Download Dividing Integers

Additional
Whole Number
Operations
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All rights reserved.
2.1 and 2.2 Dividing Whole Numbers
Objectives
1.
2.
3.
4.
5.
6.
7.
8.
9.
Write division problems in three ways.
Identify the parts of a division problem.
Divide 0 by a number.
Recognize that a number cannot be divided by 0.
Divide a number by itself.
Divide a number by 1.
Use short division.
Use multiplication to check the answer to a division
problem.
Use tests for divisibility.
Copyright © 2010 Pearson Education, Inc. All rights reserved.
Slide 1.5- 2
Just as 4  5, 4 × 5, and (4)(5) are different ways of
indicating multiplication, there are several ways to
write 20 divided by 4.
Being divided
20 ÷ 4 = 5
Divided by
Divided by
5
4 20
Being divided
20
5
4
Being divided
Divided by
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Slide 1.5- 3
Parallel
Example 1
Using Division Symbols
Write the division problem 21 ÷ 7 = 3 using two
other symbols.
This division can also be written as shown below.
3
7 21
21
3
7
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Slide 1.5- 4
In division, the number being divided is the
dividend, the number divided by is the divisor,
and the answer is the quotient.
dividend ÷ divisor = quotient
quotient
divisor dividend
dividend
 quotient
divisor
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Slide 1.5- 5
Parallel
Example 2
Identifying the Parts of a Division
Problem
Identify the dividend, divisor, and quotient.
a. 36 ÷ 9 = 4
36 ÷ 9 = 4
dividend
quotient
divisor
b. 75  3
dividend
25
75
3
25
quotient
divisor
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Slide 1.5- 6
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Slide 1.5- 7
Parallel
Example 3
Dividing 0 by a Number
Divide.
a. 0 ÷ 21 = 0
b. 0 ÷ 1290 = 0
c.
0
0
275
d.
0
130 0
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Slide 1.5- 8
Parallel
Example 4
Changing Division Problems to
Multiplication
Change each division problem to a multiplication
problem.
27
a. 3  9
becomes 3  9 = 27 or 9  3 = 27
7
b. 8 56
becomes 8  7 = 56 or 7  8 = 56
c. 90 ÷ 9 = 10 becomes 9  10 = 90 or 10  9 = 90
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Slide 1.5- 9
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Slide 1.5- 10
Parallel
Example 5
Dividing 0 by a Number
All the following are undefined.
8
a.
0
is undefined
b. 0 12 is undefined
c. 25 ÷ 0 is undefined
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Slide 1.5- 11
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Slide 1.5- 12
Parallel
Example 6
Dividing a Nonzero Number by
Itself
Divide.
a. 25 ÷ 25 = 1
1
b. 49 49
c.
82
 1
82
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Slide 1.5- 13
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Slide 1.5- 14
Parallel
Example 7
Dividing Numbers by 1
Divide.
a. 15 ÷ 1 = 15
72
b. 1 72
34
c. 1  34
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Slide 1.5- 15
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Slide 1.5- 16
Parallel
Example 11
Checking Division by Using
Multiplication
Check each answer.
135 R2
a. 4 542
(divisor  quotient) + remainder = dividend
(4

135)
540
+ 2
+ 2
= 542
Matches
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Slide 1.5- 17
Parallel
Example 11
Checking Division by Using
Multiplication
Check each answer.
154 R3
b. 8 1236
(divisor  quotient) + remainder = dividend
(8

