Algebra I System of Linear Equations

Transcription

1 1

2 Algebra I System of Linear Equations

3 Table of Contents Click on the topic to go to that section Solving Systems by Graphing Solving Systems by Substitution Solving Systems by Elimination Choosing your Strategy Writing Systems to Model Situations Teacher Note Standards 3

4 Solving Systems by Graphing Return to Table of Contents 4

5 System of Equations A system of linear equations is comprised of 2 or more linear equations. The solution of the system will be the values of the variables which make all the equations true. 5

6 Solve by Graphing Example: y = 2x - 4 The graph of the line that represents the solutions to the above equation is shown y It represents all the points whose x and y values make the above equation true x The line is easy to find from the equation since the equation is in slopeintercept form

7 Solve by Graphing Example: y y = x Similarly, this graph is of the line that represents the solutions to this equation. It represents all the points whose x and y values make the above equation true x -10 7

8 Solve by Graphing Example: y = 2x - 4 y = x y Here are the lines that represent the solutions to both those equations. Each line shows the infinite set of solutions for each equation. What must be true about the point at which they cross? x DISCUSS. 8

9 Solve by Graphing Example: y = 2x - 4 y = x y At the point they cross, both equations must be true, since that point is on both lines x They appear to cross at (3, 2). Let's check that in both equations

10 Solve by Graphing Example: Substitute x = 3 and y = 2 into both equations and see if both equations are true. y = 2x y (2) = 2(3) = 2 correct x y = x + 5 (2) = -(3) = 2 correct 10

11 System of Equations Not all systems have solutions...and some have an infinite number of solutions. Let's see how to figure out whether there are solutions, how many, and what they are. Click here to watch a music video that introduces what we will learn about systems. 11

12 The Number of Solutions When graphing two lines there are three possibilities. They meet in one point: the point of intersection. They never meet: they are parallel. They meet at all their points: they are the same line. 12

13 The Number of Solutions So, systems of equations can have either: 1 solution, if the lines meet at one point 0 solutions, if they never meet Infinite solutions, if they are the same line 13

14 Type 1: One Solution The two lines intersect in exactly ONE place. y 10 The solution is the point at which they intersect. 5 The slopes of the lines must be different, or they would never cross x

15 Type 1: One Solution y y = 2x - 4 y = x This is the example we started with. 5 As we confirmed there is one solution to this system of equations: (3, 2) x

16 Type 2: No Solution The lines never meet. There is no solution true for both lines. The lines are parallel y They must have the same slope, since they are parallel x But, they must have different intercepts, or they would be the same line

17 Type 2: No Solution y = 2x + 6 y = 2x + 2 Both are written in slope intercept form y 10 y = mx + b to make it easy to compare slopes and y-intercepts. The slope for both lines is 2 (the coefficient of x). So, the lines are parallel x The y-intercepts are different, +6 and +2, so the lines never cross

18 Type 3: Infinite Solutions The lines overlap at all points. y 10 They are different equations for the same line. 5 The lines are parallel. So, they must have the same slopes x The intercepts are the same, since all their points are the same

19 Type 3: Infinite Solutions y = 2x + 2 y = 2x + 2 y Both are written in slope intercept form 10 y = mx + b to make it easy to compare slopes and y- intercepts. The slope for both lines is 2 (the coefficient of x). So, the lines are parallel x The y-intercepts for both lines are +2, so the lines overlap everywhere

20 Type 3: Infinite Solutions y = 2x + 2 y = 2x + 2 In slope intercept form, the fact that these are the same line is obvious. y 10 But, if the equations were written as below, it would be less obvious: 5 2y - 4x = 4-6x = -3y x That's why it's always a good idea to put equations into slope-intercept form...they're easier to read, graph and compare

21 The Number of Solutions First, put the equations into slope-intercept form by solving for y. Then, decide on the number of solutions. After that, solutions can be found in three different ways. 21

22 How can you quickly decide the number of solutions a system has? 1 Solution No Solution Math Practice Infinitely Many 22

23 Solving both Equations for y Let's solve this system of equations y = -5x x + 2y = 6 The equation on the left is in slope-intercept form. Do you see that the slope is -5 and its y-intercept is +4? The equation on the right is not in slope-intercept form, so we can't see it's slope or y-intercept. So, we can't tell yet how many solutions will satisfy both equations. Let's solve the second equation for y. 23

