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# Section 2.3

Section 2.3. Introduction to Problem Solving. Page 111. Steps for Solving a Problem.

## Section 2.3

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1. Section 2.3 • Introduction to Problem Solving

2. Page 111 Steps for Solving a Problem • Step 1: Read the problem carefully and be sure that you understand it. (You may need to read the problem more than once.) Assign a variable to what you are being asked to find. If necessary, write other quantities in terms of this variable. • Step 2: Write an equation that relates the quantities described in the problem. You may need to sketch a diagram or refer to known formulas. • Step 3: Solve the equation. Use the solution to determine the solution(s) to the original problem. Include any necessary units. • Step 4: Check your solution in the original problem. Does it seem reasonable?

3. Example Page 112 • Translate the sentence into an equation using the variable x. Then solve the resulting equation. • Six times a number plus 7 is equal to 25. • The sum of one-third of a number and 9 is 18. • Twenty is 8 less than twice a number. • Solution • a. 6x + 7 = 25 c. 20 = 2x − 8 6x = 18 28 = 2x 14 = x

4. Algebraic Translations of English Phrases Consecutive Integers Page 111

5. Numbers Example Page 113 • The sum of three consecutive integers is 126. Find the three numbers. • Solution • Step 1: Assign a variable to an unknown quantity. • n:smallest of the three integers • n + 1:next integer • n + 2:largest integer • Step 2: Write an equation that relates these unknown quantities. • n + (n + 1) + (n + 2) = 126

6. Numbers Example (cont) Page 113 • Step 3: Solve the equation in Step 2. • n + (n + 1) + (n + 2) = 126 (n + n + n) + (1 + 2) = 126 3n + 3 = 126 3n = 123 n = 41 So the numbers are 41, 42, and 43. Step 4: Check your answer. The sum of these integers is 41 + 42 + 43 = 126. The answer checks.

7. Numbers Example 2 Page 113 • The sum of three odd integers is 129. Find the three numbers. • Solution • Step 1: Assign a variable to an unknown quantity. • n:smallest of the three integers • n + 2:next integer • n + 4:largest integer • Step 2: Write an equation that relates these unknown quantities. • n + (n + 2) + (n + 4) = 129

8. Numbers Example 2 (cont) Page 113 • Step 3: Solve the equation in Step 2. • n + (n + 2) + (n + 4) = 129 (n + n + n) + (2 + 4) = 129 3n + 6 = 129 3n = 123 n = 41 So the numbers are 41, 43, and 45. Step 4: Check your answer. The sum of these integers is 41 + 43 + 45 = 129. The answer checks.

9. Other types Four subtracted from six times a number is 68. Find the number? Add 4 Divide 6 A taxi charges \$2.00 to turn on the meter plus \$0.25 for each eighth of a mile. If you have \$10.00, how many eighths of a mile can you go? How many miles is that? Sub 2 Divide .25 32 eighths4 miles

10. Page 114 Note: To write x% as a decimal number, move the decimal point in the number x two places to the left and then remove the % symbol. Slide 10

11. Percent Example Page 114 • Convert each percentage to fraction and decimal notation. • 47% b. 9.8% c. 0.9% • Solution • Fraction Notation Decimal Notation • a. • b. • c.

12. Percent Example Page 114 • Convert each real number to a percentage. • 0.761 b. c. 6.3 • Solution • a. Move the decimal point two places to the right and then insert the % symbol to obtain 0.761 = 76.1% • b. • c. Move the decimal point two places to the right and then insert the % symbol to obtain 6.3 = 630%. Note that percentages can be greater than 100%.

13. Percent Example Page 115 • The price of an oil change for an automobile increased from \$15 to \$24. Calculate the percent increase. • Solution A = p times B Base = 15amount = 24-15=9

14. Using the Percent Formula Amount = Percent times Base or a=pb What is 9% of 50? 9 is 60% of what? 18 is what percent of 50?

