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Week 2 Due for this week… Homework 2 (on MyMathLab – via the Materials Link) Monday night at 6pm. Read Chapter 3 and 8.1, 8.2 Do the MyMathLab Self-Check for week 2. Learning team planning for week 5. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 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 The Linear Equation It has one variable : x ax + b = 0 Where a and b are real numbers and a is not equal to 0. A photo album of linear equations 2x+3=0 x+8=0 3x=7 2x+5=9-5x 3+5(x-1)=-7+x 3=8x+5 Etc. (say cheese!) How to legally play with linear equations. Most of the magic we work with are really obvious tricks. Remember what happens if we multiply by things that equal 1? e.g. 5/5 10/10 ? Adding the SAME THING to both sides of an equation does not change the equation… The Addition Property of Equality So if you start with a = b You don’t change anything if you take a new number or variable and add it to both sides! a+c = b+c The GOAL We want x all by itself on one side and the rest of the junk on the other (so we can use our calculator or fingers). The technical term for this is: solving for x Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 9 EXAMPLE Using the addition property of equality Solve each equation. a. x + 21 = 9 b. n – 8 = 17 Solution a. x + 21 = 9 x + 21 + (−21) = 9 + (−21) b. n – 8 = 17 n – 8 + 8 = 17 + 8 x + 0 = −12 n + 0 = 25 x = −12 n = 25 The solution is −12. The solution is 25. ** Try exercises 17-26 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 10 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 x = 19 The solution is 19. Check: −7 + x = 12 ? −7 + 19 = 12 12 = 12 The answer checks. ** Try exercises 27-34 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 11 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 12 EXAMPLE Using the multiplication property of equality Solve each equation. 1 a. x 8 6 b. −11a = 33 Solution a. 1 x 8 6 1 6 x 86 6 1 x 48 x 48 The solution is 48. b. −11a = 33 11a 33 11 11 a 3 The solution is −3. ** Try exercises 41-48 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 13 EXAMPLE Solving and checking a solution 2 2 Solve the equation x and then check the 5 9 solution. Solution 2 2 x 5 9 9 2 2 9 x 2 5 9 2 9 1 x 5 9 x 5 9 The solution is . 5 ** Try exercises 49-58 Check: 2 2 x 5 9 2 ? 2 9 5 9 5 2 2 5 5 The answer checks. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley- Slide 14 EXAMPLE 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. Solution a. 146 + x = 153 b. 146 + x = 153 146 + (−146) + x = 153 − 146 x=7 The cat weighs 7 lbs. ** Try exercises 61-70 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 15 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 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 Slide 17 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 Solution b. 5x3 +9=0 5 c. 7 0 x 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. ** Try exercises 13-26 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 18 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 4x − 6 −18 −1 0 1 2 3 Solution −3 x 4x − 6 −2 −1 0 1 2 3 −18 −14 −10 −6 −2 2 6 From the table 4x – 6 = −2 when x = 1. Thus the solution to 4x – 6 = −2 is 1. ** Try exercises 27-30 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 19 Slide 20 Rules for Happiness First do whatever adding and subtracting you can do… THEN do whatever multiplication and division you can do. THEN you should be done! EXAMPLE Solving linear equations Solve each linear equation. a. 12x − 15 = 0 b. 3x + 19 = 5x + 5 Solution a. 12x − 15 = 0 12x − 15 + 15 = 0 + 15 12x = 15 12 x 15 12 12 15 5 x 12 4 ** Try exercises 39-42 3x + 19 = 5x + 5 b. 3x − 3x + 19 = 5x − 3x + 5 19 = 2x + 5 19 − 5 = 2x + 5 − 5 14 = 2x 14 2x 2 2 7x Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 22 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 6 x 16 6 6 16 x 6 8 3 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 23 EXAMPLE Check continued x + 5 (x – 1) = 11 8 8 5 1 11 3 3 33 11 3 8 8 3 5 11 3 3 3 8 5 5 11 3 3 8 25 11 3 3 11 11 The answer checks, the 8 solution is . 3 ** Try exercises 43-52 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 24 EXAMPLE Clearing fractions from linear equations Solve the linear equation. 1 1 x x 5 2 6 Solution 1 1 x x 5 2 6 1 1 6 x x 56 6 2 3x x 30 2x 30 x 15 The solution is 15. ** Try exercises 55-58 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 25 EXAMPLE Clearing decimals from linear equations Solve the linear equation. 5.3 0.8x 7 Solution 5.3 0.8x 7 10 5.3 0.8x 7 10 53 8x 70 53 53 8x 70 53 8x 17 8 x 17 8 8 ** Try exercises 53-54 17 x 8 17 The solution is . 