Simplifying and Solving Equations



Objective To simplify and solve equations.

www.everydaymathonline.com

ePresentations

eToolkit

Algorithms Practice

EM Facts Workshop Game™

Teaching the Lesson

Family Letters

Assessment Management

Common Core State Standards

Ongoing Learning & Practice

Key Concepts and Skills

Applying the Distributive Property

• Convert between fractions and decimals. 

Math Journal 2, p. 337 Students use the distributive property to solve number stories.

[Number and Numeration Goal 5]

• Add and subtract decimals and signed numbers.  [Operations and Computation Goal 1]

• Add and subtract fractions.  [Operations and Computation Goal 3]

• Use a method to solve equations.  [Patterns, Functions, and Algebra Goal 2]

• Use distributive strategies to simplify algebraic expressions.  [Patterns, Functions, and Algebra Goal 4]

Curriculum Focal Points

Math Boxes 9 5 

Math Journal 2, p. 340 Geometry Template Students practice and maintain skills through Math Box problems.

Study Link 9 5 

Math Masters, p. 295 Students practice and maintain skills through Study Link activities.

• Use inverse operations and properties of equality to find equivalent equations.  [Patterns, Functions, and Algebra Goal 4]

Interactive Teacher’s Lesson Guide

Differentiation Options READINESS

Revisiting Pan-Balance Problems Math Masters, p. 296 Students review a systematic method for solving equations. ENRICHMENT

Generating Unsimplified Equations to Find Equivalent Names Math Masters, pp. 297A and 297B Students use the distributive property to generate equivalent names for numbers. ENRICHMENT

Writing and Solving Equations Math Masters, p. 297 Students translate word sentences into equations and solve them.

Key Activities Students simplify equations by eliminating parentheses and combining like terms. They solve the simplified equations using the equivalent-equations method learned in Lesson 6 11.

EXTRA PRACTICE

Solving Equations Math Masters, p. 298 Students simplify and solve equations.

Ongoing Assessment: Recognizing Student Achievement Use journal page 339.  [Patterns, Functions, and Algebra Goal 2]

Key Vocabulary equivalent equations  simplify an equation

Materials Math Journal 2, pp. 338 and 339 Student Reference Book, pp. 251 and 252 Study Link 94

Advance Preparation For the optional Readiness activity in Part 3, fill in problems on Math Masters, page 296 before making copies. See Part 3 for suggested problems.

Teacher’s Reference Manual, Grades 4–6 pp. 289–294

810

Unit 9

More about Variables, Formulas, and Graphs

Mathematical Practices

SMP1, SMP2, SMP3, SMP6, SMP8

Getting Started

Content Standards

6.NS.4, 6.EE.4, 6.EE.5

Mental Math and Reflexes

Math Message

Remind students that one way to check if two expressions are equivalent is to substitute a value for the variable. Write pairs of expressions. Students show thumbs-up if the expressions are equivalent and thumbs-down if they are not. Suggestions:

Read pages 251 and 252 in your Student Reference Book. Explain why equations a–c are equivalent. a. _ y + 4 = 10 3 b. 2y - (-12) = 30 c. 150 = 60 + 10y 2

16x ; 4x + 12x thumbs-up 3y ; y + y + y thumbs-up

Study Link 9 4 Follow-Up 

2(z - 2); 4z - 2 thumbs-down

Go over the answers as a class. Resolve disagreements by asking students to substitute at least two different values for each of the variables.

8n + 12n; 4(2n + 3n) thumbs-up 12g ; 120g ÷ 10g thumbs-down 4(m - 2); -2(4 - 2m) thumbs-up

1 Teaching the Lesson ▶ Math Message Follow-Up (Student Reference Book, pp. 251 and 252)

WHOLE-CLASS DISCUSSION SOLVING SO S OLVING OL O LV VIN IN NG

Algebraic Thinking Use the same steps outlined in the example on page 251 of the Student Reference Book to solve each equation. Because the solution ( y = 9) is the same for all three equations, they are equivalent equations. Go over the example on page 252. Discuss the general outline of the solution strategy. The equation is first transformed into an equivalent equation that looks like the equations students have been solving.

