1. Slide 6 - 1 Copyright © 2009 Pearson Education, Inc. Slide 6 - 1 Copyright © 2009 Pearson Education, Inc. Welcome to MM150 – Unit 4 Seminar Unit 4 Seminar – Instructor: Larry Musolino Email: lmusolino@kaplan.edulmusolino@kaplan.edu 4.3 – Graphing Linear Equations 4.1 – Variation 4.2 – Linear Inequalities 1

2. Slide 6 - 2 Copyright © 2009 Pearson Education, Inc. Slide 6 - 2 Copyright © 2009 Pearson Education, Inc. 4.3 Graphing Linear Equations 2

3. Slide 6 - 3 Copyright © 2009 Pearson Education, Inc. Different Type of Equation In previous unit, we examined linear equation in one variable such as: 2x + 4 = 14 Notice there is only one variable in this equation, namely “x”. When we solve this equation, we arrive at one solution, namely: x = 5 Now, in Unit 4, we examine a different type of equation, linear equation with two variables, x and y. y = 3x + 2 In this case, there will be an infinite number of solutions and so we cannot list out all the infinite solutions. Instead we use a graph as means to show all the infinite solutions to this type of equation.

4. Slide 6 - 4 Copyright © 2009 Pearson Education, Inc. Finding Solutions In order to find a solution to an equation such as: y = 3x + 2 We will need to find a replacement value for x, and a replacement value for y such that when we substitute these two values in the given equation, a true equation results. Thus a solution to this type of equation will consist of a pair of values, one value for x, and another value for y. We call this an ordered pair, and we will write this solutions in the form (x, y) 4

5. Slide 6 - 5 Copyright © 2009 Pearson Education, Inc. How to Graph a Linear Equation in two variables A graph is really a visual way to show all the solutions to the a given equation such as: y = 3x + 2 Notice in this equation, we have two variables, “x” and “y” and thus any solution must have two replacement values, one for “x” and one for “y”. When we determine a pair of numbers x, and y which represents a solution, we will write it in the form (x, y) 5

6. Slide 6 - 6 Copyright © 2009 Pearson Education, Inc. Finding Solutions Consider the equation: y = 3x + 2 Is x = 4 and y = 14 a solution? To answer this we replace x and y with given values in equation, and check if we arrive at a true equation. Thus, x = 4 and y = 14 is a valid solution to this equation since when we substitute these values into the given equation, a true equation results: y = 3x + 2 14 = 3(4) + 2 14 = 12 + 2 14 = 14 (True equation!) 6

7. Slide 6 - 7 Copyright © 2009 Pearson Education, Inc. Finding Solutions Consider the equation: y = 3x + 2 Is x = 5 and y = 20 a solution? To answer this we replace x and y with given values in equation, and check if we arrive at a true equation. Thus, x = 5 and y = 20 is not a valid solution to this equation since when we substitute these values into the given equation, a false equation results: y = 3x + 2 20 ?= 3(5) + 2 20 ?= 15 + 2 20 ≠ 17 (False equation!) 7

8. Slide 6 - 8 Copyright © 2009 Pearson Education, Inc. Finding Solutions Consider the equation: y = 3x + 2 We know that x = 4 and y = 14 is a valid solution to this equation (from Slide 6). We write this solution in the following format. (4, 14) this is called an ordered pair Remember we always write the solution in the form: (x, y ) x-value is first, y-value is second.

9. Slide 6 - 9 Copyright © 2009 Pearson Education, Inc. You Try It #1 Determine if x = 3 and y = 5 is a solution to equation: y = 4x + 1 9

10. Slide 6 - 10 Copyright © 2009 Pearson Education, Inc. You Try It #1 (Solution) Determine if x = 3 and y = 5 is a solution to equation: y = 4x + 1 Solution: Substitute given value of x and given value of y in equation and determine if true equation results: y = 4x + 1 5 ?= 4(3) + 1 5 ?= 12 + 1 5 ≠ 12 Therefore we know that x= 3 and y = 5 is not a solution to the given equation. 10

11. Slide 6 - 11 Copyright © 2009 Pearson Education, Inc. You Try It #2 Determine if (3, -10) is a solution to equation: y = -2x - 4 11

12. Slide 6 - 12 Copyright © 2009 Pearson Education, Inc. You Try It #2 (Solution): Determine if (3, -10) is a solution to equation: y = -2x – 4 Solution: Substitute given value of x and given value of y in equation and determine if true equation results. Here we know that x = 3 and y = -10 y = -2x - 4 -10 ?= -2(3) - 4 -10 ?= -6 - 4 -10 = -10 √ checks Therefore we know that (3, -10) is a solution to the given equation.

