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Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Introduction to Equations Basic Concepts The Addition Property of Equality.

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Presentation on theme: "Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Introduction to Equations Basic Concepts The Addition Property of Equality."— Presentation transcript:

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2 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Introduction to Equations Basic Concepts The Addition Property of Equality The Multiplication Property of Equality 2.1

3 Slide 3 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

4 Slide 4 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Using the addition property of equality Solve each equation. a. x + 21 = 9 b. n – 8 = 17 b. a. x + 21 = 9 x (21) = 9 + (21) x + 0 = 12 x = 12 n – 8 = 17 n – = n = 25 n + 0 = 25 The solution is 12. The solution is 25.

5 Slide 5 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Solving and checking a solution Solve the equation 7 + x = 12 and then check the solution. Check: Isolate x by adding 7 to each side. 7 + x = x = x = 19 x = = 12 The solution is 19. The answer checks. 7 + x = = 12 ?

6 Slide 6 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

7 Slide 7 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Using the multiplication property of equality Solve each equation. a. b. 11 a = 33 b. a. 11 a = 33 The solution is 48. The solution is 3.

8 Slide 8 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Solving and checking a solution Solve the equation and then check the solution. Check: The solution is The answer checks. ?

9 Slide 9 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Application A veterinary assistant holds a cat and steps on a scale. The scale reads 153 lbs. Alone the assistant weighs 146 lbs. a.Write a formula to show the relationship of the weight of the cat, x, and the assistant. b.Determine how much the cat weighs. a x = x = (146) + x = b. x = 7 The cat weighs 7 lbs.

10 Slide 10 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

11 Linear Equations Basic Concepts Solving Linear Equations Applying the Distributive Property Clearing Fractions and Decimals Equations with No Solutions or Infinitely Many Solutions 2.2

12 Slide 12 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley If an equation is linear, writing it in the form ax + b = 0 should not require any properties or processes other than the following: using the distributive property to clear parentheses, combining like terms, applying the addition property of equality.

13 Slide 13 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Determine whether an equation is linear Determine whether the equation is linear. If the equation is linear, give values for a and b. a. 9x + 7 = 0 b. 5x = 0c. b. This equation is not linear because it cannot be written in the form ax + b = 0. The variable has an exponent other than 1. a. The equation is linear because it is in the form ax + b = 0 with a = 9 and b = 7. c. This equation is not linear because it cannot be written in the form ax + b = 0. The variable appears in the denominator of a fraction.

14 Slide 14 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Using a table to solve an equation Complete the table for the given values of x. Then solve the equation 4x – 6 = 2. From the table 4x – 6 = 2 when x = 1. Thus the solution to 4x – 6 = 2 is 1. x x 618 x x

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16 Slide 16 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Solving linear equations Solve each linear equation. a. 12x 15 = 0 b. 3x + 19 = 5x + 5 b. a. 12x 15 = 0 12x = x = 15 3x + 19 = 5x + 5 3x 3x + 19 = 5x 3x = 2x = 2x = 2x

17 Slide 17 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Applying the distributive property Solve the linear equation. Check your solution. x + 5 (x – 1) = 11 6x 5 = 11 x + 5 (x – 1) = 11 x + 5 x – 5 = 11 6x = x = 16

18 Slide 18 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE continued Check x + 5 (x – 1) = 11 The answer checks, the solution is

19 Slide 19 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Clearing fractions from linear equations Solve the linear equation. The solution is 15.

20 Slide 20 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Clearing decimals from linear equations Solve the linear equation. The solution is

21 Equations with No Solutions or Infinitely Many Solutions An equation that is always true is called an identity and an equation that is always false is called a contradiction. Slide 21

22 Slide 22 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Determining numbers of solutions Determine whether the equation has no solutions, one solution, or infinitely many solutions. a.10 – 8x = 2(5 – 4x) b.7x = 9x + 2(12 – x) c.6x = 4(x + 5) a.10 – 8x = 2(5 – 4x) 10 – 8x = 10 – 8x – 8x = – 8x 0 = 0 Because the equation 0 = 0 is always true, it is an identity and there are infinitely many solutions.