15. Using the Percent Formula Amount = Percent times Base or a=pb A television regularly sells for \$940. The sale price is \$611. Find the percent decrease in the television’s price? Base = 940amount = 940-611=329 Decrease = 940-611=329

16. Percent Example Page 115 • Number 72 on page 122, Tuition Increase:Tuition is currently \$125 per credit. There are plans to raise tuition by 8% for next year. What will the new tuition be per credit? • Solution New tuition. Tuition increase. A = p times B Base = \$125percent = 108% Base = \$125percent = 8%

17. Percent Example Page 115 • Number 69 on page 122, Voter Turnout:In the 1980 election for president there were 86.5 million voters, whereas in 2008 there were 132.6 million voters. Fine the percent change in the number of voters. • Solution A = p times B Base = 86.5Amount = 132.6 – 86.5 = 46.1

18. Percent Example Page 115 • A car salesman sells a total of 85 cars in the first and second quarter of the year. In the second quarter, he had an increase of 240% over the previous quarter. How many cars did the salesman sell in the first quarter? • Solution • Step 1:Assign a variable. • x: the amount sold in the first quarter. • Step 2:Write an equation. • x + 2.4x = 85 (Note: A=pB which is A=240%*B)

19. Percent Example (cont) Page 115 • Step 3: Solve the equation in Step 2. • x + 2.4x = 85 • 3.4x = 85 • x = 25 • In the first quarter the salesman sold 25 cars. • Step 4: Check your answer. • An increase of 240% of 25 is 2.4 × 25 = 60. Thus the amount of cars sold in the second quarter would be 25 + 60 = 85.

20. Percent Example After a 40% price reduction, an exercise machine sold for \$564. What was the exercise machine’s price before this reduction? Divide .6

21. A Formula for Motion Motion Formula d = r · t d = distance r = rate t = time Distance equals rate times time.

22. Motion Example Page 117 • A truck driver travels for 4 hours and 30 minutes at a constant speed and travels 252 miles. Find the speed of the truck in miles per hour. • Solution • Step 1: Let r represent the truck’s rate, or speed, in miles. • Step 2: The rate is to be given in miles per hour, so change the 4 hours and 30 minutes to 4.5 or 9/2 hours. • d = rt • distance = 252 • time = 4 hours 30 minutes

23. Motion Example (cont) Page 117 • Step 3: Solve the equation. • Step 4: d = rt The speed of the truck is 56 miles per hour. The answer checks.

24. Motion Example Page 117 Number 90 on page 123, Finding Speeds:Two cars pass on a straight highway while traveling in opposite directions. One car is traveling 6 miles per hour faster than the other car. After 1.5 hours the two cars are 171 miles apart. Find the speed of each car. 1.5r 1.5(r+6) r r+6 1.5 1.5 171

25. Motion Example, cont Page 117 Number 90 on page 123, Finding Speeds:Two cars pass on a straight highway while traveling in opposite directions. One car is traveling 6 miles per hour faster than the other car. After 1.5 hours the two cars are 171 miles apart. Find the speed of each car. • Solve the equation and answer the question. 1.5r + 1.5(r+6) = 171 1.5r + 1.5r + 9 = 171 Distributive property 3r + 9 = 171 Combine like terms. 3r = 162 Subtract 9 r = 54 Divide both sides by 3 The first car will be 54mph, second car will be 60mph.

26. Motion Problem Motion Problem:Two cars leave from the same place, traveling in opposite directions. The rate of the faster car is 55 miles per hour. The rate of the slower car is 50 miles per hour. In how many hours will the cars be 420 miles apart? 55 mph 50 mph X X 55x 50x 171

27. Mixture Example Page 118 • A chemist mixes 100 mL of a 28% solution of alcohol with another sample of 40% alcohol solution to obtain a sample containing 36% alcohol. How much of the 40% alcohol was used? • Solution • Step 1: Assign a variable. • x:milliliters of 40% • x + 100:milliliters of 36% • . 28 .4x 0.36x+36 x x+100

28. Mixture Example (cont) • A chemist mixes 100 mL of a 28% solution of alcohol with another sample of 40% alcohol solution to obtain a sample containing 36% alcohol. How much of the 40% alcohol was used? • Step 2: Write an equation. • 0.28(100) + 0.4x = 0.36(x + 100) Column three.