Student Page Algebra

Review the procedure used in Lesson 9-4 to simplify an equation: 1. Eliminate all parentheses. 2. Combine like terms on each side of the equal sign. 3. Solve. Check the solution using substitution.

A Systematic Method for Solving Equations Many equations with just one unknown can be solved using only addition, subtraction, multiplication, and division. If the unknown appears on both sides of the equals sign, you must change the equation to an equation with the unknown appearing on one side only. You may also have to change the equation to one with all the constants on the other side of the equals sign.

Note A constant is just a number, such as 3 or 7.5 or π. Constants don’t change, or vary, the way variables do.

The operations you use to solve an equation are similar to the operations you use to solve a pan-balance problem. Remember— you must always perform the same operation on both sides of the equals sign. Solve 3y + 10 = 7y - 6

Demonstrate how to use the distributive property to remove the parentheses in the first step of the procedure above. 5(b + 3) = (5 ∗ b) + (5 ∗ 3) = 5b + 15 4(b - 1) = (4 ∗ b) - (4 ∗ 1) = 4b - 4

Step

Operation

1. Remove the unknown term (the variable term) from the left side of the equation.

Subtract 3y from each side. (S 3y)

2. Remove the constant term from the right side of the equation.

Add 6 to both sides. (A 6)

3. Change the 4 y term to a 1y term. (Remember, 1 y and 1 ∗ y and y all mean the same thing.)

Divide both sides by 4. (D 4)

Equation 3 y + 10 = 7 y - 6 -3 y -3 y 10 = 4 y - 6 10 = +6 16 =

4 y-6 +6

4y

16 = 16  4 = 4 =

4y 4 y 4 y

Check: Substitute the solution, 4, for y in the original equation: 3y + 10 = 3 ∗ 4 + 10 = 12 + 10 = 22 =

7y - 6 7∗4-6 28 - 6 22

Since 2 2 = 2 2 is true, the solution, 4, is correct. So, y = 4. Each step in the above example produced a new equation that looks different from the original equation. But even though these equations look different, they all have the same solution (which is 4). Equations that have the same solution are called equivalent equations.

Student Reference Book, p. 251 237_254_EMCS_S_SRB_G6_ALG_576523.indd 251

3/15/11 11:05 AM

Lesson 9 5 

811

Student Page

▶ Simplifying and

Algebra Like terms are terms that have exactly the same unknown or unknowns. The terms 4x and 2x are like terms because they both contain x. The terms 6 and 15 are like terms because they both contain no variables; 6 and 15 are both constants.

Equations that include the square of a variable, like 3x 2  2x  1, can be difficult to solve.

If an equation has parentheses, or if the unknown or constants appear on both sides of the equals sign, here is how you can simplify it.

♦ If an equation has parentheses, use the distributive property or other properties to write an equation without parentheses.

♦ If an equation has two or more like terms on one side of the

In 1145, Abraham bar Hiyya Ha-Nasi gave a complete solution to any equation that can be written as ax 2  bx  c  0. The solutions are:

equals sign, combine the like terms. To combine like terms means to rewrite the sum or difference of like terms as a single term. For example, 4y  7y  11y and 4y  7y  3y.

2 4苶 ]/2a 苶ac) x[b 兹(b 2 4苶 ]/2a 苶ac) x[b兹(b

INDEPENDENT ACTIVITY

Solving Equations

PROBLEM PRO P RO R OBL BLE B L LE LEM EM SO S SOLVING OL O L LV VIN V ING

(Math Journal 2, pp. 338 and 339)

Algebraic Thinking Work through Problem 2 on journal page 338 as a class. Write all steps and operations for solving the problem on the board.

Solve 5(b  3)  3b  5  4(b  1). Reminder: 5(b  3) means the same as 5 * (b  3).

Equation

Operation 1. Use the distributive property to remove the parentheses.