154)
1232
+ 3
+ 3
= 1235
Does not match original
dividend.
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Slide 1.5- 18
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Slide 1.5- 19
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Slide 1.5- 20
Parallel
Example 12
Testing for Divisibility by 2
Are the following numbers divisible by 2?
a.
128
Because the number ends in 8, which is an even
number, the number is divisible by 2.
Ends in 8
b.
4329
The number is NOT divisible by 2.
Ends in 9
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Slide 1.5- 21
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Slide 1.5- 22
Parallel
Example 13
Testing for Divisibility by 3
Are the following numbers divisible by 3?
a.
2128
Add the digits.
2 + 1 + 2 + 8 = 13
Because 13 is not divisible by 3, the number
is not divisible by 3.
b. 27,306
Add the digits.
2 + 7 + 3 + 0 + 6 = 18
Because 18 is divisible by 3, the number is
divisible by 3.
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Slide 1.5- 23
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Slide 1.5- 24
Parallel
Example 14
Testing for Divisibility by 5
Are the following numbers divisible by 5?
a. 17,900
The number ends in 0 and is divisible by 5.
b. 5525
The number ends in 5 and is divisible by 5.
c. 657
The number ends in 7 and is not divisible by 5.
Copyright © 2010 Pearson Education, Inc. All rights reserved.
Slide 1.5- 25
Parallel
Example 15
Testing for Divisibility by 10
Are the following numbers divisible by 10?
a. 18,240
The number ends in 0 and is divisible by 10.
b. 3225
The number ends in 5 and is not divisible by 10.
c. 248
The number ends in 8 and is not divisible by 10.
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Slide 1.5- 26
2.3 Solving Application Problems
Objectives
1. Find indicator words in application problems.
2. Solve application problems.
3. Estimate an answer.
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Slide 1.10- 27
As you read an application problem, look for
indicator words that help you.
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Slide 1.10- 28
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Slide 1.10- 29
Parallel
Example 1
Applying Division
A group of foreign language students plan a
summer trip to Germany. The cost of the
entire trip is $21,413. If seven students
equally pay for the trip, what is the cost of the
trip for each student?
Step 1 Read. A reading of the problem shows that
seven students divide the cost of $21,413
equally.
Step 2 Work out a plan. The indicator words, divided
equally, show that the amount each pays can
be found by dividing $21,413 by 7.
Copyright © 2010 Pearson Education, Inc. All rights reserved.
Slide 1.10- 30
Parallel
Example 1
continued
Applying Division
A group of foreign language students plan a summer trip to
Germany. The cost of the entire trip is $21,413. If seven students
equally pay for the trip, what is the cost of the trip for each
student?
Step 3 Estimate. Round $21,413 to $21,000. Then
$21,000 ÷ 7 = $3000, so a reasonable answer
would be a little greater than $3,000 each.
Step 4 Solve. Find the actual answer by dividing
$21,413 by 7.
3059
7 21413
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Slide 1.10- 31
Parallel
Example 1
continued
Applying Division
A group of foreign language students plan a summer trip to
Germany. The cost of the entire trip is $21,413. If seven students
equally pay for the trip, what is the cost of the trip for each
student?
Step 5 State the answer. Each student will pay
$3059.
Step 6 Check. The exact answer of $3059 is
reasonable, as $3059 is close to the
estimated answer of $3000. Check the
answer by multiplying.
$3059
×
7
$21,413
Amount each student will pay.
Number of students
Total raised, matches amount
given in problem.
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Slide 1.10- 32
Parallel
Example 2
Applying Addition
A salesman takes a business trip and travels
the following miles each day. On Monday he
travels 385 miles, Tuesday 412 miles,
Wednesday 394 miles, Thursday 403 miles,
and Friday 431 miles. Find the total miles
traveled for the week.
Step 1 Read. Reading the problem shows that each
day’s mileage is given and the total mileage
for the week must be found.
Step 2 Work out a plan. Add the daily mileage to
arrive at the weekly total.
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Slide 1.10- 33
Parallel
Example 2
continued
Applying Addition
A salesman takes a business trip and travels the following miles each
day. On Monday he travels 385 miles, Tuesday 412 miles, Wednesday
394 miles, Thursday 403 miles, and Friday 431 miles. Find the total
miles traveled for the week.
Step 3 Estimate. Because each day the salesman
travels about 400 miles for a week of five
days, a reasonable estimate would be
5 ∙ 400 = 2000 miles.
Step 4
Solve. Find the exact answer.
Number of miles for the week
Copyright © 2010 Pearson Education, Inc. All rights reserved.
385
412
394
403
+ 431
2025 Slide 1.10- 34
Parallel
Example 2
continued
Applying Addition
A salesman takes a business trip and travels the following miles each
day. On Monday he travels 385 miles, Tuesday 412 miles, Wednesday
394 miles, Thursday 403 miles, and Friday 431 miles. Find the total
miles traveled for the week.
Step 5 State the answer. The salesman traveled
2025 miles for the week.
Step 6 Check. The exact answer of
2025
2025 miles is close to the
385
412
estimate of 2000 miles, so it is
394
reasonable. Add up the columns
403
to check the exact answer.
+ 431
Check by adding up. 2025
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Slide 1.10- 35
Parallel
Example 3
Determining Whether Subtraction
is Necessary
A delivery company delivered 4876 fewer packages
in January than in December. The number of
packages delivered in December was 27,495. Find
the number of packages delivered in January.
Step 1 Read. The number of packages decreased