24 Solving for y 10x + 2y = 6-10x -10x Subtract 10x from both sides 2y = -10x + 6 Divide both sides by 2 2y = -10x y = -5x + 3 This is now in slope-intercept form. We can see the slope and y-intercept m = -5 b = 3 24

25 Solving both Equations for y y= -5x + 4 Original Equations 10x + 2y = 6 y= -5x + 4 Slope Intercept Form y = -5x + 3 m = -5 b = 4 Slopes and Intercepts m = -5 b = 3 The slopes are the same but the y-intercepts are different. How many solutions are there? 25

26 Solving both Equations for y Let's solve this system of equations y = 2x + 5 6x + 2y = 4 The equation on the left is in slope-intercept form and we can see the slope is +2 and the y-intercept is +5. The equation on the right is not in slope-intercept form, let's solve that equation for y. 26

27 Solving for y 6x + 2y = 4-6x -6x Subtract 6x from both sides 2y = -6x + 4 Divide both sides by 2 2y = -6x y = -3x + 2 This is now in slope-intercept form. m = -3 b = +2 27

28 Solving both Equations for y y= -5x + 4 Original Equations 6x + 2y = 4 y= -5x + 4 Slope Intercept Form y = -3x + 2 m = -5 b = 4 Slopes and Intercepts m = -3 b = +2 The slopes are different. How many solutions are there? 28

29 1 How many solutions does this system have: y = 2x - 7 y = 3x + 8 A 1 solution Answer B C no solution infinitely many solutions 29

30 2 How many solutions does this system have: 3x - y = -2 y = 3x + 2 Answer A B C 1 solution no solution infinitely many solutions 30

31 3 How many solutions does this system have: 3x + 3y = 8 1 y = x 3 Answer A B C 1 solution no solution infinitely many solutions 31

32 4 How many solutions does this system have: y = 4x 2x - 0.5y = 0 Answer A B C 1 solution no solution infinitely many solutions 32

33 5 How many solutions does this system have: 3x + y = 5 6x + 2y = 1 Answer A B C 1 solution no solution infinitely many solutions 33

34 Consider this... Suppose you are walking to school. Your friend is 5 blocks ahead of you. You can walk two blocks per minute and your friend can walk one block per minute. How many minutes will it take for you to catch up with your friend? Math Practice 34

35 Solution First, make a table to represent the problem. Time (min.) Friend's distance from your start (blocks) Your distance from your start (blocks) 35

36 Solution Continued Next, plot the points on a graph. Blocks Time (min.) Friend's distance from your start (blocks) Your distance from your start (blocks) Time (min.) 36

37 Solution Continued The point where the lines intersect is the solution to the system. Blocks (5, 10) is the solution In the context of this problem this means after 5 minutes, you will meet your friend at block Time (min.) 37

38 Example Solve this system of equations graphically: y = 2x - 3 y = x y Answer x

39 Example Solve the system of equations graphically: y = -3x + 4 y y = x Answer x

40 Checking Your Work Given the graph below, what is the point of intersection? y = -3x - 1 y = 4x y x

41 Checking Your Work Now take the ordered pair we just found and substitute it into the equations to prove that it is a solution for BOTH lines. (-1, 2) y = -3x - 1 (2) = -3(-1) = = 2 y = 4x + 6 (2) = 4(-1) = = 2 Math Practice 41

42 6 Solve the following system by graphing: y y = -x + 4 y = 2x + 1 A (3, 1) 10 5 Answer B (1, 3) Click for answer choices AFTER students have graphed the system C (-1, 3) x D (1, -3)

43 7 Solve the following system by graphing: 1 y = x y = x 1 2 A (0,-1) 10 5 y Answer B (0,0) Click for answer choices AFTER students have graphed the system C (-1, 0) x D (0, 1)

44 8 Solve the following system by graphing: y = x y = x A (0, 4) 10 5 y Answer B (-4, 2) Click for answer choices AFTER students have graphed the system C (5, 6) x D (2, 5)

45 Graphing Quickly Transforming linear equations into slope-intercept form usually saves time in the end. It also makes it easy to check your work. 45

