1. 3.1 Systems of Linear Equations

2. Using graphs and tables to solve systems Using substitution and elimination to solve systems Using systems to model data Value, interests, and mixture problems Using linear inequalities in one variable to make predictions

3. Using Two Models to Make a Prediction When will the life expectancy of men and women be equal? –L = W(t) = 0.114t + 77.47 –L = M(t) = 0.204t + 69.90 Years since 1980 Years of Life 60 80 100 20 406080100120 (84.11, 87.06) Equal at approximately 87 years old in 2064.

4. System of Linear Equations in Two Variables (Linear System) Two or more linear equations containing two variables y = 3x + 3 y = -x – 5

5. Solution of a System An ordered pair (a,b) is a solution of a linear system if it satisfies both equations. The solution sets of a system is the set of all solutions for that system. To solve a system is to find its solution set. The solution set can be found by finding the intersection of the graphs of the two equations.

6. Graph both equations on the same coordinate plane –y = 3x + 3 –y = -x – 5 Verify –(-3) = 3(-2) + 3 –-3 = -6 + 3 –-3 = -3 –(-3) = -(-2) – 5 –-3 = 2 – 5 –-3 = -3 Only one point satisfies both equations (-2,-3) is the solution set of the system Find the Ordered Pairs that Satisfy Both Equations Solutions for y = 3x + 3 Solutions for y = -x – 5 Solution for both (-2,-3)

7. Example ¾x + ⅜y = ⅞ y = 3x – 5 Solve first equation for y –¾x + ⅜y = ⅞ –8(¾x + ⅜y) = 8(⅞) –24x + 24y = 56 48 8 –6x -6x + 3y = 7 – 6x –3y = -6x + 7 3 3 3 –y = -2x + 7/3

8. (1.45,-.6) y = 3x – 5 -.6 = 3(1.45) – 5 -.6 = 4.35 – 5 -.6 ≈ -.65 y = -2x + 7/3 -.6 = -2(1.45) + 7/3 -.6 = -2.9 + 7/3 -.6 ≈ -.57

9. Inconsistent System A linear system whose solution set is empty –Example…Parallel lines never intersect no ordered pairs satisfy both systems

10. Dependent System A linear system that has an infinite number of solutions –Example….Two equations of the same line All solutions satisfy both lines y = 2x – 2 -2x + y = -2 -2x +2x + y = -2 +2x y = 2x – 2

11. One Solution System There is exactly one ordered pair that satisfies the linear system –Example…Two lines that intersect in only one point

12. Solving Systems with Tables x01234 y = 4x – 6-6-22610 y = -6x + 141482-4-10 Since (2,2) is a solution to both equations, it is a solution of the linear system.

13. 3.2 Using Substitution

14. Isolate a variable on one side of either equation Substitute the expression for the variable into the other equation Solve the second equation Substitute the solution into one of the equations 3.2 Using Substitution

15. 1.y = x – 1 2.3x + 2y = 13 3x + 2(x – 1) = 13 3x + 2x – 2 = 13 5x – 2 +2 = 13 +2 5x = 15 5 5 x = 3 y = 3 – 1 y = 2 Solution set for the linear system is (3,2). Example 1

16. Example 2 1.2x – 6y = 4 2.3x – 7y = 8 2x – 6y +6y = 4 +6y 2x = 6y + 4 2 2 2 x = 3y + 2 3(3y + 2) – 7y = 8 9y + 6 – 7y = 8 2y + 6 -6 = 8 -6 2y = 2 2 y = 1 x = 3(1) + 2 x = 5 The solution set is (5, 1).

17. Using Elimination Adding left and right sides of equations If a=b and c=d, then a + c = b + d –Substitute a for b and c for d a + c = a + c both sides are the same

18. Using Elimination Multiply both equations by a number so that the coefficients of one variable are equal in absolute value and opposite sign. Add the left and right sides of the equations to eliminate a variable. Solve the equation. Substitute the solution into one of the equations and solve.

19. Example 1 5x – 6y = 9 + 2x + 6y = 12 7x + 0 = 21 7x = 21 7 7 x = 3 2(3) + 6y = 12 6 -6 + 6y = 12 -6 6y = 12 6 6 y = 2 Solution set for the system is (3,2)

20. Example 2 1.3x + 7y = 29 2.6x – 12y = 32 -2(3x + 7y) = -2(29) -6x – 14y = -58 + 6x – 12y = 32 0 – 26y = -26 -26 -26 y = 1 3x + 7(1) = 29 3x + 7 -7 = 29 -7 3x = 22 3 3 x = 22/3 Solution (22/3, 1)

