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

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

Solution EXAMPLE Solving and checking a solution Solve the equation −7 + x = 12 and then check the solution. Solution Isolate x by adding 7 to each side. −7 + x = 12 −7 + 7 + x = 12 + 7 0 + x = 19 Check: −7 + x = 12 −7 + 19 = 12 ? x = 19 12 = 12 The solution is 19. The answer checks. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5

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

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

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

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2.2 Linear Equations Basic Concepts Solving Linear Equations Applying the Distributive Property Clearing Fractions and Decimals Equations with No Solutions or Infinitely Many Solutions Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

using the distributive property to clear parentheses, 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. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution EXAMPLE 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. 5x3 + 9 = 0 c. Solution a. The equation is linear because it is in the form ax + b = 0 with a = 9 and b = 7. 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. 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. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 13

Solution EXAMPLE Using a table to solve an equation Complete the table for the given values of x. Then solve the equation 4x – 6 = −2. x −3 −2 −1 1 2 3 4x − 6 −18 Solution x −3 −2 −1 1 2 3 4x − 6 −18 −14 −10 −6 6 From the table 4x – 6 = −2 when x = 1. Thus the solution to 4x – 6 = −2 is 1. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 14

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Solution EXAMPLE Solving linear equations Solve each linear equation. a. 12x − 15 = 0 b. 3x + 19 = 5x + 5 Solution a. 12x − 15 = 0 b. 3x + 19 = 5x + 5 3x − 3x + 19 = 5x − 3x + 5 12x − 15 + 15 = 0 + 15 19 = 2x + 5 12x = 15 19 − 5 = 2x + 5 − 5 14 = 2x Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 16

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

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

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

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

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

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

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

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Introduction to Problem Solving 2.3 Introduction to Problem Solving Steps for Solving a Problem Percent Problems Distance Problems Other Types of Problems Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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Solution EXAMPLE Translating sentences into equations 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 b. c. 20 = 2x − 8 6x = 18 28 = 2x 14 = x Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 28

Solution EXAMPLE Solving a number problem 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 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 29

Step 3: Solve the equation in Step 2. EXAMPLE continued 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. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 30

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

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

Solution EXAMPLE Converting to percent notation 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%. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 33

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

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

Step 3: Solve the equation in Step 2. EXAMPLE continued 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. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 36

Solution EXAMPLE 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. 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 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 37

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

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

Step 3: Solve the equation in Step 2. EXAMPLE continued 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. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 40

Step 4: Check your answer. 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) = 80 + 28 = 108 of pure alcohol. The concentration is or 36%. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 41

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2.4 Formulas Basic Concepts Formulas from Geometry Solving for a Variable Other Formulas Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution EXAMPLE 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. Solution The distance traveled is 59,715 – 59, 478 = 237 miles and the number of gallons used is G = 8.6. Thus, Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 44

Solution EXAMPLE Calculating area of a region A residential lot is shown. Find the area of this lot. 205 ft 372 ft 116 ft Solution The area of the rectangle: The area of the triangle: Total area = 76,260 + 21,576 = 97,836 square feet. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 45

Solution EXAMPLE 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. Solution 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°. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 46

Solution EXAMPLE Finding the volume and surface area of a box Find the volume and the surface area of the box shown. 12 cm 5 cm 6 cm Solution The volume of the box is V = LHW V = 12 ∙ 6 ∙ 5 V = 360 cm3 The surface area of the box is Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 47

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

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

Other Formulas To calculate a student’s 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

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

Solution EXAMPLE 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. Solution = −5°C Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 52

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2.5 Linear Inequalities Solutions and Number Line Graphs The Addition Property of Inequalities The Multiplication Property of Inequalities Applications 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. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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

Interval Notation 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: Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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

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

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

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

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

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

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

Translating words to inequalities 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 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Calculating revenue, cost, and profit 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 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Calculating revenue, cost, and profit 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 – 175,000 115x – 175,000 > 0 115x > 175,000 x > 1521.74 Must sell at least 1522 cases. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley