solved the equation to find that 0 is a solution.

How can you tell when a linear equation has no solution?

26x 2 30 5 6x 2 30

26x 2 30 2 6x 5 6x 2 30 2 6x

212x 2 30 5 230

212x 2 30 1 30 5 230 1 30

212x 5 0

212x ····· 212 5 0 ···· 212

x 5 0

No; There is one solution to this equation, x 5 0.



10 Write a number in the box so that the equation will have the type of solution(s) shown.

no solution

1 ·· 3 x 1 5 5 1 ·· 3 x 1

infinitely many solutions

1 ·· 3 x 1 5 5 1 ·· 3 x 1

one solution

1 ·· 3 x 1 5 5 x 1 5

11 What is the solution to the equation 3(x 2 4) 5 2(x 2 6)?

A x 5 0

B x 5 1

C There are infinitely many solutions.

D There is no solution.

Brian chose D as the correct answer. How did he get that answer?

How could you explain the correct response to Brian?

Pair/Share

What do you notice about the solution to equations that have the same variable term on each side of the equation?

Pair/Share

How can you simplify this equation to justify the correct answer?

What is the difference between an equation with no solution and an equation with infinitely many solutions?

L14: Solutions of Linear Equations138©Curriculum Associates, LLC Copying is not permitted.

Part 1: Introduction Lesson 14

AT A GLANCEStudents discover that some linear equations have more than one solution.

STEP BY STEP• Tell students that this page models how to test

whether a value is a solution to a given equation.

• Explain that a solution is any value of x that makes an equation true. Ask, Is the value 4 a solution to the equation 2x 1 3 5 7? Explain. [No; 4 is not a solution because 8 1 3 Þ 7.] Ask, Is the value 2 a solution to the equation? [Yes; 2 is a solution because 4 1 3 5 7.]

• Have students read the problem at the top of the page.

• Work through the first three parts of Explore It as a class.

• Point out how the distributive property is used to expand the expression 3(x 2 2) to 3x 2 6.

• Ask students to answer the last question on their own. Then invite volunteers to explain their answer.

SMP Tip: The last question provides an opportunity for students to refine their communication skills (SMP 6). A model response includes using appropriate terminology. As a class, develop a model response. For example: Yes. An equation can have more than one solution. The values 6, 2, and 0 are all solutions to the equation.

Use the distributive property to simplify.

• Write: 1 ··

2 (2x 1 4) on the board.

• Remind students to expand the expression over addition.

1 ··

2 (2x 1 4) 5 1

··

2 (2x) 1 1

··

2 (4) 5 x 1 2

• Point out that the distributive property can also expand over subtraction.

1 ··

2 (2x 2 4) 5 1

··

2 (2x) 2 1

··

2 (4) 5 x 2 2

• Point out that the expression within the parentheses can also be viewed as “2x 1 (24).”

Concept Extension

On the board, write: 2x 1 4 5 14.

• What is the value of x in the equation? Justify your answer.

5, because 2(5) 1 4 5 14, which simplifies to 14 5 14.

• Does any number other than 5 make this equation true? Explain.

Guide students to realize that any other value would make the left side of the equation greater or less than 14.

Replace the right side of the equation with 2(x 1 6) 2 8.

• Make a prediction. Is there one or more than one solution to the new equation? Support your answer.

Elicit opinions from the class. Guide the discussion so students realize that because 2x 1 4 5 2x 1 4, any value could be a solution. If necessary, test different values.

Mathematical Discourse


Solutions of Linear EquationsLesson 14 Part 1: Introduction

You’ve learned how to solve linear equations and how to check your solution. In this lesson, you’ll learn that not every linear equation has just one solution. Take a look at this problem.

Jason and his friend Amy are arguing. Jason says that a linear equation always has just one solution. Amy says that some linear equations have more than one solution. Who’s right? Amy asked Jason to explore solutions to the following equation.

2x 1 1 1 x 5 3(x 2 2) 1 7

Explore It

Use the math you already know to solve this problem.

Remember that a solution to an equation is a number that makes the equation true. To check to see if a number is a solution to this equation, replace x with its value.




Can an equation have more than one solution? Explain. Who is right—Jason or Amy?