21. Example 3 1.2x + 5y = 14 2.7x – 3y = 8 3(2x + 5y) = 3(14) 6x + 15y = 42 5(7x – 3y) = 5(8) 35x – 15y = 40 6x + 15y = 42 + 35x – 15y = 40 41x + 0 = 82 41 41 x = 2 2(2) + 5y = 14 4 -4 + 5y = 14 -4 5y = 10 5 5 y = 2 Solution (2,2)

22. Using Elimination with Fractions 2x – 5y -1 3 9 3 3x – 2y 7 15 3 5 9( ) = ( )9 15( ) = ( )15 -2(6x – 5y) = (-3)-2 -12x + 10y = 6 +___________ -9x + 0 = 27 -9 -9 x = -3 3x – 10y = 21 6x – 5y = -3 6(-3) – 5y = -3 -18 +18 – 5y = -3 +18 -5y = 15 -5 -5 y = -3

23. Inconsistent Systems y = 3x + 3 y = 3x – 2 3x + 3 = 3x – 2 3x -3x + 3 = 3x -3x – 2 3 ≠ - 2 False y = 3x + 3 -1(y) = -1(3x + 3) -y = -3x – 3 + y = 3x – 2 0 = 0 – 2 0 ≠ -2 False If the result of substitution or elimination of a linear system is a false statement, then the system is inconsistent.

24. Dependent Systems 1.y = 3x – 4 2.-9x + 3y = -12 -9x + 3(3x – 4) = -12 -9x + 9x -12 = -12 -12 = -12 True y -3x = 3x -3x – 4 -3(-3x + y) = (-4)-3 9x + -3y = 12 + -9x + 3y = -12 0 = 0 True If the result of applying substitution or elimination to a linear system is a true statement, then the system is dependent.

25. Solving systems in one variable To solve an equation A = B, in one variable, x, where A and B are expressions, Solve, graph or use a table of the system –y = A –y = B Where the x-coordinates of the solutions of the system are the solutions of the equation A = B

26. Systems in one variable y = 4x – 3 y = -x + 2 4x – 3 = -x + 2 4x +x – 3 = -x +x + 2 5x – 3 +3 = 2 +3 5x = 5 5 x = 1 y = 4(1) – 3 y = 1 (1,1)

27. Graphing to solve equations in one variable -2x + 6 = 5/4x – 3 –(3,1) -2x + 6 = -4 –(6,-4) (3,1) (6,-4)

28. Using Tables to Solve Equations in One Variable -2x + 7 = 4x – 5 Solution (2,3) xyy -211-13 9-9 07-5 15 233 317 4 11

29. 3.3 Systems to Model Data

30. Predict when the life expectance of men and women will be the same. L = W(t) =.114t + 77.47 L = M(t) =.204t + 69.90.114t -.114t + 77.47 =.204t + 69.90 -.114t 77.47 -69.90 =.09t + 69.90 -69.90 7.57 =.09t t ≈ 84.11

31. Solving to Make Predictions In 1950, there were 4 women nursing students at a private college and 84 men. If the number of women nursing students increases by 13 a year and the number of male nursing students by 6 a year. What year will the number of male and female students be the same? A = W(t) = 13t + 4 A = M(t) = 6t + 84

32. Using substitution 13t + 4 = 6t + 84 13t + 4 -4 = 6t + 84 -4 13t -6t = 6t -6t + 80 7t = 80 7 7 t ≈ 11.43 Equal in 1961~1962

33. 3.4 Value, Interest and Mixture

34. Five Step Method 1.Define each variable 2.Write a system of two equations 3.Solve the system 4.Describe the results 5.Check

35. Value Problems Total-Value Formula –If n objects each have a value v, then their total value T is give by T=vn –Total value equals the value of the object times the number of objects

36. Value Problem Example A medical store is selling rolls of cloth bandages for $4/roll and wrist splints for $16/each. If the store sells a combination of 72 items in a week for a total revenue of $552, how many of each item were sold? 1.Define the variable Let r be the number of bandage rolls sold Let s be the number of splits sold

37. 2.Write a system of two equations T = 4r + 16s Substitute Total revenue to get first equation 552 = 4r + 16s Store sold 72 items total r + s = 72 3.Solve the system 4r + 16s = 552 r + s = 72 Multiply the second equation by -4

38. 4r + 16s = 552 -4(r + s = 72) -4r – 4s = -288 4r + 16s = 552 + -4r – 4s = -288 0 + 12s = 264 12 12 s = 22 r + (22) = 72 r + 22 -22 = 72 -22 r = 50

39. 4.Describe each result The store sold 50 rolls of cloth bandages and 22 wrist splints. 5.Check 50 + 22 = 72 4(50) + 16(22) = 552