13. Slide 6 - 13 Copyright © 2009 Pearson Education, Inc. How to find Solutions How can we find solutions to an equation such as: y = 3x + 2 One simple method: – Step 1: Pick an arbitrary value for “x” – Step 2: Substitute this value of “x” in the given equation and solve for y. – Step 3: The corresponding solution will then be (x, y). Example #1 – Step 1: Pick an arbitrary value for “x”, say x = 1 – Step 2: Substitute this value of “x” in the given equation and solve for y: y = 3(1) + 2 = 3 + 2 = 5 – Step 3: The corresponding solution will then be (1, 5). 13

14. Slide 6 - 14 Copyright © 2009 Pearson Education, Inc. How to find Solutions How can we find solutions to an equation such as: y = 3x + 2 One simple method: – Step 1: Pick an arbitrary value for “x” – Step 2: Substitute this value of “x” in the given equation and solve for y. – Step 3: The corresponding solution will then be (x, y). Example #2 – Step 1: Pick an arbitrary value for “x”, say x = 5 – Step 2: Substitute this value of “x” in the given equation and solve for y: y = 3(5) + 2 = 15 + 2 = 17 – Step 3: The corresponding solution will then be (5, 17). 14

15. Slide 6 - 15 Copyright © 2009 Pearson Education, Inc. How to find Solutions How can we find solutions to an equation such as: y = 3x + 2 One simple method: – Step 1: Pick an arbitrary value for “x” – Step 2: Substitute this value of “x” in the given equation and solve for y. – Step 3: The corresponding solution will then be (x, y). Example #3 – Step 1: Pick an arbitrary value for “x”, say x = 0 – Step 2: Substitute this value of “x” in the given equation and solve for y: y = 3(0) + 2 = 0 + 2 = 2 – Step 3: The corresponding solution will then be (0, 2). 15

16. Slide 6 - 16 Copyright © 2009 Pearson Education, Inc. How to find Solutions How can we find solutions to an equation such as: y = 3x + 2 One simple method: – Step 1: Pick an arbitrary value for “x” – Step 2: Substitute this value of “x” in the given equation and solve for y. – Step 3: The corresponding solution will then be (x, y). Example #4 – Step 1: Pick an arbitrary value for “x”, say x = -2 – Step 2: Substitute this value of “x” in the given equation and solve for y: y = 3(-2) + 2 = -6 + 2 = -4 – Step 3: The corresponding solution will then be (-2, -4). 16

17. Slide 6 - 17 Copyright © 2009 Pearson Education, Inc. How to find Solutions So we just found four different solutions to equation: y = 3x + 2 Here is the summary: Notice we could continue to generate as many solutions as we want, by generating more arbitrary values for “x”. Then, for an equation such as y = 3x + 2 there will be an infinite number of solutions. Remember, we cannot list out all these infinite solutions, so we use a graph as a convenient way to display all these infinite solutions. xy(x, y) -2-4(-2, -4) 02(0, 2) 15(1, 5) 517(5, 17) 17

18. Slide 6 - 18 Copyright © 2009 Pearson Education, Inc. Rectangular Coordinate System The horizontal line is called the x-axis. The vertical line is called the y-axis. The point of intersection of the x- axis and y-axis is called the origin. x-axis y-axis origin Quadrant I Quadrant II Quadrant IIIQuadrant IV 18

19. Slide 6 - 19 Copyright © 2009 Pearson Education, Inc. Plotting Points (x, y) Tells us the horizontal distance to travel Tells us the vertical distance to travel Always start at the origin !!! 19