Develop Skills and Strategies

CCSS8.EE.C.7a

Yes.

Yes.

Yes.

2(6) 1 1 1 6 5 3(6 2 2) 1 7

12 1 1 1 6 5 3(4) 1 7

19 5 19

2(22) 1 1 1 (22) 5 3(22 2 2) 1 7

24 1 1 1 (22) 5 3(24) 1 7

25 5 212 1 7

25 5 25

2(0) 1 1 1 0 5 3(0 2 2) 1 7

0 1 1 1 0 5 3(22) 1 7

1 5 26 1 7

1 5 1

Yes, 6, 22, and 0 are all solutions to Amy’s equation. She was right.

L14: Solutions of Linear Equations 139©Curriculum Associates, LLC Copying is not permitted.

Part 1: Introduction Lesson 14

AT A GLANCEStudents simplify a linear equation and relate it to an image of the simplified equation on a balance scale.

STEP BY STEP• Read Find Out More as a class.

• Be certain students understand where the distributive property is applied in the first step. That is, 3(x 2 2) 5 3x 2 6.

• In the next step, be sure students understand how to combine like terms. Terms with the same variable raised to the same power with numeric coefficients are like terms.

• Have student pairs answer the Reflect question using appropriate terminology. If necessary, define infinitely many solutions.

Point to the balance scale. If necessary, model how it works. Explain that the pans (areas where terms are placed) are even when both sides of the equation have the same value.

ELL Support

SMP Tip: The balance scale model is a way to visually represent the simplified equation in the problem (SMP 4).

Model equations using a balance scale.

Materials: pencil, paper, colored pencils

• Write the equation 3x 1 4 1 x 5 2(2x 1 2).

• Work as a class to simplify. [4x 1 4 5 4x 1 4]

• Sketch the simplified equation on a balance scale. Show the variable in one color and the constants in another.

• Ask, How does the sketch show that the equation has infinitely many solutions? [If I take away 4 from each side, I have 4x 5 4x. Since x can be any value, there are infinitely many solutions.]

Hands-On Activity

Discuss real-life situations that involve linear equations. Have volunteers share any suggestions they have. As a class, write possible equations to model the situations.

Examples: temperature conversion 1 C 5 F 2 32 ······ 1.8 2 , length 1 ft 5 in. ··· 12 2 , travel (distance 5 rate 3 time), time (hr 5 60 3 min), weight (lb 5 16 3 oz), currency exchange rates, cell phone plans, tips

Real-World Connection


Lesson 14Part 1: Introduction

Find Out More

Look at how you could solve Amy’s equation.

2x 1 1 1 x 5 3(x 2 2) 1 7

First, simplify each side:

Use the distributive property. 2x 1 1 1 x 5 3x 2 6 1 7

Combine like terms. 3x 1 1 5 3x 1 1

Once both sides of an equation are in simplest form, you can say a lot about the solution without actually solving the equation. You can just look at the structure.

Think about how you might solve this equation with pictures. Look at the pan balance below. The left pan represents 3x 1 1. So does the right pan.

x x x x x x

If you take away 3x from both sides, you end up with 1 5 1, a true statement. If you take away 1 from both sides, you end up with 3x 5 3x, a true statement. You can replace x with any number and you will always get a true statement. The pan will remain balanced. This equation has infinitely many solutions.

You’ve seen that a linear equation can have one solution or, in a case like this, infinitely many solutions. You will also see that a linear equation can have no solution.

Reflect

1 Once both sides of an equation are in simplest form, how can you tell if it has infi nitely many solutions?

If the variable terms and the constant on each side of the equation are the same,

the equation has infinitely many solutions.

140 L14: Solutions of Linear Equations©Curriculum Associates, LLC Copying is not permitted.

Lesson 14Part 2: Modeled Instruction

AT A GLANCEStudents explore how to identify an equation with infinitely many solutions, no solution, or only one solution.

STEP BY STEP• Read the problem at the top of the page as a class.

• Direct students’ attention to the small squares on the balance scales. Ask, What does each small square represent? [the constant 11]

• Relate the first Model It equation to the equation students solved in Find Out More on the previous page. If you subtract 3 from both sides, you have 2x 5 2x, which is true. Substituting any value for x will also be true.