46 Example Solve the following system of linear equations by graphing: 2x + y = 5 -x + y = 2 Step 1: Rewrite the linear equations in slope-intercept form 2x + y = 5-2x -2x y = -2x + 5 -x + y = 2 +x +x y = x

47 Solution Continued Step 2: Plot the y-intercept and use the slope to plot the second point y = -2x + 5 y-intercept = (0, 5) slope = -2 slope= (down 2, right 1) 10 5 y y = x x y-intercept = (0, 2) -5 slope = 1 slope= (up 1, right 1)

48 Solution Continued Step 3: Locate the point of intersection and check your work: (1, 3) y y = -2x + 5 (3) = -2(1) = = y = x x (3) = (1) =

49 Example Solve the system of equations graphically: 2x + y = 3 x - 2y = 4 Step 1: Rewrite in slope-intercept form 2x + y = 3-2x -2x y = -2x + 3 x - 2y = 4 -x -x -2y = -x + 4-2y = -x y = x

50 Solution Continued Step 2: Plot y-intercept and use slope to plot second point y-intercept = (0, 3) y slope = slope= (down 2, right 1) 5 y-intercept = (0, -2) 1 slope = 2 slope= (up 1, right 2) x

51 Solution Continued Step 3: Locate the Point of Intersection and check your work: (2, -1) y y = -2x + 3 (-1) = -2(2) = = y = x ( 1) = (2) = x 1 =

52 9 What is the solution of the system of linear equations provided on the graph? A (0, 1) B (1, 0) 10 5 y Answer C (2, 3) D (3, 2) x

53 10 Which graph below represents the solution to the following system of linear equations: -x + 2y = 2 3y = x + 6 Answer A C B D 53

54 11 Solve the following system by graphing: x 3y = 3 y = x 7 10 y A (3, 4) B (9, 2) Click for answer choices AFTER students have C graphed infintely the system many D no solution x Answer

55 Example Solve the system of equations graphically: y = 3x + 6 9x - 3y = -18 Step 1: Rewrite in slope-intercept form y = 3x + 6 9x - 3y = -18-9x -9x -3y = -9x -18-3y = -9x y = 3x

56 Solution Continued Step 2: Plot y-intercept and use slope to plot second point y = 3x + 6 y-intercept = (0, 6) slope = 3 slope= (up 3, right 1) y = 3x + 6 y-intercept = (0, 6) slope = 3 slope= (up 3, right 1) 56

57 Solution Continued Step 3: Locate the Point of Intersection and check your work: infinite amount of points: infinite solutions y = 3x + 6 9x - 3y =

58 Example Solve the system of equations graphically: 4x - 2y = 10 8x - 4y = 12 Step 1: Rewrite in slope-intercept form 4x - 2y = 10-4x -4x -2y = -4x y = -4x y = 2x - 5 8x - 4y = 12-8x 8x -4y = -8x y = -8x y = 2x

59 Solution Continued Step 2: Plot y-intercept and use slope to plot second point y = 2x - 5 y-intercept = (0, -5) slope = 2 slope= (up 2, right 1) y = 2x -3 y-intercept = (0, -3) slope = 2 slope= (up 2, right 1) 59

60 Solution Continued Step 3: Locate the Point of Intersection and check your work: no point of intersection: no solution y = 2x - 5 y = 2x

61 12 Solve the this system by graphing: y = 3x + 4 y 4y = 12x Answer 5 A (2, 4) B (0.4, 2.2) Click for answer choices AFTER students have graphed the system C D infinitely many solutions no solution x 61

62 13 Solve the this system by graphing: y = 3x + 4 4y = 12x y Answer A (3,4) 5 B (-3,-4) C Click for answer choices AFTER students have graphed the system infinitely many x D no solution

63 Solving Systems by Substitution Return to Table of Contents 63

64 Example Solve the system of equations graphically. y = x y y = -2x Teacher Note Why was it difficult to solve Click this for Additional system by graphing? Question x

65 Substitution Explanation Graphing can be inefficient or approximate. Another way to solve a system of linear equations is to use substitution. Substitution allows you to create a one variable equation. 65

