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

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

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

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

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

Section 3.2

Introduction to Problem Solving

Page 2: Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving.

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

Objectives

• Solving a Formula for a Variable

• Steps for Solving a Problem

• Percentages

Page 3: Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving.

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

Example

A parallelogram has an area of 35 cm2 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.Solutiona. We must solve A = bh for h.

b. Substitute 35 for A and 7 for b into this formula.

A bhA

hb

35

7h 5 cm

Page 4: Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving.

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

532

9C F

= −5°C

532

9C F

25

329

3C

59

9C

Page 5: Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving.

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) = −8xSolution

−5x +10y = −8x

−5(x – 2y) = −8x

−5x + 5x 10y = −8x + 5x

10y = −3x3

10y x

Thus 3.

10f x x

Page 6: Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving.

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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

Page 7: Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving.

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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

Page 8: Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving.

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Example

The sum of three consecutive integers is 126. Find the three numbers.SolutionStep 1: Assign a variable to an unknown quantity.

n: smallest of the three integersn + 1: next integern + 2: largest integer

Step 2: Write an equation that relates these unknown quantities.

n + (n + 1) + (n + 2) = 126

Page 9: Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving.

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Example (cont)

Step 3: Solve the equation in Step 2.n + (n + 1) + (n + 2) = 126(n + n + n) + (1 + 2) = 126

3n + 3 = 1263n = 123

n = 41So 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.

Page 10: Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving.

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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. SolutionStep 1: Assign a variable to an unknown quantity.

x: width of the courtx + 44: length of the court

Step 2: Write an equation that relates these unknown quantities. x + (x + 44) + x + (x + 44) = 288

Page 11: Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving.

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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.

Page 12: Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving.

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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. Solutionx: speed of the slower truckx + 20: speed of the faster truck

Rate Time Distance

Slower Truck x 4 4x

Faster Truck x + 20 4 4 (x + 20)

Total 320

4x + 4(x + 20) = 320

Page 13: Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving.

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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) = 3204x + 4x + 80 = 320

8x + 80 = 3208x = 240x = 30

Check: Distance traveled by the slower truck 30 × 4 = 120Distance traveled by the faster truck 50 × 4 = 200 120 + 200 = 320 miles

Page 14: Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving.

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Example

Convert each percentage to fraction and decimal notation.a.47% b. 9.8% c. 0.9%Solution

Fraction Notation Decimal Notationa.

b.

c.

4747%

100 47% 0.47

9.8 98 49 2 499.8%

100 1000 500 2 500

9.8% 0.098

0.9 90.9%

100 1000 0.9% 0.009

Page 15: Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving.

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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? SolutionStep 1: Assign a variable.

x: the amount sold in the first quarter.

Step 2: Write an equation. x + 2.4x = 85

Page 16: Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving.

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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.

Page 17: Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving.

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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?SolutionLet x = milliliters of 40%Let x + 100 = milliliters of 36%

Concentration Solution Amount (milliliters)

Pure alcohol

0.28 100 28

0.40 x 0.4x

0.36 x + 100 0.36x + 36

Page 18: Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving.

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)

Concentration Solution Amount (milliliters)

Pure alcohol

0.28 100 28

0.40 x 0.4x

0.36 x + 100 0.36x + 36

Page 19: Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving.

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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 = 800x = 200

200 mL of 40% alcohol solution was added to the 100 mL of the 28% solution.

Page 20: Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving.

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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%.

1080.36,

300