29. Mixture Example (cont) • Step 3: Solve the equation in Step 2. • 0.28(100) + 0.4x = 0.36(x + 100) • 28(100) + 40x = 36(x + 100) • 2800 + 40x = 36x + 3600 • 2800 + 4x = 3600 • 4x = 800 • x = 200 • 200 mL of 40% alcohol solution was added to the 100 mL of the 28% solution.

30. Mixture Example (cont) • Step 4: Check your answer. • If 200 mL of 40% solution are added to the 100 mL of 28% solution, there will be 300 mL of solution. • 200(0.4) + 100(0.28) = 80 + 28 = 108 of pure alcohol. • The concentration is or 36%.

31. Solving a Mixture Problem Mixture Problem: How many ounces of a 50% alcohol solution must mixed with a 20% alcohol solution to make 240 ounces of 40% alcohol solution?. x 240-x .5x .2(240-x) 96

32. Solving a Mixture Problem Mixture Problem continued: How many ounces of a 50% alcohol solution must mixed with a 20% alcohol solution to make 240 ounces of 40% alcohol solution?. Distributive property Combine like terms. Subtract 48 Divide both sides by .3 Answer: 160 ounces of 50% and 80 ounces of 20%

33. Problems Involving Simple Interest Suppose you invested \$25,000, part at 9% simple interest and the remainder at 12%. If the total yearly interest from these investments was \$2550, find the amount invested at each rate. x 25000-x .09x .12(25000-x) x = the amount in the first account. 25000 – x = amount in the second account. .09x +.12(25000 – x) = 2550

34. Problems Involving Simple Interest (continued) Suppose you invested \$25,000, part at 9% simple interest and the remainder at 12%. If the total yearly interest from these investments was \$2550, find the amount invested at each rate. 4. Solve the equation and answer the question .09x + .12(25000 – x) = 2550 .09x + .12(25000)-.12x = 2550 distributive property -.03x + 3000 = 2550 combine like terms .09x-.12x -3000 -3000 subtract 3000 -.03x = -450 divide by -.03 x = 15000 Answer is: \$15,000 at 9% and \$10,000 at 12%

35. Problems Involving Simple Interest Teaching Example 12 on page 119, College Loans:A student borrows two amounts of money. The interest rate on the first amount is 4% and the interest rate on the second amount is 6%. If he borrowed \$5000 more at 4% than at 6%, then the total interest for one year is \$950. How much does the student owe at each rate? .04x .06(x-5000) x x-5000 x = the amount in the first loan. x-5000 = amount in the second loan. Calculate interest

36. Problems Involving Simple Interest Teaching Example 12 on page 119, College Loans (continued): A student borrows two amounts of money. The interest rate on the first amount is 4% and the interest rate on the second amount is 6%. If he borrowed \$5000 more at 4% than at 6%, then the total interest for one year is \$950. How much does the student owe at each rate? • .04x +.06(x-5000) = 950 .04x +.06(x-5000) = 950 .04x +.06x-300 = 950 Distribute. .10x -300 = 950 Combine like terms. .10x = 1250 Add 300 from both sides. x = 12500 Divide both sides by .1. Invest \$12500 in the 4% account and \$7500 in the 6% account. x x-5000 .04x .06(x-5000)

37. Solving a Mixture Problem Mixture Problem: How many pounds of cashews selling for \$8.96 per pound must be mixed with 12 pounds of chocolates selling for \$4.48 per pound to create a mixture that sells for \$7.28 per pound?. x x+12 8.96x 4.48(12) 7.28(x+12)

38. Solving a Mixture Problem (cont) Mixture Problem (cont): How many pounds of cashews selling for \$8.96 per pound must be mixed with 12 pounds of chocolates selling for \$4.48 per pound to create a mixture that sells for \$7.28 per pound? Answer: 20 pounds

39. DONE

40. Objectives • Steps for Solving a Problem • Percent Problems • Distance Problems • Other Types of Problems

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