66 Solving by Substitution Step 1: If you are not given a variable already alone, find the EASIEST variable to solve for (get it alone) Step 2: Substitute the expression into the other equation and solve for the variable Step 3: Substitute the numerical value you found into EITHER equation and solve for the other variable. Write the solution as (x, y) 66

67 Example Solve the system using substitution: y = x y = -2x Step 1: Choose an equation from the system and substitute it into the other equation y = x First Equation y = -2x Second Equation x = -2x Substitute First Equation into Second Equation 67

68 Step 2: Solve the new equation x = -2x x +2x Add 2x to both sides 3x = Subtract 6.1 from both sides 3x = Solution Continued 3x = Divide both sides by 3 3 x = -2.5 This is the value of x for our solution...now we have to find y. 68

69 Solution Continued Step 3: Substitute the solution x = -2.5 into either equation and solve. y = x y = (-2.5) y = 3.6 y = -2x y = -2(-2.5) y = y = 3.6 The solution to the system of linear equations is (-2.5, 3.6). We only had to plug the x value into one of the equations to get this. The second one just provides a check. If it comes out the same, our solution must be correct. 69

70 Good Practice After you evaluate the solution, it is good practice is to double check your work by substituting the solution into both equations. CHECK: See if (-2.5, 3.6) satisfies both equations y = -2x (3.6) = -2(-2.5) = = 3.6 y = x (3.6) = (-2.5) = 3.6 Math Practice If your checks end in true statements, the solution is correct. 70

71 Example Solve the system using substitution: 2x - 3y = -1 y = x - 1 Step 1: Substitute one equation into the other equation. Since one equation is already solved for y, I'll substitute that into the other equation. 2x - 3y = -1 y = x - 1 2x - 3(x - 1) = -1 71

72 Solution Continued Step 2: Solve the new equation 2x - 3(x - 1) = -1 2x - 3x + 3 = -1 x = 4 Step 3: Substitute the solution into either equation and solve 2x - 3y = -1 2(4) - 3y = y = -1-3y = -9 y = 3 (4, 3) You end with the correct answer with either equation you use for this step. y = x - 1 y = (4) - 1 y = 3 (4, 3) 72

73 Example Continued Check: See if (4, 3) satisfies both equations 2x - 3y = -1 2(4) - 3(3) = = -1-1 = -1 y = x - 1 (3) = (4) = 3 Math Practice The ordered pair satisfies both equations so the solution is (4, 3) 73

74 14 Solve by substitution: y = x - 3 y = -x + 5 A (4, 9) B (-4, -9) Click for answer choices AFTER students have solved the system C (4, 1) Answer D (1, 4) 74

75 15 Solve by substitution: y = x y = -3x 7 A (2, -8) Answer B (-3, 2) Click for answer choices AFTER students have C infinitely solved the system many solutions D no solutions 75

76 16 Solve by substitution. y = 4x x + 3y = -1 A (4, 5) B (5, 4) Click for answer choices AFTER students have C solved infintely the system many solutions D no solutions Answer 76

77 17 Solve by substitution. y = 8x x + 3y = 0 A (-2, -2) B (-2, 2) Click for answer choices AFTER students have solved the system C (2, -2) Answer D (2, 2) 77

78 18 Solve by substitution. 8x + 3y = -9 y = 3x + 14 A (-8, 5) B (7, 5) Click for answer choices AFTER students have solved the system C (-3, 5) Answer D (-7, 5) 78

79 Choosing a Variable Sometimes, a variable is not already isolated. When this occurs, you need to determine which one you would prefer to isolate and finish solving the system with the substitution method. Examine each system of equations. Which variable would you choose to substitute? y = 4x y = -2x + 9 Why? -y + 4x = -1 x - 4y = 1 2x + 4y = -10-8x - 3y = -12 Teacher Note 79

80 19 Examine this system of equations. Which variable could quickly be solved for and substituted into the other equation? y = -2x + 5 2y = 10-4x Answer A B x y 80

81 20 Examine this system of equations. Which variable could quickly be solved for and substituted into the other equation? 2y - 8 = x y + 2x = 4 A x Answer B y 81

82 21 Examine this system of equations. Which variable could quickly be solved for and substituted into the other equation? x - y = 20 2x + 3y = 0 Answer A B x y 82