5b  15  3b  5  4b  4

2. Combine like terms.

Problem 2:

2b  20  4b  4

3. Subtract 2b from both sides. (S 2b) 4. Add 4 to both sides. (A 4)

88 – p

Add 1p (A 1p)

6p + 28

=

88

Subtract 28. (S 28)

6p

=

60

Divide by 6. (D 6)

p

=

10

24  2b 24 / 2  2b / 2 12  b

1. Check that 12 is the solution of the equation in the example above. Solve. 2. 5x  7  1  3x

=

Operation

20  2b  4 4  4 24  2b

5. Divide both sides by 2. (D 2)

5p + 28

2b  20  4b  4  2b  2b 20  2b  4

3. 5 * (s  12)  10 * (3  s)

4. 3(9  b)  6(b  3)

Check your answers on page 423.

Student Reference Book, p. 252

Remind students to check the solution by substituting it for the variable in the original equation. 5(10) + 28 = 88 - (10) 78 = 78 Assign the remaining problems on journal pages 338 and 339. Allow enough time to go over the answers as a class.

Student Page Date

Student Page Date

Time

Simplifying and Solving Equations

9 5 䉬

250–252

Simplify each equation. Then solve it. Record the operations you used for each step. 1.

Time

LESSON

LESSON

6y  2y  40

2.

9 5 䉬

11.

Simplifying and Solving Equations

8v  25  v  80

3.

y  10

8d  3d  65

Solution 4.

250–252

p  10

13.

Solution

v  15

1

3n  2 n  42

e2

Solution

Solution 7.

6.

n  12

3(1  2y )  y  2y  4y

3m  1  m  6  2  9

Solution 8.

14.

z6

16  3s  2s  24  2s  20

12e  19  7  e

夹 15.

g3

m  3

8  12x  6 º (1  x )

Solution

Are the following 2 equations equivalent? 5y  3  6y  4  12y

5.

Solution

g  3g  32  27  5g  2

Solution

d  13

s  12

Yes

5y  3  6y  4(1  3y)

Sample answer: They have the same solution, y  1.

Explain your answer.

夹 16.

Are the following 2 equations equivalent? 5(f  2)  6  16

Yes

f 1  3

Sample answer: They have the same solution, f  4.

Explain your answer.

Solution 9.

y3

Solution

4.8  b  0.6b  1.8  3.6b

10.

x  19

4t  5  t  7

Try This 17.

2z  4 Solve 5  z  1

Solution

(Hint: Multiply both sides by 5.)

Solution

b  3.3

Solution

t4

Math Journal 2, p. 338

812

Unit 9

continued

3z  6z  60  z

5p  28  88  p

Solution

Solution

12.

More about Variables, Formulas, and Graphs

Math Journal 2, p. 339

z3

Student Page

Ongoing Assessment: Recognizing Student Achievement

Journal page 339 Problems 15 and 16



Date

Time

LESSON

Number Stories and the Distributive Property

9 5 䉬

Solve each problem mentally. Then record the number model you used. 248 249

Use journal page 339, Problems 15 and 16 to assess students’ abilities to simplify equations and recognize equivalent equations. Students are making adequate progress if their explanations for Problems 15 and 16 indicate that each equation has the same solution (y = -1); (f = 4).

A carton of milk costs $0.60. John bought 3 cartons of milk one day and 4 cartons the next day.

1.

Number model

During a typical week, Karen runs 16 miles and Jacob runs 14 miles.

2.

[Patterns, Functions, and Algebra Goal 2]

About how many miles in all do Karen and Jacob run in 8 weeks? Number model

Point out that the fraction bar in the Try This problem acts as a grouping symbol. The numerator and the denominator in the 2z + 4 expression _ can each be treated as if there were parentheses 5 around them. The expression may be written as the division expression (2z + 4) ÷ 5 or as the multiplication expression 1 ∗ (2z + 4). One way to start is to multiply both sides of the _ 5 equation by 5. 2z + 4 _ =z-1

$4.20 0.60(3  4)  n, or (0.60 º 3)  (0.60 º 4)  n

How much did he spend in all?

About 240 miles (8 º 16)  (8 º 14)  m, or 8(16  14)  m

Mark bought 6 CDs that cost $12 each. He returned 2 of them.