40. Value Problem Hospital has 1,000 rooms, 300 private rooms and 700 shared rooms. If the private rooms cost $100 more than the shared rooms, what is the price for each type of room so the total revenue when all rooms are full is $250,000? –private room price is $100 more than shared p = s + 100 –total revenue is $250,000 300p + 700s = 250,000

41. p = s + 100 300p + 700s = 250,000 300(s + 100) + 700s = 250,000 300s + 30,000 + 700s = 250,000 1,000s + 30,000 -30,000 = 250,000 -30,000 1,000s/1,000 = 220,000/1,000 s = 220 p = (220) + 100 p = 320

42. Model a Value Situation The Monera Hospital has 1,200 rooms. private rooms cost $240/night and public rooms $160/night. Let x and y be room cost respectively. Assume all rooms are filled. 1.Let R = f(x) be the total revenue per night. Find the equation of f.

43. Total revenue R = 240x + 160y Total number of rooms x + y = 1,200 y = 1,200 – x Substitute R = 240x + 160(1,200 – x) R = 240x + 192,000 – 160x R = 80x + 192,000 Equation of f f(x) = 80x + 192,000

44. Graph the equation x, 0 ≤ x ≤ 1,200 What is the slope? –80 What does this represent? –If 1 more room is reserved as private and one less as public, then the revenue will increase by $80.

45. What is f(800)? f(800) = 80(800) + 192,000 = 64,000 + 192,000 = 256,000 What does this represent? –If 800 rooms are filled as private rooms, then the total revenue of the night would be $256,000. Find f(15,000) –f(15,000) = 80(15,000) + 192,00 = 1,392,000 –Since there are only 1,200 rooms a model breakdown occurs.

46. To make a total profit of $200,000 how many of each room must be filled? 200,000 = 80x + 192,000 8,000 = 80x x = 100 y = 1,200 – 100 = 1,100 100 private rooms and 1,100 public rooms must be booked to get a total profit of $200,000.

47. Interest Problems Principal – money deposited in an account such as a savings account, certificate of deposit, or mutual fund Interest – percentage of the principal earned when a person invests money Annual Simple Interest Rate – percentage of principal that equals the interest earned per year. –Invest $100 and get $6.5 yr, then the simple interest rate is 6.5%

48. Interest from Investment If a person invests $2800 at 3.5% simple annual interest, how much interest will they earn in a year?.035(2800) = 98 The person will earn $98 in interest.

49. Interest Problems A person is investing in two different accounts, he invests twice as much in an account at 3.9% than he does in a second account at 7.2% annual interest. Both have 3 year averages. How much will a person have to invest in each account to earn a total of $395?

50. If he invests twice as much at 3.9% x = 2y Since the total interest is $395,.039x +.072y = 395 System x = 2y.039x +.072y = 395 Solve the system.039(2y) +.072y = 395x = 2(2633.33).078y +.072y = 395x = 5266.66.15y = 395 y = 2633.33

51. Modeling an Interest Problem A person invests $2500 in a mutual fund with a five year annual interest of 5.5% and a CD account at 3.25% annual interest. Let x and y be the money invested in each account respectively. Let I = f(x) be the total earned from investing $2500 for one year

52. Total interest is: I =.055x +.0325y I in terms of just x: x + y = 2500 x = 2500 – y Substitute I =.055x +.0325(2500 – y) =.055x + 81.25 -.0325y =.0225x + 81.25 f(x) =.0225x + 81.25

53. Graph the equation between 0 ≤ x ≤ 2500 What is the slope? –.0225 What does this mean? –If one more dollar is invested at 5.5% and one less at 3.25%, the total interest will increase by 2.25 cents

54. Use a graphing calculator to create a table. How can this help to determine how much to invest in each account? If a person doesn’t know how much they want to risk by investing in each account, a table could help give a clearer idea of how much to invest in each account.

55. How much should be invested in each account to earn $124 in interest? 124 =.0225x + 81.25 42.75 =.0225x 1900 = x y = 2500 – 1900 = 600 A person should invest $1900 at 5.5% and $600 at 3.25%.

56. Mixture Problems For an x% solution of two substances that are mixed, x% of the solution is one substance and (100 – x)% is the other substance. –If 2 oz of analgesic is diluted in 10 oz of water, 2/10 =.2 = 20% of the solution is analgesic the remaining 8/10 =.8 = 80% is water.

