1. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving

2. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Objectives Solving a Formula for a Variable Steps for Solving a Problem Percentages

3. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example A parallelogram has an area of 35 cm 2 and a base of 7 cm. a. Write a formula to find the height h of the parallelogram with known area A and base b. b. Use the formula to find h. Solution a. We must solve A = bh for h. b. Substitute 35 for A and 7 for b into this formula.

4. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example The formula is used to convert degrees Fahrenheit to degrees Celsius. Use this formula to convert 23°F to an equivalent Celsius temperature. Solution = 5°C

5. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example Solve the equation for y and write a formula for a function f defined by y = f(x). 5(x – 2y) = 8x Solution 5x +10y = 8x 5(x – 2y) = 8x 5x + 5x 10y = 8x + 5x 10y = 3x Thus

6. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Step 1: Read the problem carefully to 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, make a table, or refer to known formulas. STEPS FOR SOLVING A PROBLEM

7. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Step 3: Solve the equation and determine the solution. Step 4: Look back and check your answer. Does it seem reasonable? Did you find the required information? STEPS FOR SOLVING A PROBLEM

8. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example 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

9. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example (cont) 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.

10. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example The perimeter of a basketball court is 288 ft. The width is 44 ft less than the length. Find the dimensions of the court. Solution Step 1: Assign a variable to an unknown quantity. x: width of the court x + 44: length of the court Step 2: Write an equation that relates these unknown quantities. x + (x + 44) + x + (x + 44) = 288

11. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example (cont) The perimeter of a basketball court is 288 ft. The width is 44 ft less than the length. Find the dimensions of the court. Step 3: Solve x + (x + 44) + x + (x + 44) = 288 4x + 88 = 288 4x = 200 x = 50 The width is 50 ft and the length is 50 + 44 = 94 ft. Step 4: Check your answer. The perimeter is 50 + 94 + 50 + 94 = 288 feet. It checks.

12. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example Two trucks leave Perrysburg at the same time and travel in opposite directions. After 4 hours, they are 320 miles apart. If one truck is traveling 20 miles per hour faster than the other truck, find the speeds of the two trucks. Solution x: speed of the slower truck x + 20: speed of the faster truck RateTimeDistance Slower Truckx44x4x Faster Truckx + 2044 (x + 20) Total320 4x + 4(x + 20) = 320

13. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example (cont) Solve. The speed of the slower truck is 30 mph and the speed of the faster truck is 20 miles per hour faster, or 50 mph. 4x + 4(x + 20) = 320 4x + 4x + 80 = 320 8x + 80 = 320 8x = 240 x = 30 Check: Distance traveled by the slower truck 30 × 4 = 120 Distance traveled by the faster truck 50 × 4 = 200 120 + 200 = 320 miles

14. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example Convert each percentage to fraction and decimal notation. a. 47%b. 9.8%c. 0.9% Solution Fraction NotationDecimal Notation a. b. c.

15. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example 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

16. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example (cont) 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.

17. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example 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 Let x = milliliters of 40% Let x + 100 = milliliters of 36% ConcentrationSolution Amount (milliliters) Pure alcohol 0.2810028 0.40x0.4x 0.36x + 1000.36x + 36

18. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 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) ConcentrationSolution Amount (milliliters) Pure alcohol 0.2810028 0.40x0.4x 0.36x + 1000.36x + 36

19. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 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.

20. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 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%.