1. U1 - Algebra Notes 1

2. Introduction To get the full value, we recommend that you play this presentation and work through each problem as it comes up on a slide. You can also print these slides if you prefer to work on paper. 2

3. Algebra Review – Part 1 Blanks filled in Examples solved 3

4. Explanation Given: -5m+7x 2 y+1 The ‘-5’ and the ‘7’ are called the Numerical coefficients or just coefficients. The ‘-5m’, ‘7x 2 y’, and ‘1’ are called terms. The ‘1’ is called a constant term since it contains no unknowns. Terms are separated by either addition or subtraction. When two or more terms are combined by addition or subtraction we get a polynomial. 4

5. Now you try! a) 21a + 7a + 8b - 10b + 5 - 8 5

6. Answer = 28a - 2b - 3 6

7. Now you try! b) 3d 2 – 11d – 6d 2 – 7d 7

8. Answer = 3d 2 – 18d 8

9. Now you try! c) -4(2x + 1) – (5 – 2x) 9

10. Answer = -8x – 4 – 5 + 2x = -6x – 9 10

11. Now you try! d) 3a (3a – 7) + 5a (7a – 4) 11

12. Solution = 9a 2 – 21a + 35a 2 – 20a = 44a 2 – 41a 12

13. Now you try! e) 3(2x + 4y) – 5(7x – 3y) – (8x – 2y) 13

14. Solution = 6x + 12y – 35x + 15y – 8x + 2y = -37x + 29y 14

15. Now you try! f) 6 – 3y (4x+5) + y (5 – 2x) 15

16. Solution = 6 – 12xy – 15y + 5y – 2xy = -10y – 14 xy + 6 16

17. Now you try! g) 17

18. Solution 18

19. Polynomials Polynomials with exactly two terms are called binomials. Polynomials with exactly three terms are called trinomials. Single terms are called monomials. 19

20. Evaluating Expressions and Word Problems Expand and simplify the following expressions then evaluate for the value(s) of the variable(s). for x = – 1, and y = – 2 20

21. Evaluating Expressions and Word Problems Substituting for the given values, = 15(-1) 2 -10(-1)(-2) – 12(-2) 2 + 4(-1)(-2) = 15(1) – 20 – 12(4) + 4(2) = 15 –20 – 48 +8 = –45 21

22. b) Evaluate the expression for 22

23. Solution continued for b) for x=-2, 23

24. Solution to b) continued 24

25. Solution continued 25

26. Simplifying c) 26 Now you try for y= -3

27. Solution c) 27

28. Solution to c) continued c) Substituting for y=-3 28

29. Solution to c) continued c) 29

30. Question 2 A car brakes and decelerates uniformly. The distance, d metres, that it travels in t seconds is given by the formula, where u is its speed, in metres per second, before the brakes are applied. Determine how far the car travels under the following conditions on the next slide. 30

31. Question 2 (con’t) a) It is traveling 25 m/s before the brakes are applied. It travels for 4 s after the brakes are applied. 31

32. Question 2 a) Solution Let u = 25, t = 4 into Therefore, it takes 44 m to stop! 32

33. Question 2 b) b) It is traveling 35 m/s before the brakes are applied. It travels for seconds after the brakes are applied. 33

34. Question 2 b) Solution Let u = 35, t = It takes 86.625 m to stop! 34

35. Algebra Review – Part 2 Solving Linear Equations Examples with Solutions 35

36. Part 2 Solve each of the following equations. When the answer is not an integer, express the answer first as an exact fraction and then rounded to two decimal places. 36

37. a) Solve the following equation 37

38. Solution a) Solution: 38

39. b) Solve the following equation 39

40. Solution b) Solution: 40

41. c) Solve the following equation 41

42. Solution c) Solution: 42

43. d) Solve the following equation 43

44. Solution d) Solution: 44

45. e) Solve the following equation 45

46. Solution- note you may omit the first two lines e) Solution: 46

47. f) Solve the following equation 47

48. Solution f) Solution: 48

49. Part 2 - Systems of Equations Notes and Applications 49

50. Explanation A system of equations refers to a number of equations, which must be satisfied simultaneously. Usually, the number of equations in a system will equal the number of variables (or unknowns). Further, in this course, we will mostly solve systems involving two equations in two unknowns. 50