57. Mixture Problem Example A pharmacist needs 20oz of 15% expectorant cough medicine, but only has a 12% and 24% solution. How many ounces of the 12% should he mix with the 24% solution to make 20oz of 15% solutions? x + y = 20.12x +.24y =.15(20)

58. System x + y = 20.12x +.24y = 3 Solve y = 20 – x.12x +.24(20 – x) = 3.12x + 4.8 -.24x = 3 -.12x + 4.8 = 3 -.12x = -1.8 x = 15 y = 20 – 15 = 5 To make 20oz 15% solution, she would need 15 oz of the 12% solution and 5 oz of the 24% solution.

59. Mixture Example A chemist needs 6 quarts of 20% alcohol solution, but only has a 30% alcohol solution. How much water should he mix with the solution to form 6 quarts of 20% solution? x + y = 6.30x =.20(6).3x = 1.2 x = 4 4 + y = 6 y = 2 Chemist needs to mix 4 quarts of 30% solution with 2 quarts of water.

60. 3.5 Using Linear Inequalities to Make Predictions

61. You are debating between two car rental companies, one charges $48 per day and $.15/mile. The second charges $24 per day and $.55/mile. C = F(d) =.15d + 48 C = S(d) =.55d + 24 Graph the equation to see for how many miles the first company is cheaper.

62. d > 60 then the first company is cheaper

63. aba+cb+c Addition Property If a < b, then a + c < b + c –Similar properties hold for ≤, > and ≥. –This holds for subtraction as well Subtracting is same as adding negative a +c ab a+cb+c +(-c) 5 < 8 5 + 2 < 8 + 2 7 < 10 true 5 < 8 5 - 2 < 8 - 2 3 < 6 true

64. Multiplication Property For a positive number c, –if a < b, then ac < bc. For a negative number c, –if a bc. Similar properties hold for ≤, > and ≥. This holds for division as well –dividing by a nonzero number is the same as multiplying by its reciprocal **When multiplying or dividing both sides of an inequality by a negative number, reverse the inequality symbol.

65. Multiplication Property (cont) 5(2) < 8(2) 10 < 16 true 5(-2) < 8(-2) -10 < -16 false -10 > -16 4/2 < 8/2 2 < 4 true 4/(-2) < 8/(-2) -2 < -4 false -2 > -4

66. Linear inequalities in One Variable A linear inequality in one variable is an inequality that can be put into one of these forms: –mx + b < 0 –mx + b ≤ 0 –mx + b > 0 –mx + b ≥ 0 m and b are constants and m ≠ 0 A number satisfies an inequality, if the inequality becomes a true statement

67. Solutions to Inequalities 6x – 14 < 16 –Does 3 satisfy the equation? 6(3) – 14 < 16 18 – 14 < 16 4 < 16 true Yes, 3 satisfies the equation –Does 8 satisfy the equation? 6(8) – 14 < 16 48 – 14 < 16 34 < 16 false No, 8 doesn’t satisfy the equation

68. Inequality Examples 6x – 14 < 16 –Find all solutions of the inequality 6x – 14 < 16 6x < 30 x < 5 -5x + 21 > 11 -5x /(-5) > -10/(-5) x < 2 **Remember to change the sign when dividing or multiplying by a negative number.

69. Inequalities on timelines x < 2(-∞,2) x ≤ 2(-∞,2] x > 2(2,∞) x ≥ 2[2,∞) Interval – set of real numbers represented by the number line or by an unbroken portion of it Interval notation –( or ) represents –[ or ] represents ≤ or ≥ –All real numbers is represented by (- ∞,∞) 0 2 x 0 x 0 x 0 x 2 2 2

70. Solving a Linear Inequality -2(3x – 6) < 20 – 2x -6x + 12 -12 < 20 – 2x -12 -6x +2x < 8 – 2x +2x -4x /(-4) < 8 /(-4) x > 2 (2,∞) 02 x

71. Three-part Inequalities 2 < x < 4(2,4) 2 ≤ x ≤ 4[2,4] 2 < x ≤ 4(2,4] 2 ≤ x < 4[2,4) 024 x 024 x 024 x 024 x

72. Solving Three-part Inequalities -2 ≤ -6x – 8 < 16 -2 +8 ≤ -6x – 8 +8 < 16 +8 6 /(-6) ≤ -6x /(-6) < 24 /(-6) -1 ≥ x > -4 (-4,-1] -4-20 x

73. Using Inequalities to Model Comparisons between two quantities You are debating between two car rental companies, one charges $48 per day and $.15/mile. The second charges $24 per day and $.55/mile. –C = F(d) =.2d + 48 –C = S(d) =.55d + 24 For how many miles the first company is cheaper? –F(d) < S(d)

74. .15d + 48 <.55d + 24.15d + 48 -24 <.55d + 24 -24.15d -.15d + 24 <.55d -.15d 24 <.40d 60 < d The first company offers the lower price if the car is driven over 60 miles. (Same as original example)