3.

$48 (6 º 12)  (2 º 12)  c, or (6  2) º 12  c

How much did he spend in all? Number model

Max collects stamps. He had 9 envelopes, each containing 25 stamps. He sold 3 envelopes to another collector.

4.

150 stamps (9 º 25)  (3 º 25)  s, or (9  3) º 25  s

How many stamps did he have left? Number model

Jean is sending party invitations to her friends. She has 8 boxes with 12 invitations in each box. She has already mailed 5 boxes of invitations.

5.

36 invitations (8 º 12)  (5 º 12)  p, or (8  5) º 12  p

How many invitations are left? Number model

5

2z + 4 = 5 ∗ (z - 1) 5 ∗ _  5

2z + 4 = 5z - 5

Math Journal 2, p. 337

2 Ongoing Learning & Practice ▶ Applying the

INDEPENDENT ACTIVITY

Distributive Property (Math Journal 2, p. 337)

Algebraic Thinking Students use the distributive property to solve number stories.

▶ Math Boxes 9 5 

INDEPENDENT ACTIVITY

(Math Journal 2, p. 340)

Student Page Date

Time

LESSON

Math Boxes

9 5 䉬

1.

Mixed Practice Math Boxes in this lesson are paired with Math Boxes in Lessons 9-1 and 9-3. The skills in Problems 4 and 5 preview Unit 10 content.

The area of the shaded part of the rectangle is 20 units2.

Solve.

2.

1 a. f 3

12

 6  8 Solution

h 16

b.

30  b  6  11b

c.

4g  4  2g  36

Write a number sentence to find the value of h.

Solution

Number sentence: 20  h(16  12)

Solution

Solve for h.

5

h 3.

units

Length of I苶N 苶

12 15

b  2 g  16 250 251

248 249

I

Triangles THG and TIN are similar.

a.

f  6

5

m

T

b.

Length of H 苶I苶 

c.

 The size-change factor:  triangle TIN 

m

H 3m

4mG

N

12 m

m triangle THG

1

:

4 179

4.

I am a quadrangle with 2 pairs of congruent adjacent sides. One of my diagonals is also my only line of symmetry. How many sides do I have?

5.

The polygon below is a regular polygon. Find the measure of angle X without using a protractor.

4

C

O

E

V

X

Use your Geometry Template to draw this polygon in the space provided at the right.

169

m⬔X 

N

120 

233

Math Journal 2, p. 340

Lesson 9 5 

813

Study Link Master Name

Date

STUDY LINK

▶ Study Link 9 5

Time



Each equation in Column 2 is equivalent to an equation in Column 1. Solve each equation in Column 1. Write Any number if all numbers are solutions of the equation.

(Math Masters, p. 295)

251 252

Home Connection Students solve equations and find equivalent equations.

Match each equation in Column 1 with an equivalent equation in Column 2. Write the letter label of the equation in Column 1 next to the equivalent equation in Column 2. Column 1

Column 2

A 4x - 2 = 6

Solution

x=2

B 3s = -6 Solution s = -2 C 3y - 2y = y

Solution

Any number

D 5a = 7a

a =0

Solution

C A B C A A C A D B B A D

INDEPENDENT ACTIVITY



Equivalent Equations

95

6j + 8 = 8 + 6j 2c - 1 = 3 6w = -12

3 Differentiation Options

2h _ =1 2h 3q _ - 6 = -4 3

3(r + 4) = 18 2(5x + 1) = 10x + 2 -5x - 5(2 - x) = 2(x - 7)

INDEPENDENT ACTIVITY

READINESS

s=0

▶ Revisiting

5b - 3 - 2b = 6b + 3 1 _t + 3 = 2_ 4 2

5–15 Min

Pan-Balance Problems

6z = 12 2a = (4 + 7)a

(Math Masters, p. 296)

Practice Write each product or quotient in exponential notation. 1.

25

22 ∗ 23

2.

104 _ 102

102

3.

52 ∗ 52

54

4.