51. Example 1 Solve the system: 3a+ b= 10 and 2a-5b = 1 The method we used to solve this system is called substitution. 51

52. Example 2 Find the point of intersection of the lines with equations 2x + 3y = -1 and 3x-4y=7 using the method of substitution. 52

53. Example 2 (con’t) 53

54. Example 2 (con’t) There is a second method for solving systems of equations, called elimination, which is often used when isolating a variable produces fractions (most equations). Let’s resolve this system using elimination. Remember that both of these equations are conditions, so we can manipulate them and then either add or subtract them. 2x+3y=-1 and 3x-4y=7 54

55. Example 2 (con’t) 55

56. Example 3 Solve using elimination: 4p + 5n= 3 and 8p + 3n = -1 Mostly, we are interested in using equations to solve applied problems. 56

57. Example 4 In a parking lot, there is space for both cars and buses. When 7 cars and 10 buses are parked for a day, the total revenue is $305. When 11 cars and 18 buses are parked for the day, the total revenue is $525. How much does it cost to park a bus for the day? 57

58. Example 4 (con’t) 58

59. Example 5 John has $10000 to invest. He invests part of it in bonds paying 6% per annum and the rest in a 2 nd mortgage (to a friend) for 8% per annum. After one year, the total interest from these investments was $690. How much did he invest in bonds? 59

60. Example 5 (con’t) 60

61. Example 6 Two types of expensive coffee beans were blended to make a delicious blend. Individually, they cost $23 per kg and $32 per kg. How much of each type were mixed to make 180 kg of coffee which is sold at $30 per kg. Assume the pricing of the blend is based on the pricing of the original types of coffee. 61

62. Example 6 (con’t) 62

63. Example 7 Renting a car has a daily rental cost and a charge per km for the distance driven. Renting a car for three days and driving 1000 km costs $217. Renting the same car for seven days and driving 2000 km costs $473. Find the daily rental charge and the cost per km to drive the rented car. 63

64. Example 7 (con’t) 64

65. Example 8 The total number of dollars raised to a fundraiser was $3975. There were a total of 475 tickets sold, some student tickets for $5 each, the remainder to profs at $10 each. How many students attended the fundraiser? 65

66. Example 8 (Solution) Try this again using two variables Solution: Let x be the number of student tickets Let 475-x be the number of prof tickets Type of ticket price/ticket NumberValue $ $5x5x $10475-x10(475-x) Total475$3975 66

67. Example 8 Solution (con’t) 5x+10(475-x)=3975 5x+4750-10x=3975 -5x=-775 x=155 So, 155 student tickets were sold and 475-155=320 tickets were sold to profs. 67

68. Example 9 Shannon collects $29.70 in only nickels, dimes and quarters from the pop machine near the cafeteria, and finds that there are 14 more dimes than nickels, and 38 more quarters than nickels. How many are there of each type of coin? 68

69. Example 9 (Solution) Solution: Type of CoinNumberValue Nickel (0.05)x0.05x Dime (0.10)x+140.10(x+14) Quarter (0.25)x+38.25(x+38) 69

70. Example 9 (solution con’t) 0.05x + 0.10x+1.4 + 0.25x+ 9.5=29.70 0.4x+10.9=29.70 0.4x=29.70-10.9 0.4x=18.8 x=18.8/0.4 = 47 So, there were 47 nickels, 47+14=61 dimes and 47+38=85 quarters Check: 0.05(47) +0.1(61) +0.25(85) = 29.70 70

71. A final thought on solving systems of equations You can also think of a combination method which uses both aspects of elimination and substitution. Here is an example of how to do this: Solve the system 71

72. Solution This is a simpler method for some. 72

73. Solution continued Next, we realize that both are equal to 6y 73