1. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1 1 Section 2.5 An Introduction to Problem Solving Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1

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3. 3 The following tables summarize many of the algebraic expressions that we have used in previous sections for modeling the conditions. Getting Started – A Strategy and Some Terminology

4. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 4 English PhraseAlgebraic Phrase The sum of a number and 4x + 4 2 more than a numberx + 2 5 added to a numberx + 5 A number increased by 7x + 7 A number plus 3x + 3 Algebraic Translations of English Phrases Addition

5. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 5 Subtraction English PhraseAlgebraic Phrase The difference between a number and 2x – 2 The difference between 2 and a number2 – x 2 fewer than a numberx – 2 5 less than a numberx – 5 A number decreased by 7x – 7 8 minus a number8 – x A number minus 3x – 3 Algebraic Translations of English Phrases

6. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 6 Multiplication English PhraseAlgebraic Phrase Six times a number The product of 4 and a number Twice a number 25% of a number Algebraic Translations of English Phrases

7. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 7 Division English PhraseAlgebraic Phrase A number divided by 8 The quotient of 5 and a number The quotient of a number and 5 The reciprocal of a number Algebraic Translations of English Phrases

8. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 8 8 Objective #1: Examples

9. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 9 9 Objective #1: Examples

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13. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 13 Solving Word Problems Strategy for Solving Word Problems Step 1 Read the problem carefully several times until you can state in your own words what is given and what the problem is looking for. Let x (or any variable) represent one of the unknown quantities in the problem. Step 2 If necessary, write expressions for any other unknown quantities in the problem in terms of x. Step 3 Write an equation in terms of x that translates, or models, the conditions of the problem. Step 4 Solve the equation and answer the problem’s question. Step 5 Check the solution in the original wording of the problem, not in the equation obtained from the words.

14. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 14 Solving Word ProblemsEXAMPLE SOLUTION A rectangular soccer field is twice as long as it is wide. If the perimeter of the soccer field is 300 yards, what are its dimensions? STEP 1: Let x represent one of the quantities. x = the width of the soccer field. STEP 2: Represent other unknown quantities in terms of x. Since the field is twice as long as it is wide, then 2x = the length of the soccer field.

15. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 15 Solving Word ProblemsCONTINUED STEP 3: Write an equation in x that describes the conditions. The soccer field is in the shape of a rectangle and therefore has a perimeter equal to twice the length plus twice the width. This can be expressed as follows: 2(2x) + 2(x) = 300

16. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 16 Solving Word ProblemsCONTINUED STEP 4: Solve the equation and answer the question. 4x + 2x = 300Multiply 6x = 300Add like terms x = 50 Divide both sides by 6 2(2x) + 2(x) = 300 Therefore the width of the soccer field is 50 yards. The length of the field is twice the width, and 2x = 2(50) = 100. Therefore, the length of the soccer field is 100 yards.

17. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 17 Solving Word ProblemsCONTINUED The problem states that the perimeter of the soccer field is 300 yards. Let’s use this information to verify our answer. The formula for the perimeter of a rectangle is repeated as follows: STEP 5: Check the proposed solution in the original wording of the problem. 2(2x) + 2(x) = 300 2(2(50)) + 2(50) = 300 200 + 100 = 300 300 = 300 So, the dimensions of the soccer field are 50 yards by 100 yards. Replace x with 50 Multiply Add

18. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 18 The product of 5 and a number is 135. Find the number. 1.Let x represent one of the quantities. x = the number 2.Represent other quantities in terms of x. There are no other unknown quantities. 3.Write an equation in x that describes the conditions. 5x = 135 4.Solve the equation and answer the question. x = 27 Divide both sides by 5. The number is 27. 5.Check the solution. The product of 5 and 27 is 135. Solving Word ProblemsEXAMPLE

19. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 19 Consecutive Integers English PhraseAlgebraic Expression Example Two consecutive integersx, x + 15, 6 Three consecutive integersx, x + 1, x + 28, 9, 10 Two consecutive even integersx, x + 26, 8 Two consecutive odd integersx, x + 2–5, –3 Three consecutive even integersx, x + 2, x + 42, 4, 6 Three consecutive odd integersx, x + 2, x + 43, 5, 7 Algebraic Translations of English Phrases

20. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 20 EXAMPLE EXAMPLE The sum of three consecutive odd integers is 45. Find the integers. 1.Let x represent one of the quantities. x = the number 2.Represent other quantities in terms of x. x + 2 = the second number, x + 4 = the third number 3.Write an equation in x that describes the conditions. x + x + 2 + x + 4 = 45 4.Solve the equation and answer the question. 3x + 6 = 45 Combine like terms. 3x = 39Subtract 6 from both sides. x = 13 x + 2 = 15 Use x to find next two consecutive odd integers. x + 4 =17 The numbers are 13, 15 and 17. 5.Check the solution. The sum of 13, 15 and 17 is 45. Solving Word Problems

21. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 21 EXAMPLE EXAMPLE The sum of three consecutive odd integers is 45. Find the integers. 1.Let x represent one of the quantities. x = the number 2.Represent other quantities in terms of x. x + 2 = the second number, x + 4 = the third number 3.Write an equation in x that describes the conditions. x + x + 2 + x + 4 = 45 4.Solve the equation and answer the question. 3x + 6 = 45 Combine like terms. 3x = 39Subtract 6 from both sides. x = 13 x + 2 = 15 Use x to find next two consecutive odd integers. x + 4 =17 The numbers are 13, 15 and 17. 5.Check the solution. The sum of 13, 15 and 17 is 45. Solving Word Problems

22. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 22 EXAMPLE : EXAMPLE : A field is six times as long as it is wide. If the perimeter is 420 feet, find the dimensions of the field. 1.Let w represent one of the quantities. w = the width of the field 2.Represent other quantities in terms of w. 6w = length 3.Write an equation in w that describes the conditions. Recall: 2w + 2l = P 2w + 2(6w) = 420 4.Solve the equation and answer the question. 2w + 12w = 420 Multiply. 14w = 420Combine like terms. w = 30 and Solve. 6w = 180 Answer: The width is 30 feet and the length is 180 feet. Check the solution. The perimeter is 2(30) + 2(180) = 60 +360 = 420 feet. Solving Word Problems

23. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 23 EXAMPLE : EXAMPLE : A field is six times as long as it is wide. If the perimeter is 420 feet, find the dimensions of the field. 1.Let w represent one of the quantities. w = the width of the field 2.Represent other quantities in terms of w. 6w = length 3.Write an equation in w that describes the conditions. Recall: 2w + 2l = P 2w + 2(6w) = 420 4.Solve the equation and answer the question. 2w + 12w = 420 Multiply. 14w = 420Combine like terms. w = 30 and Solve. 6w = 180 Answer: The width is 30 feet and the length is 180 feet. Check the solution. The perimeter is 2(30) + 2(180) = 60 +360 = 420 feet. Solving Word Problems

24. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 24 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 24 Objective #2: Examples

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31. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 31 Objective #2: Examples