23 Slide 23 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE continued b.7x = 9x + 2(12 – x) Because the equation 0 = 24 is always false, it is a contradiction and there are no solutions. c. 6x = 4(x + 5) 7x = 9x + 24 – 2x 7x = 7x = 24 6x = 4x x = 20 x = 10 Thus there is one solution.

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25 Slide 25 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

26 Introduction to Problem Solving Steps for Solving a Problem Percent Problems Distance Problems Other Types of Problems 2.3

27 Slide 27 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

28 Slide 28 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Translating sentences into equations Translate the sentence into an equation using the variable x. Then solve the resulting equation. a.Six times a number plus 7 is equal to 25. b.The sum of one-third of a number and 9 is 18. c.Twenty is 8 less than twice a number. a. 6x + 7 = 25 6x = 18 b.c. 20 = 2x 8 28 = 2x 14 = x

29 Slide 29 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Solving a number problem The sum of three consecutive integers is 126. Find the three numbers. Step 1: Assign a variable to an unknown quantity. n + (n + 1) + (n + 2) = 126 n + 1: next integer n + 2: largest integer Step 2: Write an equation that relates these unknown quantities. n: smallest of the three integers

30 Slide 30 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE continued Step 3: Solve the equation in Step 2. n + (n + 1) + (n + 2) = 126 Step 4: Check your answer. The sum of these integers is = 126. The answer checks. (n + n + n) + (1 + 2) = 126 3n + 3 = 126 3n = 123 n = 41 So the numbers are 41, 42, and 43.

31 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 31

32 Slide 32 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Converting percent notation Convert each percentage to fraction and decimal notation. a.47%b. 9.8%c. 0.9% a. Fraction Notation: Decimal Notation: b. Fraction Notation: Decimal Notation: c. Fraction Notation: Decimal Notation:

33 Slide 33 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Converting to percent notation Convert each real number to a percentage. a.0.761b. c. 6.3 a. Move the decimal point two places to the right and then insert the % symbol to obtain = 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%.

34 Slide 34 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Calculating a percent increase The price of an oil change for an automobile increased from $15 to $24. Calculate the percent increase.

35 Slide 35 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Solving a percent problem 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? Step 1: Assign a variable. x: the amount sold in the first quarter. Step 2: Write an equation. x + 2.4x = 85

36 Slide 36 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE continued Step 3: Solve the equation in Step 2. 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 = 85. In the first quarter the salesman sold 25 cars. x + 2.4x = x = 85 x = 25

37 Slide 37 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Solving a distance problem 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. Step 1: Let r represent the trucks 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

38 Slide 38 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE continued Step 3: Solve the equation. The speed of the truck is 56 miles per hour. Step 4: d = rt The answer checks.

39 Slide 39 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Mixing chemicals 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? ConcentrationSolution Amount (milliliters) Pure alcohol x0.4x 0.36x x + 36 Step 1: Assign a variable.x: milliliters of 40% x + 100: milliliters of 36% Step 2: Write an equation. 0.28(100) + 0.4x = 0.36(x + 100)

40 Slide 40 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE continued Step 3: Solve the equation in Step mL of 40% alcohol solution was added to the 100 mL of the 28% solution. 0.28(100) + 0.4x = 0.36(x + 100) 28(100) + 40x = 36(x + 100) x = 36x x = x = 800 x = 200

41 Slide 41 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE continued 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) = = 108 of pure alcohol. The concentration is or 36%.

42 Slide 42 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

43 Formulas Basic Concepts Formulas from Geometry Solving for a Variable Other Formulas 2.4

44 Slide 44 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Calculating mileage of a trip A tourist starts a trip with a full tank of gas and an odometer that reads 59,478 miles. At the end of the trip, it takes 8.6 gallons of gas to fill the tank, and the odometer reads 59,715 miles. Find the gas mileage for the car. The distance traveled is 59,715 – 59, 478 = 237 miles and the number of gallons used is G = 8.6. Thus,

45 Slide 45 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Calculating area of a region A residential lot is shown. Find the area of this lot. The area of the rectangle: 205 ft 372 ft 116 ft The area of the triangle: Total area = 76, ,576 = 97,836 square feet.

46 Slide 46 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Finding angles in a triangle In a triangle, the smaller angles are equal in measure and are one-third of the largest angle. Find the measure of each angle. Let x represent the measure of each of the two smaller angles. Then the measure of the largest angle is 3x, and the sum of the measures of the three angles is given by The measure of the largest angle is 3x, thus 36 3 = 108°. The measure of the three angles are 36°, 36°, and 108°.

