1. 1 Equations and Inequalities © 2008 Pearson Addison-Wesley. All rights reserved Sections 1.1–1.4

2. 1.1 Linear Equations 1.2 Applications and Modeling with Linear Equations 1.3 Complex Numbers 1.4 Quadratic Equations Equations and Inequalities 1

3. Linear Equations 1.1 Basic Terminology of Equations ▪ Solving Linear Equations ▪ Identities, Conditional Equations, and Contradictions ▪ Solving for a Specified Variable (Literal Equations)

4. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-4 Solve. 1.1 Example 1 Solving a Linear Equation (page 85) Solution set: {6} Distributive property Combine terms. Add 4 to both sides. Add 12x to both sides. Combine terms. Divide both sides by 4.

5. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-5 Solve. 1.1 Example 2 Clearing Fractions Before Solving a Linear Equation (page 85) Solution set: {–10} Multiply by 10, the LCD of all the fractions. Distributive property Combine terms. Add –4s and –6 to both sides. Combine terms. Divide both sides by –6.

6. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-6 Decide whether the equation is an identity, a conditional equation, or a contradiction. Give the solution set. 1.1 Example 3(a) Identifying Types of Equations (page 86) This is a conditional equation. Solution set: {11} Add –4x and 9 to both sides. Combine terms. Divide both sides by 2.

7. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-7 Decide whether the equation is an identity, a conditional equation, or a contradiction. Give the solution set. 1.1 Example 3(b) Identifying Types of Equations (page 86) This is a contradiction. Solution set: ø Distributive property Subtract 14x from both sides.

8. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-8 Decide whether the equation is an identity, a conditional equation, or a contradiction. Give the solution set. 1.1 Example 3(c) Identifying Types of Equations (page 86) This is an identity. Solution set: {all real numbers} Distributive property Combine terms. Add x and –3 to both sides.

9. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-9 Solve for the specified variable. 1.1 Example 4 Solving for a Specified Variable (page 87) Solve for the specified variable. (a) d = rt, for t (b), for k Divide both sides by r. Factor out k. Divide both sides by

10. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-10 Solve for the specified variable. 1.1 Example 4 Solving for a Specified Variable (cont.) Solve for the specified variable. (c), for y Distributive property Subtract 8y and 8 from both sides. Divide both sides by 3.

11. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-11 Caden borrowed $2580 to buy new kitchen appliances for his home. He will pay off the loan in 9 months at an annual simple interest rate of 6.0%. How much interest will he pay? 1.1 Example 5 Applying the Simple Interest Formula (page 88) Caden will pay $116.10 in interest.

12. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-12 1.2 Solving Applied Problems ▪ Geometry Problems ▪ Motion Problems ▪ Mixture Problems ▪ Modeling with Linear Equations Applications and Modeling with Linear Equations

13. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-13 1.2 Example 1 Find the Dimensions of a Square (page 91) The length of a rectangle is 2 in. more than the width. If the length and width are each increased by 3 in., the perimeter of the new rectangle will be 4 in. less than 8 times the width of the original rectangle. Find the dimensions of the original rectangle. Assign variables: Let x = the length of the original rectangle. Then, x − 2 = the width of the original rectangle. x + 3 = the length of the new rectangle (x − 2) + 3 = x + 1 = the width of the new rectangle.

14. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-14 1.2 Example 1 Find the Dimensions of a Square (cont.) The perimeter of the new rectangle is The perimeter of the new rectangle is 4 in. less than 8 times the width of the original rectangle, so we have

15. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-15 1.2 Example 1 Find the Dimensions of a Square (cont.) Distributive property. Combine terms. Add –4x and 20 to both sides. Divide both sides by 4. The length of the original rectangle is 7 in. The width of the original rectangle is 7 – 2 = 5 in.

16. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-16 1.2 Example 2 Solving a Motion Problem (page 92) Krissa drove to her grandmother’s house. She averaged 40 mph driving there. She was able to average 48 mph returning, and her driving time was 1 hr less. What is the distance between Krissa’s house and her grandmother’s house? Let x = the distance between Krissa’s house and her grandmother’s house Use d = rt

17. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-17 1.2 Example 2 Solving a Motion Problem (cont.) The driving time returning was 1 hour less than the driving time going, so Multiply by 40·48. Distributive property. Subtract 48x. Divide by –8. The distance from Krissa’s home to her grandmother’s home is 240 mi.

18. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-18 1.2 Example 3 Solving a Mixture Problem (page 93) How many liters of a 25% anti-freeze solution should be added to 5 L of a 10% solution to obtain a 15% solution? Let x = the amount of 25% solution The number of gallons of pure antifreeze in the 25% solution plus the number of gallons of pure antifreeze in the 10% solution must equal the number of gallons of pure antifreeze in the 15% solution.

19. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-19 1.2 Example 3 Solving a Mixture Problem (cont.) Create a table to show the relationships in the problem. Write an equation:

20. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-20 1.2 Example 3 Solving a Mixture Problem (cont.) 2.5 liters of the 25% solution should be added. Distributive property. Subtract.15x and.5. Divide by.1.

21. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-21 1.2 Example 4 Solving an Investment Problem (page 94) Last year, Owen earned a total of $1456 in interest from two investments. He invested a total of $28,000, part at 4.8% and the rest at 5.5%. How much did he invest at each rate? Let x = amount invested at 4.8%. Then 28,000 − x = amount invested at 5.5%.

22. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-22 1.2 Example 4 Solving an Investment Problem (cont.) Create a table to show the relationships in the problem. The amount of interest from the 4.8% account plus the amount of interest from the 5.5% account must equal the total amount of interest.

23. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-23 1.2 Example 4 Solving an Investment Problem (cont.) Owen invested $12,000 at 4.8% and $28,000 − $12,000 = $16,000 at 5.5%. Distributive property Combine terms. Subtract 1540. Divide by –.007.

24. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-24 1.2 Example 5 Modeling the Prevention of Indoor Pollutants (page 95) A range hood removes contaminants at a rate of F liters of air per second. The percent P of contaminants that are also removed from the surrounding air can be modeled by the linear equation where 10 ≤ F ≤ 75. What flow F must a range hood have to remove 70% of the contaminants from the air?

25. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-25 1.2 Example 5 Modeling the Prevention of Indoor Pollutants (cont.) The flow rate must be approximately 59.26 L per second to remove 70% of the contaminants. Substitute P = 70 into and solve for F. Subtract 7.18. Divide by 1.06.

26. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-26 1.2 Example 6(a) Modeling Health Care Costs (page 96) The projected per capita health care expenditures in the United States, where y is in dollars, x years after 2000, are given by the linear equation What were the per capita health care expenditures in 2007? Let x = 2007 – 2000 = 7, then find the value of y. In 2007, per capita health care expenditures were $7440.

27. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-27 1.2 Example 6(b) Modeling Health Care Costs (page 96) When will the per capita expenditures reach $8220? Let y = 8220, then solve for x. Per capita health care expenditures are projected to reach $8220 nine years after 2000, or in 2009. Subtract 4710. Divide by 390.

28. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-28 Complex Numbers 1.3 Basic Concepts of Complex Numbers ▪ Operations on Complex Numbers

29. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-29 1.3 Example 1 Writing √– a as i √ a (page 104) Write as the product of a real number and i. (a) (b) (c)

30. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-30 1.3 Example 2 Finding Products and Quotients Involving Negative Radicands (page 105) Multiply or divide. Simplify each answer. (a) (b)

31. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-31 1.3 Example 2 Finding Products and Quotients Involving Negative Radicands (cont.) Multiply or divide. Simplify each answer. (c) (d)

32. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-32 1.3 Example 3 Simplifying a Quotient Involving a Negative Radicand (page 105) Write in standard form a + bi. Factor.

33. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-33 1.3 Example 4 Adding and Subtracting Complex Numbers (page 106) Find each sum or difference. (a) (4 – 5i) + (–5 + 8i) = –1 + 3i = [4 + (–5)] + (–5i + 8i) (b) (–6 + 3i) + (12 – 9i)= 6 – 6i (c) (–10 + 7i) – (5 – 3i) = –15 + 10i (d) (15 – 8i) – (–10 + 4i)= 25 – 12i = (–10 – 5) + [7i + (3i)]

34. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-34 1.3 Example 5 Multiplying Complex Numbers (page 107) Find each product. (a) (5 + 3i)(2 – 7i) (b) (4 – 5i) 2 = 5(2) + (5)(–7i) + (3i)(2) + (3i)(–7i) = 10 – 35i + 6i – 21i 2 = 10 – 29i – 21(–1) = 31 – 29i = 4 2 – 2(4)(5i) + (5i) 2 = 16 – 40i + 25i 2 = 16 – 40i + 25(–1) = –9 – 40i

35. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-35 1.3 Example 5 Multiplying Complex Numbers (cont.) Find the product. (c) (9 – 8i)(9 + 8i)= 9 2 – (8i) 2 = 81 – 64i 2 = 81 – 64(–1) = 81 + 64 = 145 or 145 + 0i This screen shows how the TI–83/84 Plus displays the results in this example.

36. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-36 1.3 Example 6 Simplifying Powers of i (page 107) Simplify each power of i. Write the given power as a product involving or. (a)(b) (a)(b)

37. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-37 1.3 Example 7(a) Dividing Complex Numbers (page 108) Write in standard form a + bi. Multiply the numerator and denominator by the complex conjugate of the denominator. i 2 = –1 Combine terms. Lowest terms; standard form Multiply.

38. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-38 1.3 Example 7(b) Dividing Complex Numbers (page 108) Write in standard form a + bi. Multiply the numerator and denominator by the complex conjugate of the denominator. –i 2 = 1 Lowest terms; standard form Multiply. This screen shows how the TI–83/84 Plus displays the results in this example.

39. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-39 Quadratic Equations 1.4 Solving a Quadratic Equation ▪ Completing the Square ▪ The Quadratic Formula ▪ Solving for a Specified Variable ▪ The Discriminant

40. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-40 1.4 Example 1 Using the Zero-Factor Property (page 111) Solve. Factor. Set each factor equal to 0 and then solve for x. or

41. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-41 1.4 Example 1 Using the Zero-Factor Property (cont.) Now check. Solution set:

42. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-42 1.4 Example 2 Using the Square Root Property (page 112) Solve each quadratic equation. Generalized square root property (a) (c) (b)

43. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-43 1.4 Example 3 Using the Method of Completing the Square, a = 1 (page 113) Solve by completing the square. Rewrite the equation so that the constant is alone on one side of the equation. Square half the coefficient of x, and add this square to both sides of the equation.

44. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-44 1.4 Example 3 Using the Method of Completing the Square, a = 1 (cont.) Factor the resulting trinomial as a perfect square and combine terms on the other side. Use the square root property to complete the solution. Solution set:

45. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-45 1.4 Example 4 Using the Method of Completing the Square, a ≠ 1 (page 113) Solve by completing the square. Rewrite the equation so that the constant is alone on one side of the equation. Square half the coefficient of x, and add this square to both sides of the equation. Divide both sides of the equation by a, 4.

46. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-46 1.4 Example 4 Using the Method of Completing the Square, a ≠ 1 (cont.) Factor the resulting trinomial as a perfect square and combine terms on the other side. Use the square root property to complete the solution. Solution set:

47. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-47 1.4 Example 5 Using the Quadratic Formula (Real Solutions) (page 115) Solve. a = 1, b = 6, c = –3 Write the equation in standard form. Quadratic formula

48. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-48 1.4 Example 5 Using the Quadratic Formula (Real Solutions) (cont.) Solution set:

49. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-49 1.4 Example 6 Using the Quadratic Formula (Nonreal Complex Solutions) (page 115) Solve. a = 4, b = –3, c = 5 Write the equation in standard form. Quadratic formula

50. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-50 1.4 Example 6 Using the Quadratic Formula (Nonreal Complex Solutions) (page 115) Solution set:

51. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-51 1.4 Example 7 Solving a Cubic Equation (page 116) Solve. Factor as a difference of cubes. Zero-factor property or Quadratic formula with a = 1, b = 5, c = 25 Solution set:

52. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-52 1.4 Example 8(a) Solving for a Quadratic Variable in a Formula (page 116) Solve for r. Use ± when taking square roots. Goal: Isolate r. Multiply by 3. Divide by πh. Square root property Rationalize the denominator. Simplify.

53. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-53 1.4 Example 8(b) Solving for a Quadratic Variable in a Formula (page 116) Solve for y. Use ± when taking square roots. Write in standard form. Use the quadratic formula with a = 2m, b = –n, c = –3p. Simplify.

54. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-54 1.4 Example 9(a) Using the Discriminant (page 118) Determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. a = 4, b = –12, c = 9 There is one distinct rational solution.

55. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-55 1.4 Example 9(b) Using the Discriminant (page 118) Determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. a = 3, b = 1, c = 5 There are two distinct nonreal complex solutions. Write in standard form.

56. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-56 1.4 Example 9(c) Using the Discriminant (page 118) Determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. a = 2, b = –6, c = –7 There are two distinct irrational solutions. Write in standard form.