Download Simplifying and Solving Equations

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