83 Rewriting Sometimes you need to rewrite one of the equations so that you can use the substitution method. For example: The system: 3x - y = 5 2x + 5y = -8 Solve for y: 3x - y = 5-3x 3x -y = -3x Which letter is the easiest to solve for? The "y" in the first equation because Click to discuss which letter there is only a "-1" as the coefficient. So, the original system is equivalent to: y = 3x - 5 Click to see 2x + 5y = -8 y = 3x

84 Solution Continued Now Substitute and Solve: y = 3x - 5 2x + 5 y = -8 2x + 5(3x - 5) = -8 2x + 15x - 25 = -8 17x - 25 = -8 17x = 17 x = 1 84

85 Solution Continued Substitute x = 1 into one of the equations. 2(1) + 5y = y = -8 5y = -10 y = -2 The ordered pair (1, -2) satisfies both equations in system. Math Practice 3x - y = 5 3(1) - (-2) = = 5 5 = 5 2x + 5y = -8 2(1) + 5(-2) = = -8-8 = -8 85

86 22 Solve using substitution. 6x + y = 6-3x + 2y = -18 A (-6, 2) B (6, -2) Click for answer choices AFTER students have C solved (-6 the, -2) system Answer D (2, -6) 86

87 23 Solve using substitution. 2x - 8y = 20 -x + 6y = -12 A (6, -1) Answer B (-6, 5) Click for answer choices AFTER students have solved the system C (5, 5) D (-6, -1) 87

88 24 Solve using substitution. -3x - 3y = 12-4x - 7y = 7 Answer A (-3, -7) B (-7, 3) Click for answer choices AFTER students have solved the system C (3, 7) D (7, 3) 88

89 25 Solve the system by substitution: y = x - 6 y = -4 Answer A (-10, -4) B Click (-4, for 2) answer choices AFTER students have C (2, -4) solved the system D (10, 4) 89

90 26 Solve the system by substitution: y + 2x = -14 y = 2x + 18 A (1, 20) B Click (1, for 18) answer choices AFTER C students (8, -2) have solved the D (-8, system 2) Answer 90

91 Example Solve this system using substitution: x + y = 6 5x + 5y = 10 x + y = 6 x = 6 - y 5(6 - y) + 5y = y + 5y = = 10 - solve the first equation for x - substitute 6 - y for x in 2nd equation - solve for y - This is FALSE! Since 30 = 10 is a false statement, the system has no solution. Answer: NO SOLUTION 91

92 Example Solve the following system using substitution: x + 4y = -3 2x + 8y = -6 x + 4y = -3 - solve the first equation for x x = -3-4y 2(-3-4y) + 8y = -6 - sub. -3-4y for x in 2nd equation -6-8y + 8y = -6 - solve for y -6 = -6 - This is ALWAYS TRUE! Since -6 = -6 is always a true statement, there are infinitely many solutions to the system. Answer: Infinite Solutions 92

93 27 Solve the system by substitution: 4x = -5y + 50 x = 2y - 7 A (6, 6.5) Answer B Click (5, for 6) answer choices AFTER C students (4, 5) have solved the D (6, system 5) 93

94 28 Solve the system by substitution: y = -3x y + 4x = 19 A (6, 5) Answer B Click (-7, for 5) answer choices AFTER C students (42, -103) have solved the D (6, system -5) 94

95 29 Solve the system using substitution. 3 4 x + y = 2 6x + 8y = 16 Answer A (-4, 5) B (4, Click -1) for answer choices AFTER students Chave infinitely solved many the solutions system D no solutions 95

96 30 Solve using substitution. 16x + 2y = -5 y = -8x - 6 A (-3, -1) Answer B No Solution Click for answer Cchoices Infinite AFTER Solutions students have D (-1, solved -3) the system 96

97 Solving System by Elimination Return to Table of Contents 97

98 Standard Form Recall that the Standard Form of a linear equation is: Ax + By = C When both linear equations of a system are in standard form the system can be solved by using elimination. The elimination strategy adds or subtracts the equations in the system to eliminate a variable. 98

99 Additive Inverses Let's talk about what's happening with these numbers = 3 + (-3) = -5x + 5x = 9x + (-9x) = 99

100 Choosing a Variable How do you decide which variable to eliminate? First: Look to see if one variable has the same or opposite coefficients. If so, eliminate that variable. Math Practice 100