20. Slide 6 - 20 Copyright © 2009 Pearson Education, Inc. Plotting Points Each (x, y) solution such as (2, 4) will be plotted as a single point. To plot a solution such as (2, 4) 1.We always start at the origin, this is intersection point of the x-axis and y-axis. 2.Then move horizontally the distance specified by the x-value, – Note: If the x-value is positive number then move to the right, if the x- value is negative number then move to the left. 3.From this position, move vertically the distance specified by the y-value, – Note: If the y-value is positive number then move up, if the y-value is a negative number then move down. 20

21. Slide 6 - 21 Copyright © 2009 Pearson Education, Inc. Plotting Points Each point in the xy-plane corresponds to a unique ordered pair (x, y). The x-value will be the horizontal distance to travel. The y-value will be the vertical distance to travel. Example: Plot the point (2, 4) – Start at origin – Move 2 units right (horizontal) – Then move 4 units up (vertical) 2 units 4 units Point (2, 4) Origin 21

22. Slide 6 - 22 Copyright © 2009 Pearson Education, Inc. Graphing Linear Equations How do we graph? For each (x, y) solution, plot the corresponding point on set of x-y axes. Then connect points with straight line. xy(x, y) -2-4(-2, -4) 02(0, 2) 15(1, 5) 517(5, 17) 22

23. Slide 6 - 23 Copyright © 2009 Pearson Education, Inc. You Try It #3 Find three different (x, y) solutions for the equation: y = 2x – 3 Then plot the three solutions on set of x-y axes and connect points with straight line. 23

24. Slide 6 - 24 Copyright © 2009 Pearson Education, Inc. You Try It #3 (Solution): Find three different (x, y) solutions for the equation: y = 2x – 3 Write each of the three solutions in the form (x, y). Then plot the three solutions on set of x-y axes and connect points with straight line. Solution: Note that many different solutions are possible: xy(x, y) 1(1, -1) 21(2, 1) 33(3. 3) 24

25. Slide 6 - 25 Copyright © 2009 Pearson Education, Inc. You Try It #3 (Solution) Continued: Find three different (x, y) solutions for the equation: y = 2x – 3 Write each of the three solutions in the form (x, y). Then plot the three solutions on set of x-y axes and connect points with straight line. Solution: xy(x, y) 1(1, -1) 21(2, 1) 33(3. 3) 25

26. Slide 6 - 26 Copyright © 2009 Pearson Education, Inc. Summary: To Graph Equations by Plotting Points Optional : Solve the equation for y (this means to get y by itself on one side of equation). Select at least three arbitrary values for x. For each value of x, substitute in the given equation and solve for y. Write each solution in form (x, y). Plot each of the (x, y) points. The points should be in a straight line. Draw a line through the set of points and place arrow tips at both ends of the line. 26

27. Slide 6 - 27 Copyright © 2009 Pearson Education, Inc. Graphing using Intercepts We can also graph an equation by finding two specific (x, y) solutions, namely the intercepts of the graph: – One solution is the point where the graph crosses or intercepts the horizontal x-axis (this is called the x-intercept). – Another solution is the point where the graph crosses or intercepts the vertical y-axis (this is called the y-intercept). 27

28. Slide 6 - 28 Copyright © 2009 Pearson Education, Inc. Finding Intercepts We know that for any point which falls on the x-axis, the y-value must be 0. – Some examples of x-intercepts: (3, 0), (-2, 0), (7,0) We also know that for any point which falls on the y- axis, the x-value must be 0. – Some examples of y-intercepts: (0, 4), (0, -8), (0, 9) So to find the x-intercept of graph: – Let y = 0 in the given equation and then solve for x. So to find the y-intercept of graph: – Let x = 0 in the given equation and then solve for y. 28

29. Slide 6 - 29 Copyright © 2009 Pearson Education, Inc. Graphing Using Intercepts Example: Find x-intercept and y-intercept for equation: y = -3x + 6 To find x-intercept let y = 0 and solve for x. Example: y = -3x + 6 0 = -3x + 6 -6 = -3x (Subtract 6 from both sides) 2 = x (Divide both sides by -3) To find y-intercept let x = 0 solve for y. Example: y =  3x + 6 y =  3(0) + 6 y = 0 + 6 y = 6 x-intercept: (2, 0) y-intercept: (0, 6) 29