• For the second equation, point out that the variable terms are both 2x, but the constants are 3 and 1. No matter what value you use for x, 2x 1 3 Þ 2x 1 1.

• For the third equation, point out that the variable terms are 2x and x. Students may find it easier to understand why this equation can have only one value for x that will make it true. If so, as a class, solve for x. [x 5 1]

You may need to reinforce the difference between a variable term and a constant term. Point out that in the expression 2x 1 20:

• 2x means “a number multiplied by 2.” Since the number can vary (change), 2x is a variable term.

• The value of 20 is always 20, so 20 is a constant. (The word constant means “unchanging or always the same.”)

• On the board, write: 5x 1 3. Have students identify the constant and the variable in the expression.

ELL Support• What do you notice about the balance pans in the

three models? The balance pans in the first and the last models are level, but the ones in the middle model are not.

• If the balance pans are level, what might you conclude about the equation? Do you think you can tell if the equation has one solution or infinitely many solutions?

Guide the discussion so students realize that the equation has the same value on both sides, so it has a solution. However, you cannot tell if there is one or infinitely many solutions by looking at the level of the balance pans alone.

Mathematical Discourse

Part 2: Modeled Instruction Lesson 14


Read the problem below. Then explore how to identify when an equation has one solution, infinitely many solutions, or no solution.

Yari and her friends, Alyssa and David, were each given an equation to solve.

Yari: 2(x 1 10) 2 17 5 5 1 2x 2 2

Alyssa: 5x 1 3 2 3x 5 2(x 1 3) 2 5

David: 2(x 2 3) 1 9 5 5 1 x 2 1

Whose equation has one solution? Infinitely many solutions? No solution?

Model It

You can use properties of operations to simplify each side of Yari’s equation.2(x 1 10) 2 17 5 5 1 2x 2 2

2x 1 20 2 17 5 3 1 2x

2x 1 3 5 3 1 2x

Model It

You can use properties of operations to simplify each side of Alyssa’s equation.5x 1 3 2 3x 5 2(x 1 3) 2 5

2x 1 3 5 2x 1 6 2 5

2x 1 3 5 2x 1 1

Model It

You can use properties of operations to simplify each side of David’s equation.2(x 2 3) 1 9 5 5 1 x 2 1

2x 2 6 1 9 5 4 1 x

2x 1 3 5 4 1 x

x x x x

x x x x

x x x

The variable terms are the same on both sides of the equation but the constants are different. There is no value for x that will make the equation true.

The variable terms are different. There is only one value for x that will make the equation true.

The variable terms and the constants are the same on both sides of the equation. No matter what value you choose for x, the equation will always be true.

L14: Solutions of Linear Equations 141©Curriculum Associates, LLC Copying is not permitted.

Lesson 14Part 2: Guided Instruction

AT A GLANCEStudents revisit the problems on page 126. They answer questions that are designed to reinforce their understanding of the number of solutions a linear equation will have.

STEP BY STEP• Read the first sentence in Connect It with students.

Refer back to the problems and the models on page 126.

• Have students answer problems 2–5 individually. As a class, review the answers. For problem 5, you may want to point out that when the variable terms on each side are different, you can solve the equation for x, which means that there is one solution.

• Have pairs of students work on problem 6. Then have volunteers share their answers.

TRY IT SOLUTIONS 7 Solution: one solution; Students may give the

following explanation: After substituting, the equation is 6x 1 34 5 8x 1 12. Combine the like terms 6x and 8x. Then solve for x. So there is only one value that will make the equation true.

8 Solution: no solution; Students may give the following explanation: The variable terms are the same but the constant terms are different. If you subtract 8x from both sides you have 6 5 12, which is not true. So the equation has no solution.

Make a poster about solutions to equations.

Materials: pencil, poster board or heavy paper

• As a class, create a poster that shows an equation with each of the following types of solutions: infinitely many solutions, no solution, and one solution.

• Write your own equations or use the following equations: 3(x 1 6) 2 2 5 2 1 3x 1 2, 4x 1 8 2 x 5 3(x 1 2) 1 5, and 2(x 2 4) 1 5 5 6 1 3x

• On the poster, include the original three equations, the three equations simplified, and a statement that tells how many solutions each equation has, followed by an explanation.

