Additional Whole Number Operations

Additional Whole Number Operations
© 2010 Pearson Education, Inc. All rights reserved.

2.1 and 2.2 Dividing Whole Numbers
Objectives 1. Write division problems in three ways. 2. Identify the parts of a division problem. 3. Divide 0 by a number. 4. Recognize that a number cannot be divided by 0. 5. Divide a number by itself. 6. Divide a number by 1. 7. Use short division. 8. Use multiplication to check the answer to a division problem. 9. Use tests for divisibility. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Copyright © 2010 Pearson Education, Inc. All rights reserved.
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 = 5 Being divided Being divided Divided by Being divided Divided by Divided by Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Identifying the Parts of a Division Problem
Parallel Example 2 Identifying the Parts of a Division Problem Identify the dividend, divisor, and quotient. 36 ÷ 9 = 4 b. quotient dividend divisor dividend quotient divisor Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide

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Parallel Example 4 Changing Division Problems to Multiplication Change each division problem to a multiplication problem. a. b. c. 90 ÷ 9 = 10 becomes 3  9 = 27 or 9  3 = 27 becomes 8  7 = 56 or 7  8 = 56 becomes 9  10 = 90 or 10  9 = 90 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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Parallel Example 11 Checking Division by Using Multiplication Check each answer. a. (divisor  quotient) + remainder = dividend (4  135) = 542 R2 Matches Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 11 Checking Division by Using Multiplication Check each answer. b. (divisor  quotient) + remainder = dividend (8  154) = 1235 R3 Does not match original dividend. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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Parallel Example 12 Testing for Divisibility by 2 Are the following numbers divisible by 2? 128 4329 Because the number ends in 8, which is an even number, the number is divisible by 2. Ends in 8 The number is NOT divisible by 2. Ends in 9 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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Parallel Example 13 Testing for Divisibility by 3 Are the following numbers divisible by 3? 2128 b. 27,306 Add the digits. = 13 Because 13 is not divisible by 3, the number is not divisible by 3. Add the digits. = 18 Because 18 is divisible by 3, the number is divisible by 3. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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Parallel Example 14 Testing for Divisibility by 5 Are the following numbers divisible by 5? 17,900 b c The number ends in 0 and is divisible by 5. The number ends in 5 and is divisible by 5. The number ends in 7 and is not divisible by 5. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 15 Testing for Divisibility by 10 Are the following numbers divisible by 10? 18,240 b c The number ends in 0 and is divisible by 10. The number ends in 5 and is not divisible by 10. The number ends in 8 and is not divisible by 10. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

2.3 Solving Application Problems
Objectives 1. Find indicator words in application problems. 2. Solve application problems. 3. Estimate an answer. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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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

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. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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 × $21,413 Amount each student will pay. Number of students Total raised, matches amount given in problem. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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 ∙ 400 = 2000 miles. 385 412 394 403 + 431 2025 Step 4 Solve. Find the exact answer. Number of miles for the week Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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 miles is close to the estimate of 2000 miles, so it is reasonable. Add up the columns to check the exact answer. 2025 385 412 394 403 + 431 2025 Check by adding up. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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 ,000 – 5000 = 22,000 packages. Step 4 Solve. Find the exact answer. 27,495 − 22,619 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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 27,495 Number of packages in January Decrease in packages delivered Amount of packages delivered in December. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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 $ $6000 = $1200. Step 4 Solve. Find the actual amount of deposits. $1782 × $7128 $7128 − $5931 $1197 Then subtract the expenses. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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. Slide

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. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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Parallel Example 2 Parentheses and the Order of Operations a. −5(8 – 4) – 3 b. 3 + 4(4 – 9)(20 ÷ 4) −5(4) – 3 (–5)(20 ÷ 4) −20 – 3 (–5) (5) 3 + (–20) (5) −20 + (– 3) 3 + (–100) −23 –97 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 3 Exponents and the Order of Operations a. 52 − (−2)2 b. (−7)2 − (5 − 8)2 (−4) (−7)2 − (− 3)2 (−4) 25 − 4 49 − 9(−4) 21 49 − (−36) (+36) 85 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 4 Fraction Bars and the Order of Operations Simplify. 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 Denominator Numerator Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 1 Translating Word Phrases into Expressions with Variables Write each word phrase in symbols, using x as the variable. Words Algebraic Expression A number plus nine 7 more than a number −12 added to a number 3 less than a number A number decreased by 1 14 minus a number x + 9 or 9 + x x + 7 or 7 + x −12 + x or x + (−12) x – 3 x – 1 14 – x Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 2 Translating Word Phrases into Expressions with Variables Write each word phrase in symbols, using x as the variable. Words Algebraic Expression 3 times a number Twice a number The quotient of 8 and a number A number divided by 15 The result is 3x 2x = Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 2 Translating a Sentence into an 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 is 45 8x + 13 = 45 Next, solve the equation. Check: 8x + 13 − 13 = 45 − 13 8x + 13 = 45 8x = 32 8(4) + 13 = 45 45 = 45 The solution is 4. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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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. 6 less than twice Rita’s number. The number Frankie washed. 14 = 2x – 6 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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 = 2x – 6 + 6 20 = 2x Step 5 State the answer. Rita washed 20 windows. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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 Yoshi. Amount spent by Lowell x + x + 54 = 276 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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 1 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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. Use the formula for perimeter of a rectangle. P = 2l + 2w Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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 x 78 = 4x + 6 78 – 6 = 4x + 6 – 6 72 = 4x 1 18 = x 1 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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 in. P = 78 in. checks Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Additional Whole Number Operations

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