from December to January. The number of
packages delivered in January must be found.
Step 2 Work out a plan. The indicator word, fewer,
shows that subtraction must be used to find
the number of packages delivered in January.
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Slide 1.10- 36
Parallel
Example 3
Continued
Determining Whether Subtraction
is Necessary
A delivery company delivered 4876 fewer packages in January than in
December. The number of packages delivered in December was 27,495.
Find the number of packages delivered in January.
Step 3 Estimate. Because the number of packages
in December is about 27,000 and the
decrease in January is about 5000, a
reasonable estimate would be
27,000 – 5000 = 22,000 packages.
Step 4 Solve. Find the exact answer.
Copyright © 2010 Pearson Education, Inc. All rights reserved.
27,495
− 4876
22,619
Slide 1.10- 37
Parallel
Example 3
Continued
Determining Whether Subtraction
is Necessary
A delivery company delivered 4876 fewer packages in January than in
December. The number of packages delivered in December was 27,495.
find the number of packages delivered in January.
Step 5 State the answer. The number of packages
delivered in January was 22,619.
Step 6
Check. The exact answer of 22,619 is
reasonable, as it is close to the estimate of
22,000. Check by adding.
22,619
+ 4876
27,495
Number of packages in January
Decrease in packages delivered
Amount of packages delivered in December.
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Slide 1.10- 38
Parallel
Example 4
Solving a Two-Step Problem
In the month of June, the Anderson family made
four deposits of $1782 each. They also withdrew
$5931 for expenses. What was the account balance
for the month of June?
Step 1 Read. The problem asks for account balance
for the month of June.
Step 2 Work out a plan. The wording four deposits
of $1782 each, indicates the four deposits
must be totaled. Since the deposits are all the
same, use multiplication to find the total
deposits then subtract the expenses.
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Slide 1.10- 39
Parallel
Example 4
continued
Solving a Two-Step Problem
In the month of June, the Anderson family made four deposits of $1782
each. They also withdrew $5931 for expenses. What was the account
balance for the month of June?
Step 3 Estimate. The amount of deposits is about
$1800, making the total deposited money
$1800 ∙ 4 = $7200. The expenses are about
$6000. The reasonable estimate for the
amount remaining is $7200 - $6000 = $1200.
Step 4 Solve.
Find the actual $1782
amount of
×
4
deposits.
$7128
Then subtract
the expenses. − $5931
$7128
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$1197
Slide 1.10- 40
Parallel
Example 4
continued
Solving a Two-Step Problem
In the month of June, the Anderson family made four deposits of $1782
each. They also withdrew $5931 for expenses. What was the account
balance for the month of June?
Step 5 State the answer. The balance of the account
for the month of June was $1197.
Step 6 Check. The exact answer of $1197 is
reasonable, as it is close to the estimate of
$1200. Check by adding the expenses then
dividing by the 4.
$5931
+ $1197
$7128
Matches the
deposit amounts
given in the problem.
Copyright © 2010 Pearson Education, Inc. All rights reserved.
$1782
4 7128
Slide 1.10- 41
2.4 Order of Operations
Objectives
1. Use the order of operations.
2. Use the order of operations with exponents.
3. Use the order of operations with fractions bars.
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Slide 9.4- 42
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Slide 9.4- 43
Parallel
Using the Order of Operations
Example 1
a.
18 + 20 ÷ 4
18 + 5
23
b.
3 − 24 ÷ 4 + 9
3−
6
+9
3 + (−6)
−3
+9
+9
6
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Slide 9.4- 44
Parallel
Parentheses and the Order of
Example 2
Operations
a.
−5(8 – 4) – 3
−5(4) – 3
−20 – 3
−20 + (– 3)
−23
b.
3 + 4(4 – 9)(20 ÷ 4)
3 + 4(–5)(20 ÷ 4)
3 + 4(–5) (5)
3 + (–20) (5)
3 + (–100)
–97
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Slide 9.4- 45
Parallel
Exponents and the Order of
Example 3
Operations
a.
52 − (−2)2
b.
(−7)2 − (5 − 8)2 (−4)
(−7)2 − (− 3)2 (−4)
25 − 4
21
49 − 9(−4)
49 − (−36)
49 + (+36)
85
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Slide 9.4- 46
Parallel
Fraction Bars and the Order of
Example 4
Operations
14  3  5  7 
Simplify.
6  42  8
First do the work
in the numerator.
Then do the work
in the denominator.
−14 + 3(5 – 7)
6 – 42 ÷ 8
−14 + 3(– 2)
6 – 16 ÷ 8
−14 + (– 6)
6–2
4
−20
Numerator
Denominator
20
 5
4
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Slide 9.4- 47
2.5 Using Equations to Solve Application Problems
Objectives
1. Translate word phrases into expressions
with variables.
2. Translate sentences into equations.
3. Solve application problems.
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Slide 9.8- 48
Parallel
Translating Word Phrases into
Example 1
Expressions with Variables
Write each word phrase in symbols, using x as the
variable.
Words
A number plus nine
7 more than a number
−12 added to a number
Algebraic Expression
x + 9 or 9 + x
x + 7 or 7 + x
−12 + x or x + (−12)
3 less than a number
x–3
A number decreased by 1
x–1
14 minus a number
14 – x
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Slide 9.8- 49
Parallel
Translating Word Phrases into
Example 2
Expressions with Variables
Write each word phrase in symbols, using x as the
variable.
Words
Algebraic Expression
3 times a number
3x
Twice a number
2x
The quotient of 8 and a
number
A number divided by 15
8
x
x
15
The result is
=
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Slide 9.8- 50
Parallel
Translating a Sentence into an
Example 2
Equation
If 8 times a number is added to 13, the result is 45.
Find the number.
Let x represent the unknown number.
8 times a number
added to
13
+
13
8x
Next, solve the equation.
8x + 13 − 13 = 45 − 13
8x = 32
8 x 32

