1. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Chapter 2 Equations, Inequalities, and Problem Solving 2.1 2.2 2.3 2.4 2.5 2.6 2.7

2. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 2.1 Linear Equations in One Variable Objectives: Solve linear equations using properties of equality Solve linear equations that can be simplified by combining like terms Solve linear equations containing fractions or decimals Recognize identities and equations with no solutions Key Vocabulary: Solution, equivalent equation, conditional equation, contradiction, identity

3. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Linear Equations An algebraic equation is a statement in which two expressions have equal value. Solving algebraic equations involves finding values for a variable that make the equation true. Equivalent equations are equations with the same solution set.

4. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 1 Solve for x: 3x + 7 = 16. Check: The solution or the solution set is {3}. 3x + 7 = 16 - 7 - 7

5. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 3 Solve: 4p – 11 – p = 2 + 2p – 20. 3p – 11 = 2p – 18 Combine like terms. p = – 7 Simplify. p – 11 = – 18 Simplify. – 2p – 2p. Subtract 2p from both sides. + 11 + 11 Add 11 to both sides. Of course you should check the solution. Let’s check this one using the sto→ button on the calculator

6. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 3 Solve: 12x + 30 + 8x – 6 = 10. 20x + 24 = 10 Simplify left side. 20x = –14 Simplify both sides. Divide both sides by 20.Simplify both sides. – 24 – 24 Subtract 24 from both sides. Check using the sto→ button on the calculator

7. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 4 Solve: 5(3 + z) – (8z + 9) = – 4z. 15 + 5z – 8z – 9 = – 4z Use distributive property. 6 – 3z = – 4z Simplify left side. 6 + z = 0 Simplify both sides. z = – 6 Simplify both sides. + 4z + 4z Add 4z to both sides. – 6 – 6 Subtract 6 from both sides. Check using the sto→ button on the calculator

8. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Solving a Linear Equation in One Variable 1)Clear the equation of fractions by multiplying both sides of the equation by the least common denominator (LCD) of all denominators in the equation. 2)Use the distributive property to remove grouping symbols such as parentheses. 3)Combine like terms on each side of the equation. 4)Use the addition property of equality to rewrite the equation as an equivalent equation with variable terms on one side and numbers on the other side. 5)Use the multiplication property of equality to isolate the variable. 6)Check the proposed solution in the original equation.

9. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 6 Solve for y: Check using the sto→ button on the calculator -3y 9 = 7y + 30 -30

10. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall A linear equation in one variable that has exactly one solution is called a conditional equation. An equation in one variable that has no solution is called a contradiction. An equation in one variable that has every number (for which the equation is defined) as a solution is called an identity. (infinite solutions)

11. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 8 Solve for x: The equation  7 = 3 is false no matter what value the variable x might have. Thus, the original equation has no solution. This equation is a contradiction. 3x – 7 = 3x + 3 Use distributive property. –7 = 3 Simplify both sides. –3x –3x Subtract 3x from both sides. 3x – 7 = 3(x + 1).

12. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 9 Solve for x: 5x – 5 = 2(x + 1) + 3x – 7. 5x – 5 = 2x + 2 + 3x – 7 Use distributive property. 5x – 5 = 5x – 5 Simplify the right side. Since – 5 = – 5 is a true statement for every value of x, all real numbers are solutions. The solution set is the set of all real numbers. The equation is called an identity. – 5x – 5x Subtract 5x from both sides. – 5 = – 5 Simplify.

13. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Solve: additional ex 54

14. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 2.1 summary Linear Equations in One Variable Objectives: Solve linear equations using properties of equality Solve linear equations that can be simplified by combining like terms Solve linear equations containing fractions or decimals Recognize identities and equations with no solutions Key Vocabulary: Solution, equivalent equation, conditional equation, contradiction, identity

15. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 2.2 An Introduction to Problem Solving Objectives: Write algebraic expressions that can be simplified Apply the steps for problem solving Key Vocabulary: Consecutive integers, complementary, supplementary

16. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Consecutive, even, and odd integers and their representations. Consecutive Integers: Consecutive Even Integers: Consecutive Odd Integers: xx + 1x + 2 x x + 4 xx + 2x + 4 + 1 + 2 + 4 + 2

17. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 1 Write the following as an algebraic expression. Then simplify. The sum of three consecutive odd integers, if x is the first consecutive integer. In words: Translate: first integer plusnext odd integer plusnext odd integer x+(x + 2)+(x + 4) = x + x + 2 + x + 4 = 3x + 6

18. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall General Strategy for Problem Solving 1)UNDERSTAND the problem. During this step, become comfortable with the problem. Some way of doing this are: Read and reread the problem. Propose a solution and check. Construct a drawing. Choose a variable to represent the unknown. 2)TRANSLATE the problem into an equation. 3)SOLVE the equation. 4)INTERPRET the result. Check the proposed solution in the stated problem and state your conclusion.

19. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 3 Find three numbers such that the second is 3 more than twice the first number, and the third number is four times the first number. The sum of the three numbers is 164. If we let x = the first number, then 2x + 3 = the second number 4x = the third number In words: Translate: x + (2x + 3) + 4x = 164 first number added to second number added to third number is164 7x + 3 = 164 7x = 161 x = 23 1 st # = 232 nd # = 49 3 rd # = 92

20. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 6 Kelsey Ohleger was helping her friend Benji Burnstine study for an algebra exam. Kelsey told Benji that her last three latest art history quiz scores are three consecutive even integers whose sum is 264. Help Benji find the scores. x = the first integer. Then x + 2 = the second consecutive even integer x + 4 = the third consecutive even integer. 3x + 6 = 264 3x = 258 x = 86 1 st # = 86 2 nd # = 88 3 rd # = 90

21. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Add on: find the perimeter 10 2 7y – 3 y + 1 ? The problem is miss leading on the system because they put a question mark on the picture which isn’t necessary 20 + Peri = 20 + 2(7y – 3) = 20 + 14y – 6 Peri = 14 + 14y

22. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Add on: In 2010, the population of the country was 34.6 million. This represented an increase in population of 8.1% since 2001. What was the population of the country in 2001? Round to the nearest hundredth of a million. 2001population + 8.1%of(2001population) = 2010population p + 0.081p = 34.6 1.081p = 34.6 p = 32.01 If the total bill for a TV is $456.72 and the tax rate is 8.5%, find the original price of the TV. Round to the nearest cent. $420.94 TV + 0.085TV = 456.72 1.085TV = 456.72

23. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 2.2 summary An Introduction to Problem Solving Objectives: Write algebraic expressions that can be simplified Apply the steps for problem solving Key Vocabulary: Consecutive integers, complementary, supplementary

24. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 2.3 Formulas and Problem Solving Objectives: Solve a formula for a specified variable Use formulas to solve problems

25. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Formulas A formula is an equation that states a known relationship among multiple quantities (has more than one variable in it). A = lw Area of a rectangle = length · width I = PRT Simple Interest = Principal · Rate · Time P = a + b + c Perimeter of a triangle = side a + side b + side c d = rt Distance = rate · time V = lwh Volume of a rectangular solid = length · width · height

26. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 1 & 2 Solve: V = lwh for w. Solve 4y – 5x = 9 for y.

27. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Solving Equations for a Specific Variable 1)Clear the equations of fractions by multiplying each side of the equation by the least common denominator. 2)Use the distributive property to remove grouping symbols such as parentheses. 3)Combine like terms on each side of the equation. 4)Use the addition property of equality to rewrite the equation as an equivalent equation with terms containing the specified variable on one side and all other terms on the other side. 5)Use the distributive property and the multiplication property of equality to isolate the specified variable.

28. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 3 A = P + PRT solve for T. Example 4 A = P + PRT solve for P.

29. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Compound interest formula. A = amount in account after t years P = principal or amount invested t = time in years r = annual rate of interest n = number of times compounded per year For these questions you need a scientific calculator

30. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall June Myers just received an inheritance of $15,000 and plans to place all the money in a savings account that pays 3% compounded quarterly to help her son go to college in 5 years. How much money will be in the account in 5 years? P = r = t = n = Formula: Substitute Example 5 $15,0000.035 years4 times/yr Answer: In 5 years, the account will contain $17,417.76.

31. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 6 The fastest average speed by a cyclist across the continental United States is 15.4 mph, by Pete Penseyres. If he traveled a total distance of about 3107.5 miles at this speed, find his time cycling. Write the time in days, hours, and minutes. (Source: The Guinness Book of World Records) distance formula d = rt d = r = 3107.5 miles 15.4 mph 201.7857 hours 201.7857 ÷ 24 = 8.4077 days 0.4077(24) = 9.7857 hours 0.7857(60) = 47.1429 minutes Answer: Petes cycling time was approximately 8 days, 9 hours, 47 minutes.

32. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 2.3 summary Formulas and Problem Solving Objectives: Solve a formula for a specified variable Use formulas to solve problems

33. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 2.4 Linear Inequalities and Problem Solving Objectives: Graph inequalities Solve linear inequalities Solve problems that can be modeled by linear inequalities

34. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall A solution of an inequality is a value of the variable that makes the inequality a true statement. The solution set of an inequality is the set of all solutions. The inequality x > 2 has a solution set of {x | x > 2|, read as The set of all numbers x such that x is greater than 2

35. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 1 Graph each on a number line. a. x ≥ 3 b. x < –3 c. 0.5 < x ≤ 4 2– 201345– 1– 3– 4– 52– 201345– 1– 3– 4– 52– 201345– 1– 3– 4– 5

36. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Addition Property of Inequality If a, b, and c are real numbers, then a < b and a + c < b + c are equivalent inequalities. Multiplication Property of Inequality If a, b, and c are real numbers and c is positive, then a < b and ac < bc are equivalent inequalities. If a, b, and c are real numbers and c is negative, then a bc are equivalent inequalities.

37. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 2 Solve each and graph the solutions on a number line. x – 4 < 9 + 4 + 4 x < 13 4x + 5 ≥ 3x – 7 – 3x x + 5 ≥ –7 – 5 – 5 x ≥ –12 121011131415– 12– 11– 13– 14– 15

38. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 3 Solve and graph each on a number line. 2– 201345– 1– 3– 4– 52– 201345– 1– 3– 4– 5

39. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 4 Solve and graph on a number line: 3x + 9  5(x – 1) 3x + 9  5x – 5 – 3x – 3x 9  2x – 5 14  2x 7  x + 5 + 5 54678 x ≤ 7 7(x – 2) + x > – 4(5 – x) – 12 7x – 14 + x > – 20 + 4x – 12 8x – 14 > 4x – 32 – 4x – 4x 4x – 14 > –32 + 14 + 14 4x > –18 2– 201345– 1– 3– 4– 5

40. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 7 Solve: 3(x + 4) > 3x + 5. 3(x + 4) > 3x + 5 3x + 12 > 3x + 5 Distribute on the left side. – 3x – 3x Subtract 3x from both sides. 12 > 5 Simplify. 12 > 5 is a true statement for all values of x, so this inequality and the original inequality are true for all numbers. All real numbers are solutions.

41. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 8 A salesperson earns $800 per month plus a commission of 25% of sales. Find the minimum amount of sales needed to receive a total income of at least $1800 per month. Let x = amount of sales. 800+ 25% commission of sales ≥1800 Answer: The minimum amount of sales needed for the salesperson to earn at least $1800 per month is $4000 per month.

42. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 9 In the United States, the annual consumption of cigarettes is declining. The consumption c in billions of cigarettes per year since the year 1990 can be approximated by the formula c = –9.2t + 527.33 where t is the number of years after 1990. Use this formula to predict the years that the consumption of cigarettes will be less than 200 billion per year. Answer: The annual consumption of cigarettes will be less than 200 billion more than 35.58 years after 1990, or in approximately 2026.

43. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 2.4 Summary Linear Inequalities and Problem Solving Objectives: Graph inequalities Solve linear inequalities Solve problems that can be modeled by linear inequalities

44. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 2.5 Compound Inequalities Objectives: Find the intersection and union of two sets Solve compound inequalities containing and and or

45. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Intersection of Two Sets The intersection of two sets, A and B, is the set of all elements common to both sets. A intersect B is denoted by A B A  B The solution set of a compound inequality formed by the word and is the intersection of the solution sets of the two inequalities. We use the symbol  to represent “intersection.”

46. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall The solution set of a compound inequality formed by the word or is the union of the solution sets of the two inequalities. We use the symbol  to denote “union.” Union of Two Sets The union of two sets, A and B, is the set of elements that belong to either of the sets. A union B is denoted by A A  B B

47. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall GraphsOr ( υ ) Union And ( ∩) Intersection 82 82 82 x > 8 2 < x < 8 No solution x > 2 All real #’s x 8

48. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 1 If A = {x | x is an odd number greater than 0 but less than 9} and B = {4, 5, 6, 7, 8}, find A  B. List the elements of A. A = {1, 3, 5, 7} List the elements of B. A = {4, 5, 6, 7, 8} The numbers 5 and 7 are in sets A and B. The intersection is {5, 7}.

49. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 2 Solve: x + 4 > 0 and 4x > 0. Solve each inequality separately. x + 4 > 0and4x > 0 x >  4 and x > 0 Graph the two inequalities on a number line and find their intersection. x >  4 x > 0 x >  4 and x > 0 Solution is x > 0 2– 201345– 1– 3– 4– 52– 201345– 1– 3– 4– 5

50. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 3 Solve: 5x > 0 and 3x  4 ≤  13. Solve each inequality separately. 5x > 0and3x – 4 ≤  13 x > 0 and 3x ≤  9 x ≤  3 Graph the two inequalities and find their intersection. x > 0 x ≤  3 x > 0 and x ≤  3 There is no number that is greater than 0 and less than or equal to  3. The answer is no solution. 2– 201345– 1– 3– 4– 52– 201345– 1– 3– 4– 5

51. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall To solve a compound inequality written in compact form, such as 3 < 5 – x < 9, we get x alone in the “middle part.” We must perform the same operations on all three parts of the inequality.

52. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 4 Solve and graph on a number line: 3 < 5 – x < 9. 3 < 5 – x < 9 –5 –5 –5 2– 201345– 1– 3– 4– 5

53. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 6 If A = {x | x is an odd number greater than 0 but less than 9} and B = {4, 5, 6, 7, 8}, find A  B. List the elements of A. A = {1, 3, 5, 7} B = {4, 5, 6, 7, 8} The numbers that are in either set or both sets are {1, 3, 4, 5, 6, 7, 8} This set is the union.

54. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 7 Solve: 6x – 4 ≤ 12 or x + 2 ≥ 8. Solve each inequality separately. Find the union of the graphs. The solutions are x ≤ 8/3 or x ≥ 6. 4 023567 1– 1– 2– 34 023567 1– 1– 2– 3

55. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 8 Solve:  2x – 6 <  2 or 8x < 0. Solve each inequality separately. Find the union. The solutions are all real numbers. 2– 201345– 1– 3– 4– 52– 201345– 1– 3– 4– 5

56. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 2.5 summary Compound Inequalities Objectives: Find the intersection and union of two sets Solve compound inequalities containing and and or

57. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 2.6 Absolute Value Equations Objectives: Solve absolute value equations

58. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 1 Solve:  6 + 2n  = 4. 6 + 2n = 4 or 6 + 2n =  4 2n =  2 or 2n =  10 n =  1 or n =  5 Check: To check, let n =  1 and then n =  5 in the original equation. |6 + 2(  1)| = 4 |6 + 2(  5)| = 4 |6 – 2| = 4 |6 – 10| = 4 |4| = 4 | –4| = 4 4 = 4 True 4 = 4 True The solutions are  1 and  5.

59. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 2 Solve:  2x  6 = 4 We want the absolute value expression alone on one side of the equation, so begin by adding 6 to both sides. Then apply the absolute value property.  2x  6 = 4  2x  = 10 2x = 10 or 2x =  10 x = 5 or x =  5 The solutions are  5 and 5.

60. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 3 Solve:  7x+2  = 0. 7x +2= 0 x = – 2 / 7 Solve:  3z  2  + 8 = 1. First, isolate the absolute value.  3z  2  + 8 = 1  3z  2  =  7 The absolute value of a number is NEVER negative, so this equation has no solution. Example 4

61. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 5 Solve: The absolute value of any expression is NEVER negative, so no solution exists. This equation has no solution.

62. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 6 Solve: This equation is true if the expressions inside the absolute value bars are equal to or are opposites of each other. The solutions are  1 / 3 and 1.

63. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 7 Solve: Solution: x = 5

64. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 2.6 summary Absolute Value Equations Objectives: Solve absolute value equations

65. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 2.7 Absolute Value Inequalities Objectives: Solve absolute value inequalities

66. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 1 Solve: |x| ≤ 7. The solution set of this inequality contains all numbers whose distance from 0 is less than or equal to 7. Thus 7,  7, and all numbers between 7 and  7 are in the solution set. The solutions are  7 ≤ x ≤ 7.

67. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 2 Solve for x:  x + 4  < 6. |x + 4| < 6 is equivalent to  6 < x + 4 < 6  6 < x + 4 < 6 –4 –4 –4  10 < x < 2

68. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 3 Solve for x: |x – 3| + 6 ≤ 7. First, isolate the absolute value expression by subtracting 6 from both sides. +3 +3 +3

69. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 4 Solve for x:  8x  3  <  2. The absolute value of a number is always nonnegative and can never be less than  2. Thus this absolute value inequality has no solution.

70. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 5 Solve for y: |y – 5| > 8. Solve the compound inequality. y – 5 8 +5 +5 +5 +5 y 13 The solutions are all numbers y such that y 13.

71. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 6 Solve for x:  4x + 5  + 6 > 2. The absolute value of any number is always nonnegative and thus is always greater than  4. This inequality and the original inequality are true for all values of x. The solutions are all real numbers.  4x + 5  > –4 Isolate the absolute value expression by subtracting 6 from both sides.

72. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 7 Solve  10 + 3x  + 1 > 2. Write the absolute value inequality as an equivalent compound inequality and solve. Isolate the absolute value expression by subtracting 1 from both sides. -10 -10 3x -9 x  3  10 + 3x  > 1

73. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 7 Solve Write the absolute value inequality as an equivalent compound inequality and solve. Isolate the absolute value expression by adding 1 to both sides. The solutions are x ≤  17 or x ≥ 3.

74. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 8 Solve Recall that “≤” means “less than or equal to.” The absolute value of any expression will never be less than 0, but it may be equal to 0. Thus, to solve set the expression equal to 0. The solution is –2.

75. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 2.7 summary Absolute Value Inequalities Objectives: Solve absolute value inequalities