• Include any other information or drawings that may be helpful.

Hands-On Activity

ERROR ALERT: Students who did not answer the questions correctly may have substituted into the equation incorrectly, or they may have confused the variable terms and the constant. Reviewing the three Model It examples on page 126 may be helpful.

Part 2: Guided Instruction Lesson 14


Connect It

Now you will use the models to solve this problem.

2 Look at Model It for Yari’s equation. What do you notice about both sides of the equation? What equation do you get if you subtract 2x from both sides of the equation?

3 Look at Model It for Alyssa’s equation. How is it diff erent than Yari’s equation? How is it similar?

4 Look at the pan balance for Alyssa’s equation. Is there any way to balance the pan? Explain. What equation do you get if you subtract 2x from both sides of the equation?

5 Look at the Model It for David’s equation. Are the variable terms on each side of the equation the same or diff erent? Solve David’s equation.

6 Explain how you know when an equation has one solution, no solution, or infi nitely many solutions.

Try It

Use what you just learned about equations with one solution, no solution, or infinitely many solutions. Show your work on a separate sheet of paper.

Replace c and d in the equation cx 1 d 5 8x 1 12 with the given values. Explain why the equation has one solution, no solution, or infi nitely many solutions.

7 c 5 6 and d 5 34 ___________________ c 5 8 and d 5 6 ____________________8

The variable terms are the same on both sides in both Yari’s and Alyssa’s

equation. The constant terms are the same in Yari’s equation and different in

Alyssa’s equation.

Both sides are exactly the same. 3 5 3.

No matter what you add or subtract on both sides of the equation, the pan will

always be out of balance. If you subtract 2x from both sides, you get 3 5 1.

different x 5 1

Possible answer: An equation has infinitely many solutions when you get a true

sentence like 3 5 3. An equation has no solution when you get a false statement

like 3 5 1. When the variable terms on both sides are different, the equation will

have one solution.

one solution no solution

142 L14: Solutions of Linear Equations©Curriculum Associates, LLC Copying is not permitted.

Lesson 14Part 3: Guided Practice

AT A GLANCEStudents practice identifying the number of solutions for a given equation.

STEP BY STEP• Ask students to solve the problems individually.

Some students may be able to apply a rule that they have learned. Others may need to use the work in the example as a model.

• Circulate among the class as students work and give appropriate support to students who are struggling. For the second equation in problem 9, you may need to point out that “2 25x” is the same as “1 5x.”

• When students have completed each problem, have them Pair/Share to discuss their solutions.

SOLUTIONS Ex Michelle is not correct. There are different variable

terms on each side, so there is only one solution.

9 Solution: See possible student work above. The first equation has infinitely many solutions (identical variable terms and constants), the second has one solution (different variable terms), and the third has no solution (identical variable terms but different constants. (DOK 2)

10 Solution: First equation: any number but 5. Second equation: 5. Third equation: any number but 1

··

3 .

(DOK 2)

11 Solution: A; There is one solution because there are different variable terms. Solve for x; x 5 0.

Explain to students why the other two answer choices are not correct:

B is not correct because if x 5 1, then 29 5 210, which is not true.

C is not correct because the constants are identical on both sides, but the variable terms are not. (DOK 3)



10 Write a number in the box so that the equation will have the type of solution(s) shown.

no solution

1 ·· 3 x 1 5 5 1 ·· 3 x 1

infi nitely many solutions

1 ·· 3 x 1 5 5 1 ·· 3 x 1

one solution

1 ·· 3 x 1 5 5 x 1 5

11 What is the solution to the equation 3(x 2 4) 5 2(x 2 6)?

A x 5 0

B x 5 1

C There are infinitely many solutions.

D There is no solution.

Brian chose D as the correct answer. How did he get that answer?

How could you explain the correct response to Brian?

Pair/Share

What do you notice about the solution to equations that have the same variable term on each side of the equation?

Pair/Share

How can you simplify this equation to justify the correct answer?

What is the difference between an equation with no solution and an equation with infinitely many solutions?

any number but 5

5

any number but 1 ·· 3

Brian thought that x 5 0 means that there is no solution

to the equation.