8 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 26 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 27 EXAMPLE 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) Solution a. 10 – 8x = 2(5 – 4x) 10 – 8x = 10 – 8x – 8x = – 8x Because the equation 0 = 0 is always true, it is an identity and there are infinitely many solutions. 0=0 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 28 EXAMPLE continued b. 7x = 9x + 2(12 – x) 7x = 9x + 24 – 2x 7x = 7x + 24 c. 6x = 4(x + 5) 6x = 4x + 20 2x = 20 x = 10 0 = 24 Because the equation 0 = 24 is always false, it is a contradiction and there are no solutions. Thus there is one solution. ** Try exercises 63-72 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 29 Slide 30 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 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 32 EXAMPLE 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. Solution a. 6x + 7 = 25 6x = 18 6 x 18 6 6 x 3 b. 1 x 9 18 3 1 x9 3 x 27 c. 20 = 2x − 8 28 = 2x ** Try exercises 11-18 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 28 2 x 2 2 14 = x Slide 33 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 34 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. ** Try exercises 19-28 The answer checks. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 35 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 36 EXAMPLE Converting percent notation Convert each percentage to fraction and decimal notation. a. 47% b. 9.8% c. 0.9% Solution 47 . 100 Decimal Notation: 47% 0.47. 9.8 98 49 2 49 b. Fraction Notation: 9.8% . 100 1000 500 2 500 Decimal Notation: 9.8% 0.098. 0.9 9 c. Fraction Notation: 0.9% 100 1000 ** Try exercises 35-42 Decimal Notation: 0.9% 0.009. a. Fraction Notation: 47% Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 37 EXAMPLE Converting to percent notation Convert each real number to a percentage. a. 0.761 2 b. 5 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% 2 2 b. 0.40, so 40%. 5 5 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%. ** Try exercises 43-54 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 38 EXAMPLE Calculating a percent increase The price of an oil change for an automobile increased from $15 to $24. Calculate the percent increase. Solution new value - old value 100 old value 24 - 15 100 60% 15 ** Try exercises 55-56 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 39 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 40 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. ** Try exercises 57-60 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 41 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 9 252 r 2 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 42 EXAMPLE continued Step 3: Solve the equation. 9 252 r 2 2 2 9 252 r 9 9 2 56 r The speed of the truck is 56 miles per hour. Step 4: d = rt 56 9 252 miles 2 The answer checks. ** Try exercises 71-74 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 43 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% Step 2: Write an equation. Concentration Solution Amount (milliliters) Pure alcohol 0.28 100 28 0.40 x 0.4x 0.36 x + 100 0.36x + 36 0.28(100) + 0.4x = 0.36(x + 100) Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 44 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 45 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 108 0.36, or 36%. 300 ** Try exercises 79-80 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 46 2.4 Formulas Basic Concepts Formulas from Geometry Solving for a Variable Other Formulas Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 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, D 237 27.6 miles per gallon. M G 8.6 ** Try exercises 81-82 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 48 EXAMPLE Calculating area of a region A residential lot is shown. Find the area of this lot. 205 ft 372 ft Solution 116 ft The area of the rectangle: AR LW AR 372 205 AR 76, 260 square feet The area of the triangle: AT 12 bh AT 12 116 372 AT 21,576 square feet Total area = 76,260 + 21,576 = 97,836 square feet. ** Try exercises 19-29 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 49 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 x x 3x 180 5 x 180 5 x 180 5 5 x 36 The measure of the largest angle is 3x, thus 36 ∙ 3 = 108°. The measure of the three angles are 36°, 36°, and 108°. ** Try exercises 31-38 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 50 EXAMPLE Finding the volume and surface area of a box Find the volume and the surface area of the box shown. 