47 Slide 47 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Finding the volume and surface area of a box Find the volume and the surface area of the box shown. V = LHW The volume of the box is 12 cm 5 cm 6 cm V = V = 360 cm 3 The surface area of the box is

48 Slide 48 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Calculating the volume of a soup can A cylindrical soup can has a radius of 2 ½ inches and a height of inches. Find the volume of the can. h r

49 Slide 49 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Solving for a variable Solve each equation for the indicated variable. a.b. a.b.

50 Other Formulas To calculate a students GPA, the number of credits earned with a grade of A, B, C, D, and F must be known. If a, b, c, d, and f represent these credit counts respectively, then Slide 50

51 Slide 51 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Calculating a students GPA A student has earned 18 credits of A, 22 credits of B, 8 credits of C and 4 credits of D. Calculate the students GPA to the nearest hundredth. Let a = 18, b = 22, c = 8, d = 4 and f = 0 The students GPA is 3.04.

52 Slide 52 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Converting temperature The formula is used to convert degrees Fahrenheit to degrees Celsius. Use this formula to convert 23°F to an equivalent Celsius temperature. = 5°C

53 Slide 53 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

54 Linear Inequalities Solutions and Number Line Graphs The Addition Property of Inequalities The Multiplication Property of Inequalities Applications 2.5

55 Slide 55 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solutions and Number Line Graphs A linear inequality results whenever the equals sign in a linear equation is replaced with any one of the symbols, or. x > 5, 3x + 4 < 0, 1 – y 9 A solution to an inequality is a value of the variable that makes the statement true. The set of all solutions is called the solution set.

56 Slide 56 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Graphing inequalities on a number line Use a number line to graph the solution set to each inequality. a. b. c. d.

57 Each number line graphed on the previous slide represents an interval of real numbers that corresponds to the solution set to an inequality. Brackets and parentheses can be used to represent the interval. For example: Slide 57 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Interval Notation

58 Slide 58 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Writing solution sets in interval notation Write the solution set to each inequality in interval notation. a. b.

59 Slide 59 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Checking possible solutions Determine whether the given value of x is a solution to the inequality.

60 The Addition Property of Inequalities Slide 60 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

61 Slide 61 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Applying the addition property of inequalities Solve each inequality. Then graph the solution set. a. x – 2 > 3b x 6 + x a. x – 2 > 3 x – > x > 5 b x 6 + x 4 + 2x – x 6 + x – x 4 + x 6 4 – 4 + x 6 – 4 x 2

62 Slide 62 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Multiplication Property of Inequalities

63 Slide 63 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Applying the multiplication property of inequalities Solve each inequality. Then graph the solution set. a. 4x > 12b. a. 4x > 12b.

64 Slide 64 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Applying both properties of inequalities Solve each inequality. Write the solution set in set-builder notation. a. 4x – 8 > 12b. a. 4x – 8 > 12b.

65 Slide 65 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Applications To solve applications involving inequalities, we often have to translate words to mathematical statements.

66 Slide 66 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Translating words to inequalities Translate each phrase to an inequality. Let the variable be x. a. A number that is more than 25. b. A height that is at least 42 inches. x > 25 x 42

67 Slide 67 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Calculating revenue, cost, and profit For a snack food company, the cost to produce one case of snacks is $135 plus a one-time fixed cost of $175,000 for research and development. The revenue received from selling one case of snacks is $250. a. Write a formula that gives the cost C of producing x cases of snacks. b. Write a formula that gives the revenue R from selling x cases of snacks. C = 135x + 175,000 R = 250x

68 Slide 68 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Calculating revenue, cost, and profit For a snack food company, the cost to produce one case of snacks is $135 plus a one-time fixed cost of $175,000 for research and development. The revenue received from selling one case of snacks is $250. c. Profit equals revenue minus cost. Write a formula that calculates the profit P from selling x cases of snacks. d. How many cases need to be sold to yield a positive profit? P = R – C = 250x – (135x + 175,000) = 115x – 175, x – 175,000 > 0 115x > 175,000 x > Must sell at least 1522 cases.


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