Student Model

Study the student model below. Then solve problems 9–11.

Michelle looks at the equation 26x 2 30 5 6x 2 30 and says there is no solution. Is she correct? Explain.

Look at how you can show your work.

Solution:

9 Draw lines to match each linear equation to its correct number of solutions.

Show your work.

5(4 2 x) 5 25x 1 20 no solution

25(4 2 x) 5 25x 1 20 infi nitely many solutions

5(5 2 x) 5 25x 1 20 one solution

How could you convince Michelle that there is a solution to this equation?

Pair/Share

When the variable terms on both sides of an equation are the same, what does that tell you about the solution(s) to the equation?

Pair/Share

The student solved the equation to find that 0 is a solution.

How can you tell when a linear equation has no solution?

26x 2 30 5 6x 2 30

26x 2 30 2 6x 5 6x 2 30 2 6x

212x 2 30 5 230

212x 2 30 1 30 5 230 1 30

212x 5 0

212x ····· 212 5 0 ···· 212

x 5 0

No; There is one solution to this equation, x 5 0.

20 2 5x 5 25x 1 20

220 1 5x 5 25x 1 20

25 2 5x 5 25x 1 20

Differentiated Instruction Lesson 14

L14: Solutions of Linear Equations144©Curriculum Associates, LLC Copying is not permitted.

Assessment and Remediation• Ask students to read each equation carefully and then fill in the blanks.

5(2x 1 6) 5 5(x 1 3) 1 2x 3(6 1 x) 5 2(x 1 12) 1 x___ x 1 ___ 5 ___ x 1 ___ [10, 30, 7, 15] ___ 1 ___ x 5 ___ x 1 ___ [18, 3, 3, 24]

The __________ terms are different. [variable] The _______ terms are the same, but the ________ are different. [variable; constants]

There is _____________ solution. [one] There is ___________ solution. [no]

• For students who are struggling, use the chart below to guide remediation.

• After providing remediation, check students’ understanding. Ask students to predict the number of solutions to the equation 2(3x 1 6) 5 2(x 1 6) 1 4x and explain their reasoning. [The equation simplifies to 6x 1 12 5 6x 1 12, so any value for x will be true. There are an infinite number of solutions.]

• If a student is still having difficulty, use Ready Instruction, Level 8, Lesson 13.

If the error is . . . Students may . . . To remediate . . .

no or infinitely many

not understand the relationship between the variables on each side of the equation.

Have students solve the equation for x. Students should see that because x can be isolated on one side of the equation, there is only one solution.

one or infinitely many

not understand why the same variable but different constants on each side results in no solution.

Have students try to solve the equation. Students should see that while 3x 5 3x is true, 18 5 24 is not true, so there is no solution.

Hands-On Activity Challenge ActivityDevelop a solutions matching game.

Materials: index cards, modified number cube with faces labeled 0, 0, 1, 1, M, and M (M stands for many)

Have student pairs write three equations, each one on an index card; each with a different number of solutions. On the back of the card students write the number of solutions and justify their answer. Encourage them to include at least one equation with negative integers. Partner pairs combine their cards and display them equation-side up. Then each pair takes a turn rolling the number cube and selecting an equation that matches their roll. (Students cannot select their own equation unless no other match is left.) Students review the answer on the back of the card and keep the card if it is a match. (Students can challenge equations if they disagree with the number of solutions stated.) When all the cards are selected, the pair with the greatest number of cards wins.

Use a pan balance to model equations.

Materials: actual or virtual balance scale; two sizes of weights—larger to represent x, smaller to represent the constant 1

Organize students into pairs and provide equations for them to model. As necessary, students use the distributive property, combine like terms, and use operations with integers to simplify the equations so there is one variable term and one constant on each side of the equal sign. Then they model the expression on the balance scale. If the scale balances, have students decide if there is one or an infinite number of values of x that will make the equation true. For each equation modeled, have pairs compare their results and adjust their models as necessary.

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Skills and Strategies Lesson 14 CCSS Solutions of Linear Equationsodonovanacademy.org/ourpages/auto/2013/8/16/53929464/1... · 2013. 8. 16. · Part 2: Modeled Instruction Lesson