6 cm 5 cm Solution 12 cm The volume of the box is V = LHW V = 12 ∙ 6 ∙ 5 V = 360 cm3 The surface area of the box is S 2LW 2WH 2LH S 2(12)(5) 2(5)(6) 2(12)(6) S 120 60 144 ** Try exercises 41-46 S 324 square centimeters Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 51 EXAMPLE Calculating the volume of a soup can A cylindrical soup can has a radius of 2 ½ inches and a height of 5 85 inches. Find the volume of the can. ∙ r h Solution V r 2h 2 5 45 V 2 8 1125 V 32 V 110.45 cubic inches ** Try exercises 51 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 52 EXAMPLE Solving for a variable Solve each equation for the indicated variable. yz a. 3 x for z 5 b. np nm nq for p Solution yz 5 15 x y z a. 3 x 15 x y z z 15 x y b. np nm nq for p np nq nm np n(q m) n( q m) p n p qm ** Try exercises 53-64 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 53 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 4a 3b 2c d GPA . abcd f Slide 54 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 4 18 3 22 2 8 4 158 3.04 GPA 18 22 8 4 0 52 The student’s GPA is 3.04. ** Try exercises 69-72 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 55 EXAMPLE Converting temperature The formula C 95 F 32 is used to convert degrees Fahrenheit to degrees Celsius. Use this formula to convert 23°F to an equivalent Celsius temperature. Solution 5 C F 32 9 5 C 23 32 9 5 C 9 = −5°C 9 ** Try exercises 73-80 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 56 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 Slide 58 EXAMPLE Graphing inequalities on a number line Use a number line to graph the solution set to each inequality. a. x 1 b. x 1 c. x 5 d. x 2 ** Try exercises 13-20 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 59 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: x 1 (1, ) x 1 [1, ) Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 60 EXAMPLE Writing solution sets in interval notation Write the solution set to each inequality in interval notation. a. x 6 b. y 2 (6, ) (, 2] ** Try exercises 27-32 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 61 EXAMPLE Checking possible solutions Determine whether the given value of x is a solution to the inequality. 4 x 2 8, x7 ? 4 x 2 <8 ? 4(7) 2 8 ? 28 2 8 ? 26 8 False ** Try exercises 33-42 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 62 The Addition Property of Inequalities Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 63 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 x–2+2>3+2 x>5 b. 4 + 2x ≤ 6 + x 4 + 2x – x ≤ 6 + x – x 4+x≤6 4–4+x≤6–4 x≤2 ** Try exercises 51-58 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 64 The Multiplication Property of Inequalities Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 65 Applying the multiplication property of inequalities EXAMPLE Solve each inequality. Then graph the solution set. 1 a. 4x > 12 b. 2 x 4 Solution a. 4x > 12 4 x 12 4 4 x 3 1 b. 2 x 4 1 4(2) (4) x 4 8 x ** Try exercises 59-66 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 66 EXAMPLE Applying both properties of inequalities Solve each inequality. Write the solution set in set-builder notation. a. 4x – 8 > 12 b. 4 3x 4x 5 Solution a. 4x – 8 > 12 4x 8 8 12 8 4x 20 x 5 {x | x 5} b. 4 3x 4x 5 4 3x 3x 4x 5 3x 4 x 5 4 5 x 5 5 9 x {x | x 9} ** Try exercises 71-100 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 67 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 Slide 68 EXAMPLE Translating words to inequalities Translate each phrase to an inequality. Let the variable be x. a. A number that is more than 25. x > 25 b. A height that is at least 42 inches. x ≥ 42 ** Try exercises 101-108 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 69 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. C = 135x + 175,000 b. Write a formula that gives the revenue R from selling x cases of snacks. R = 250x Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 70 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. P=R–C = 250x – (135x + 175,000) = 115x – 175,000 d. How many cases need to be sold to yield a positive profit? 115x – 175,000 > 0 115x > 175,000 x > 1521.74 Must sell at least 1522 cases. ** Try exercises 119-120 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 71 End of week 2 You again have the answers to those problems not assigned Practice is SOOO important in this course. Work as much as you can with MyMathLab, the materials in the text, and on my Webpage. Do everything you can scrape time up for, first the hardest topics then the easiest. You are building a skill like typing, skiing, playing a game, solving puzzles. NEXT TIME: Linear Equations w/2 variables and Graphing + slope and y-intercepts