30. Slide 6 - 30 Copyright © 2009 Pearson Education, Inc. Example: Graph 3x + 2y = 6 using intercepts Find the x-intercept: Let y = 0, solve for x: 3x + 2y = 6 3x + 2(0) = 6 3x = 6 x = 2 Find the y-intercept: Let x = 0, solve for y 3x + 2y = 6 3(0) + 2y = 6 2y = 6 y = 3 x-intercept: (2, 0) y-intercept: (0, 3) 30

31. Slide 6 - 31 Copyright © 2009 Pearson Education, Inc. Slope of a Line Slope measures the steepness of a line. Slope is defined as: – The ratio of the vertical change to the horizontal change for any two points on the line. 31

32. Slide 6 - 32 Copyright © 2009 Pearson Education, Inc. Types of Slope Positive slope rises from left to right. Negative slope falls from left to right. The slope of a vertical line is undefined. The slope of a horizontal line is zero. zero negativeundefined positive 32

33. Slide 6 - 33 Copyright © 2009 Pearson Education, Inc. Finding Slope To find the slope of line passing through two points, call them (x 1, y 1 ), (x 2, y 2 ): 1.Clearly label each given point as (x 1, y 1 ) and (x 2, y 2 ). 2.Substitute these values in the slope formula, and calculate a numeric result: Note: Careful when subtracting, you can use a – b = a + (-b) 33

34. Slide 6 - 34 Copyright © 2009 Pearson Education, Inc. Example of Finding Slope Find slope of line passing through the two points: (5, 4) and (7, 10) 1.First we will label each given point as (x 1, y 1 ) and (x 2, y 2 ). Thus we have: x 1 = 5, y 1 = 4 and x 2 = 7, y 2 = 10 2.Substitute these values in the slope formula: Thus, slope m = 3. Since this is positive slope it tells us this line extends uphill going from left to right. 34

35. Slide 6 - 35 Copyright © 2009 Pearson Education, Inc. Example of Finding Slope Find slope of line passing through the two points: (-1, 2) and (1, -8) 1.First we will label each given point as (x 1, y 1 ) and (x 2, y 2 ). Thus we have: x 1 = -1, y 1 = 2 and x 2 = 1, y 2 = -8 2.Substitute these values in the slope formula: Thus, slope m = -5. Since this is negative slope it tells us this line extends downhill going from left to right. 35

36. Slide 6 - 36 Copyright © 2009 Pearson Education, Inc. Example of Finding Slope Find slope of line passing through the two points: (2, -3) and (4, 5) 1.First we will label each given point as (x 1, y 1 ) and (x 2, y 2 ). Thus we have: x 1 = 2, y 1 = -3 and x 2 = 4, y 2 = 5 2.Substitute these values in the slope formula: Thus, slope m = 4. Since this is positive slope it tells us this line extends uphill going from left to right. 36

37. Slide 6 - 37 Copyright © 2009 Pearson Education, Inc. Example: Finding Slope Find the slope of the line through the points (5, -3) and (-2, -3). 37

38. Slide 6 - 38 Copyright © 2009 Pearson Education, Inc. You Try It - Slope 1.Find slope of line passing through the two points: (-1, -5) and (1, -9) 2.Does this line extend uphill or downhill going from left to right? 38

39. Slide 6 - 39 Copyright © 2009 Pearson Education, Inc. You Try It – Slope - Solution Find slope of line passing through the two points: (-1, -5) and (1, -9) 1.First we will label each given point as (x 1, y 1 ) and (x 2, y 2 ). Thus we have: x 1 = -1, y 1 = -5 and x 2 = 1, y 2 = -9 2.Substitute these values in the slope formula: Thus, slope m = -2. Since this is negative slope it tells us this line extends downhill going from left to right. 39

40. Slide 6 - 40 Copyright © 2009 Pearson Education, Inc. The Slope-Intercept Form of a Line Slope-Intercept Form of the Equation of the Line y = mx + b where m is the slope of the line and (0, b) is the y-intercept of the line. 40

41. Slide 6 - 41 Copyright © 2009 Pearson Education, Inc. Graphing Equations by Using the Slope and y-Intercept Solve the equation for y to place the equation in slope-intercept form. Determine the slope and y-intercept from the equation. Plot the y-intercept. Obtain a second point using the slope. Draw a straight line through the points. 41