8
8
x4
is
45
=
45
Check:
8x + 13 = 45
8(4) + 13 = 45
45 = 45
The solution is 4.
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Slide 9.8- 51
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Slide 9.8- 52
Parallel
Example 5
Solving an Application Problem with
One Unknown
Frankie has washed 6 less than twice as many
windows washed as Rita. If Frankie has
washed 14 windows, how many windows has
Rita washed?
Step 1 Read. The problem asks for the number of
windows that Rita has washed.
Step 2 Assign a variable. There is only one
unknown, Rita’s number of windows washed.
Step 3 Write an equation.
The number Frankie
washed.
14 = 2x – 6
6 less than twice
Rita’s number.
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Slide 9.8- 53
Parallel
Example 5
continued
Solving an Application Problem with
One Unknown
Frankie has washed 6 less than twice as many windows washed as Rita. If
Frankie has washed 14 windows, how many windows has Rita washed?
Step 4 Solve.
14 = 2x – 6
14 + 6 = 2x – 6 + 6
20 = 2x
20 2  x

2
2
10  x
Step 5 State the answer. Rita washed 20
windows.
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Slide 9.8- 54
Parallel
Example 5
continued
Solving an Application Problem with
One Unknown
Frankie has washed 6 less than twice as many windows washed as Rita. If
Frankie has washed 14 windows, how many windows has Rita washed?
Step 6 Check.
14 = 2x – 6
14 = 2(10) – 6
14 = 14
So 10 is the correct solution because it
“works” in the original problem.
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Slide 9.8- 55
Parallel
Example 6
Solving an Application Problem with
Two Unknowns
On a shopping spree, Yoshi spent $54 more
than Lowell. The total spent by them both was
$276. Find the amount spent by each person.
Step 1 Read. The problem asks for the amount
spent by each person.
Step 2 Assign a variable. There are two unknowns.
Let x represent the amounts spent by Lowell
and x + 54 be the amount spent by Yoshi.
Step 3 Write an equation.
Amount spent by
Amount spent
by Lowell
Yoshi.
x + x + 54 = 276
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Slide 9.8- 56
Parallel
Example 6
continued
Solving an Application Problem with
Two Unknowns
On a shopping spree, Yoshi spent $54 more than Lowell. The total spent by
them both was $276. Find the amount spent by each person.
Step 4 Solve. x + x + 54 = 276
2x + 54 = 276
2x + 54 − 54 = 276 − 54
2x = 222
1
2 x 222

2
2
1
x  111
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Slide 9.8- 57
Parallel
Example 6
continued
Solving an Application Problem with
One Unknown
Step 5 State the answer. The amount Lowell
spent is x, so Lowell spent $111. The amount
Yoshi spent is x + 54, so Yoshi spent $165.
Step 6 Check. Use the words in the original
problem.
Yoshi’s $165 is $54 more dollars than Lowell’s
$111, so that checks.
The total spent is $111 + $165 = $276 which
also checks.
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Slide 9.8- 58
Parallel
Example 7
Solving a Geometry Application
Problem
The length of a rectangle is 3 inches more than
the width. The perimeter is 78 inches. Find the
length and width.
Step 1 Read. The problem asks for the length and
width of the rectangle.
Step 2 Assign a variable. There are two unknowns,
length and width. Let x represent the width
and x + 3 represent the length.
Step 3 Write an equation.
P = 2l + 2w
Use the formula for
perimeter of a rectangle.
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Slide 9.8- 59
Parallel
Example 7
continued
Solving a Geometry Application
Problem
The length of a rectangle is 3 inches more than the width. The perimeter is 78
inches. Find the length and width.
Step 4 Solve.
P = 2l + 2w
78 = 2(x + 3) + 2 ∙ x
78 = 2x + 6 + 2x
78 = 4x + 6
78 – 6 = 4x + 6 – 6
72 = 4x
72 14  x

4
41
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18 = x
Slide 9.8- 60
Parallel
Example 7
continued
Solving a Geometry Application
Problem
Step 5 State the answer.
The width is x, so the width is 18 inches.
The length is x + 3, so the length is 21 inches.
Step 6 Check. Use the words in the original
problem.
The original problem says that the perimeter
is 78 inches.
P = 2 ∙ 18 in. + 2 ∙21in.
P = 36 in. + 42 in.
checks
P = 78 in.
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Slide 9.8- 61