43 _ 42

41

To review solving equations, have students generate equivalent equations and record the operations they used on Math Masters, page 296. Remind students that they have solved problems like these as pan-balance problems in the past. Use the suggested problems below or generate problems according to students’ needs. Suggestions:

Math Masters, p. 295 285-328_EMCS_B_G6_MM_U09_576981.indd 295

3/2/11 10:48 AM

 8y + (-5) = 5y + 13 y = 6  16f - 24 = 8f f = 3  11 + 9k = 71 - 3k k = 5  4r + 37 = 100 - 5r r = 7

Teaching Master

Teaching Master Name

Date

LESSON

Name

Time

“Complexifying” Equations to Find Equivalent Names

9 5 

Date

LESSON

9 5 

The equivalent name for 92 that was generated in the Example on page 297A is written in the name-collection box below. Write the two other names for 92 you found in Problems 1 and 2. Then use the same procedure to find at least three equivalent names for the other numbers and write them in the name-collection boxes.

When you solve an equation, you simplify it first. You can also apply the steps backward to “unsimplify” or “complexify” equations. This process can be used to generate interesting equivalent names for numbers.

Sample answers are given.

Example: “Complexify” an equation to generate an equivalent name for 92. Solution: Step 1: Start by writing an equation stating that 92 is equal to itself. 92 = 92 Step 2:

Next, write 92 as a sum of two whole numbers.

Step 3:

Find a common factor of the two addends on the left side, and use the distributive property to factor it out. For this example, we will use the GCF of 64 and 28, which is 4.

Time

“Complexifying” Equations to Find Equivalent Names continued

64 + 28 = 92

92

45

4 º (16 + 7) 2 ∗ (32 + 14) 2 ∗ (19 + 27)

5 ∗ (3 + 6) 15 ∗ (1 + 2) 5 ∗ (5 + 4)

4 ∗ (16 + 7) = 92 This equation shows that 4 ∗ (16 + 7) is another name for 92.

116

78

4 ∗ (25 + 4) 2 ∗ (23 + 35) 4 ∗ (15 + 14)

3 ∗ (6 + 20) 6 ∗ (3 + 10) 2 ∗ (35 + 4)

Answer the questions below to generate more names for 92. 1. a. b.

Name another common factor of 64 and 92.

2

Repeat Step 3 above using this common factor to generate a different name for 92.

p

2 ∗ (32 + 14) = 92

Repeat Step 2 above by writing 92 as the sum of two whole numbers other than 64 and 28.

g

2. a.

py g

Sample answer: 38 + 54 = 92

b.

Find the greatest common factor of your two addends and factor it out to generate another name for 92. GCF:

Sample answer: 2 Sample answer: 2 ∗ (19 + 27) = 92

Equation:

Math Masters, p. 297B

Math Masters, p. 297A 297A-297B_EMCS_B_G6_MM_U09_576981.indd 297A

814

Unit 9

3/9/11 12:11 PM

More about Variables, Formulas, and Graphs

297A-297B_EMCS_B_G6_MM_U09_576981.indd 297B

3/9/11 12:11 PM

Teaching Master ENRICHMENT

▶ Generating Unsimplified

PARTNER ACTIVITY 5–15 Min

Equations to Find Equivalent Names

Name

95 䉬

Sometimes you need to translate words into algebraic expressions to solve problems. Example: The second of two numbers is 4 times the first. Their sum is 50. Find the numbers.

4n  the second number, and n  4n  50. Because 5n  50, n  10. The first number is 10 and the second number is 4(10), or 40. For each problem, translate the words into algebraic expressions. Then write an equation and solve it. The larger of two numbers is 12 more than the smaller. Their sum is 84. Find the numbers.

1.

To extend their work with simplifying equations, remind students that they can take two steps to simplify an equation: first, use the distributive property to eliminate parentheses; then combine like terms.

They can distribute the 4: Then, they can combine like terms:

Time

Writing and Solving Equations

If n  the first number, then

(Math Masters, pp. 297A and 297B)

For example, to simplify this equation:

Date

LESSON

Equation

Smaller number

Equation

Larger number

48

g  (g  9)  29 2g  9  29; 2g  20

Number of girls

10

Sometimes it helps to label a diagram when you are translating words into algebraic expressions. The base (b) of a parallelogram is 3 times as long as an adjacent side (s). The perimeter of the parallelogram is 64 m. What is the length of the base?