42. Slide 6 - 42 Copyright © 2009 Pearson Education, Inc. Example Graph 2x - 3y = 9. Write in slope-intercept form. The y-intercept is (0,-3) and the slope is 2/3. 42

43. Slide 6 - 43 Copyright © 2009 Pearson Education, Inc. Example continued Plot a point at (0,-3) on the y-axis, then move up 2 units and to the right 3 units. 43

44. Slide 6 - 44 Copyright © 2009 Pearson Education, Inc. Horizontal Lines Graph y = -3. y is always equal to -3, for all solutions. Pick arbitrary values for x. Some solutions: (1, -3), (2, -3), (3, -3), 4(-3), etc. The graph is parallel to the x-axis. 44

45. Slide 6 - 45 Copyright © 2009 Pearson Education, Inc. Vertical Lines Graph x = -3. x always equals -3, for all solutions Pick arbitrary values for y. Some examples of solutions: (-3, 1), (-3, 2), (- 3, 3), (-3, -2), etc. The graph is parallel to the y-axis. 45

46. Slide 6 - 46 Copyright © 2009 Pearson Education, Inc. Slide 6 - 46 Copyright © 2009 Pearson Education, Inc. 4.1 Variation 46

47. Slide 6 - 47 Copyright © 2009 Pearson Education, Inc. Direct Variation Variation is an equation that relates one variable to one or more other variables. In direct variation, the values of the two related variables increase or decrease together. If a variable y varies directly with a variable x, then y = kx where k is the constant of proportionality (or the variation constant). 47

48. Slide 6 - 48 Copyright © 2009 Pearson Education, Inc. Example #1 The total charge for gasoline at the pump is directly proportional to the price per gallon of gasoline. If it costs $30.60 to purchase 12 gallons of gas, how much will it cost to purchase 15 gallons? y = kx Let “y” be total cost of gas, let “x” be number of gallons, then “k” will be the cost per gallon. $30.60 = 12x (To solve, divide both sides by 12) k = $2.55 Now we can find cost for 15 gallons: y = ($2.55)(15) = $38.25 48

49. Slide 6 - 49 Copyright © 2009 Pearson Education, Inc. Example #2 The amount of interest earned on an investment, I, varies directly as the interest rate, r. If the interest earned is $50 when the interest rate is 5%, find the amount of interest earned when the interest rate is 7%. I = rx $50 = 0.05x (To solve, divide both sides by 0.05) x = $1000 49

50. Slide 6 - 50 Copyright © 2009 Pearson Education, Inc. Example #2 continued Now we know that x = 1000, r = 7% I = rx I = 0.07(1000) I = $70 The amount of interest earned is $70. 50

51. Slide 6 - 51 Copyright © 2009 Pearson Education, Inc. You Try It #1 To use your cell phone, you are charged a fee per minute of cell phone usage. If you talk for 15 minutes and the fee is $5.25, what is the fee to talk for 23 minutes? 51

52. Slide 6 - 52 Copyright © 2009 Pearson Education, Inc. You Try It #1 (Solution) To use your cell phone, you are charged a fee per minute of cell phone usage. If you talk for 15 minutes and the fee is $5.25, what is the fee to talk for 23 minutes? Solution: – First find k. y = kx – Let “y” be the total fee and let “x” be number of minutes. – $5.25 = k(15) (To solve, divide both sides by 15) K = $0.35 – Now we can find the fee for 23 minutes. y = kx = ($0.35)(23) = $8.05 52

53. Slide 6 - 53 Copyright © 2009 Pearson Education, Inc. Inverse Variation When two quantities vary inversely, as one quantity increases, the other quantity decreases, and vice versa. If a variable y varies inversely with a variable, x, then where k is the constant of proportionality. 53

54. Slide 6 - 54 Copyright © 2009 Pearson Education, Inc. Example Suppose y varies inversely as x. If y = 12 when x = 18, find y when x = 21. First find “k”, constant of proportionality. Now substitute 216 for k, and find y when x = 21. 54

55. Slide 6 - 55 Copyright © 2009 Pearson Education, Inc. You Try It #2 In kick boxing, the amount of force needed to break a board is inversely proportional to the length of the board. (The longer the board, the easier is it to break). If it takes 2 pounds of force to break a 10 foot long board, how much force is needed to break a four foot long board? 55

56. Slide 6 - 56 Copyright © 2009 Pearson Education, Inc. You Try It #2 (Solution): In kick boxing, the amount of force needed to break a board is inversely proportional to the length of the board. If it takes 2 pounds of force to break a 10 foot long board, how much force is needed to break a four foot long board? Solution: First find “k”: Next, we use the fact that k = 20 to answer “how much force is needed to break a four foot long board?” Thus it will take 5 pounds of force to break a four foot long board. 56

57. Slide 6 - 57 Copyright © 2009 Pearson Education, Inc. Joint Variation One quantity may vary directly as the product of two or more other quantities. The general form of a joint variation, where y, varies directly as x and z, is y = kxz where k is the constant of proportionality. 57

58. Slide 6 - 58 Copyright © 2009 Pearson Education, Inc. Example The area, A, of a triangle varies jointly as its base, b, and height, h. If the area of a triangle is 48 in 2 when its base is 12 in. and its height is 8 in., find the area of a triangle whose base is 15 in. and whose height is 20 in. 58

59. Slide 6 - 59 Copyright © 2009 Pearson Education, Inc. Combined Variation A varies jointly as B and C and inversely as the square of D. If A = 1 when B = 9, C = 4, and D = 6, find A when B = 8, C = 12, and D = 5. Write the equation. 59

60. Slide 6 - 60 Copyright © 2009 Pearson Education, Inc. Combined Variation (continued) Find the constant of proportionality. Now find A. 60

61. Slide 6 - 61 Copyright © 2009 Pearson Education, Inc. Slide 6 - 61 Copyright © 2009 Pearson Education, Inc. 4.2 Linear Inequalities 61

62. Slide 6 - 62 Copyright © 2009 Pearson Education, Inc. Symbols of Inequality a < b means that a is less than b. a ≤ b means that a is less than or equal to b. a > b means that a is greater than b. a ≥ b means that a is greater than or equal to b. Find the solution to an inequality by adding, subtracting, multiplying or dividing both sides by the same number or expression. Note: Change the direction of the inequality symbol when multiplying or dividing both sides of an inequality by a negative number. 62

63. Slide 6 - 63 Copyright © 2009 Pearson Education, Inc. Example: Graphing Graph the solution set of x ≤ 4, where x is a real number, on the number line. The numbers less than or equal to 4 are all the points on the number line to the left of 4 and 4 itself. The closed circle at 4 shows that the value 4 itself is included in the solution set. 63

64. Slide 6 - 64 Copyright © 2009 Pearson Education, Inc. Example: Graphing Graph the solution set of x > 3, where x is a real number, on the number line. The numbers greater than 3 are all the points on the number line to the right of 3. The open circle at 3 is used to indicate that the value “3” itself is not included in the solution set. 64

65. Slide 6 - 65 Copyright © 2009 Pearson Education, Inc. Example: Solve and graph the solution Solve 3x – 8 < 10 and graph the solution set. The solution set is all real numbers less than 6. 65

66. Slide 6 - 66 Copyright © 2009 Pearson Education, Inc. Compound Inequality Graph the solution set of the inequality -4 < x ≤ 3 a) where x is an integer. The solution set is the integers between -4 and 3, including 3. 66

67. Slide 6 - 67 Copyright © 2009 Pearson Education, Inc. Compound Inequality continued Graph the solution set of the inequality -4 < x ≤ 3 b) where x is a real number The solution set consists of all real numbers between -4 and 3, including the 3 but not the value -4. 67

68. Slide 6 - 68 Copyright © 2009 Pearson Education, Inc. Example A student must have an average (the mean) on five tests that is greater than or equal to 85% but less than 92% to receive a final grade of B. Jamal’s grade on the first four tests were 98%, 89%, 88%, and 93%. What range of grades on the fifth test will give him a B in the course? 68

69. Slide 6 - 69 Copyright © 2009 Pearson Education, Inc. Example continued 69