3.

44 = 44

Explain that students can also apply these steps backward, starting with a simple equation like 44 = 44 and obtaining a more complex equation like 4 ∗ (9 + 2) = 44. Tell students that using these steps to “complexify” a simple equation can be a way to generate interesting names for numbers.

36

Mr. Zock’s sixth-grade class of 29 students has 9 more boys than girls. How many girls are in the class?

2.

4 ∗ (9 + 2) = 44 36 + 8 = 44

x  (x  12)  84 2x  12  84; 2x  72

s

Equation

3s

b

Label the diagram at the right. Then write an equation and solve it.

2(3s)  2s  64 6s  2s  64; 8s  64

24

Length of the base

units

Math Masters, p. 297

Have students read the example on Math Masters, page 297A. Then have them work in pairs to complete the problems on Math Masters, pages 297A and 297B.

ENRICHMENT

▶ Writing and Solving Equations

PARTNER ACTIVITY 15–30 Min

(Math Masters, p. 297)

Students translate word sentences into equations and then solve the equations.

Teaching Master Name

EXTRA PRACTICE

▶ Solving Equations

INDEPENDENT ACTIVITY 5–15 Min

LESSON

9 5 䉬

Date

Time

More Simplifying and Solving of Equations

Simplify each equation. Then solve it. Show your work. 1.

4(5t  7)  10t  2

3.

4(12  8w)  w  18

5.

7(1  4y)  13(2y  3)

2.

18(m  6)  15m  6

4.

3g  8(2g  6)  2  14g

6.

4n  5(7n  3)  9(n  5)

(Math Masters, p. 298)

To provide extra practice with equations, have students simplify and solve equations involving variable terms on both sides of the equal sign.

Solution

Planning Ahead 3 in. long) and jumbo size (1_ 13 in. Gather small #1 standard (1_ 16 16 long) paper clips for the Readiness activity in Part 3 of Lesson 9-6.

Solution

t3

Solution

w2

Solution

m 38

g  10

py g

2(6v  3)  18  3(16  3v)

Solution

v  12

Solution 8.

n  1

5  (15d  1)  2(7d  16)  d

Solution

p

7.

y  16

g

Solution

d1

Math Masters, p. 298

Lesson 9 5 

815

Name

Date

LESSON

Time

“Complexifying” Equations to Find Equivalent Names

9 5 

When you solve an equation, you simplify it first. You can also apply the steps backward to “unsimplify” or “complexify” equations. This process can be used to generate interesting equivalent names for numbers. Example: “Complexify” an equation to generate an equivalent name for 92. Solution: Step 1: Start by writing an equation stating that 92 is equal to itself. 92 = 92 Step 2:

Next, write 92 as a sum of two whole numbers. 64 + 28 = 92

Step 3:

Find a common factor of the two addends on the left side, and use the distributive property to factor it out. For this example, we will use the GCF of 64 and 28, which is 4. 4 ∗ (16 + 7) = 92

This equation shows that 4 ∗ (16 + 7) is another name for 92. Answer the questions below to generate more names for 92. 1. a. b.

b.

Repeat Step 3 above using this common factor to generate a different name for 92.

Repeat Step 2 above by writing 92 as the sum of two whole numbers other than 64 and 28.

Find the greatest common factor of your two addends and factor it out to generate another name for 92. GCF: Equation:

297A

Copyright © Wright Group/McGraw-Hill

2. a.

Name another common factor of 64 and 92.

Name LESSON

9 5 

Date

Time

“Complexifying” Equations to Find Equivalent Names continued

The equivalent name for 92 that was generated in the Example on page 297A is written in the name-collection box below. Write the two other names for 92 you found in Problems 1 and 2. Then use the same procedure to find at least three equivalent names for the other numbers and write them in the name-collection boxes.

92

45

4 º (16 + 7)

78

Copyright © Wright Group/McGraw-Hill

116

297B