101 Addition or Subtraction If the variables have the same coefficient, subtract the two equations to eliminate the variable. Same Coefficients { 3x 3x Subtract { 3x -(3x) 0x If the variables have opposite coefficients, add the two equations to eliminate the variable. Opposite Coefficients { 3x -3x Add { 3x + -3x) 0x 101

102 Example Solve the following system by elimination: 5x + y = 44-4x - y = -34 Step 1: Choose which variable to eliminate The y in both equations have opposite coefficients so they will be the easiest to eliminate Step 2: Add the two equations 5x + y = 44-4x - y = -34 x + 0y = 10 x =

103 Solution Continued Step 3: Substitute the solution into either equation and solve x = 10 5(10) + y = y = 44 y = -6 The solution to the system is (10, -6) Math Practice Check: 5x + y = 44 5(10) + (-6) = = = 44-4x - y = -34-4(10) - (-6) = = =

104 Example Solve the following system by elimination: 3x + y = 15-3x - 3y = -21 Step 1: Choose which variable to eliminate Math Practice The x in both equations have opposite coefficients so they will be the easiest to eliminate Step 2: Add the two equations 3x + y = 15-3x - 3y = -21-2y = -6 y = 3 104

105 Solution Continued Step 3: Substitute the solution into either equation and solve y = 3 3x + 3 = 15 3x = 12 x = 4 The solution to the system is (4, 3) Math Practice Check: 3x + y = 15 3(4) + 3 = = = 15-3x - 3y = -21-3(4) - 3(3) = = =

106 31 Solve the system by elimination: x + y = 6 x - y = 4 A (5, 1) Answer B Click (-5, for -1) answer choices AFTER C students (1, 5) have solved the D no system solution 106

107 32 Solve the system by elimination: 2x + y = -5 2x - y = -3 A (-2,1) Answer B Click (-1,-2) for answer choices AFTER C students (-2,-1) have solved the D infinitely systemmany 107

108 33 Solve using elimination. -2x - 8y = 10 2x - 6y = 18 A (-2, 3) B Click (4, -6) for answer choices AFTER C students (-6, 4) have solved the D (3, system -2) Answer 108

109 Multiple Methods There are 2 ways to complete the problem below using elimination. 5x + y = 17-2x + y = -4 Step 1: Choose which variable to eliminate The y in both equations have the same coefficient so they will be the easiest to eliminate Step 2: Add or Subtract the two equations First Method: Multiply one equation by -1 then add equations Second Method: Subtract equations keeping in mind that all signs change 109

110 Solution Continued First Method -1(-2x + y = -4) = 2x - y = 4 5x + y = 17 2x - y = 4 7x = 21 x = 3 Second Method 5x + y = 17 -(-2x + y = -4) 7x = 21 x = 3 Answer Why do both methods produce the same solution? 110

111 Solution Continued Step 3: Substitute the solution into either equation and solve x = 3-2(3) + y = y = -4 y = 2 The solution to the system is (3, 2) Math Practice Check: 5x + y = 17 5(3) + 2 = = = 17-2x + y = -4-2(3) + 2 = = -4-4 =

112 34 Solve the system by elimination: 2x + y = -6 A (-4, 2) 3x + y = -10 Answer B (3, 5) Click for answer choices AFTER C (4, 2) students have solved the D infinitely system many 112

113 35 Solve the system by elimination: A (2, -7) 3x + 6y = 48-5x + 6y = 32 Answer B Click (2, for 7) answer choices AFTER C students (7, 2) have solved the D infinitely system many 113

114 Common Coefficient Sometimes, it is not possible to eliminate a variable by simply adding or subtracting the equations. When this is the case, you need to multiply one or both equations by a nonzero number in order to create a common coefficient before adding or subtracting the equations. 114

115 Example Solve the following system using elimination: 3x + 4y = -10 5x - 2y = 18 The y would be the easiest variable to eliminate because 4 is a common coefficient. Multiply second equation by 2 so the coefficients are opposites. 2(5x - 2y = 18) The y coefficients are opposites, so solve by adding the equations 3x + 4y = x - 4y = 36 13x = 26 x = 2 115

116 Example Continued Solve for y, by substituting x = 2 into one of the equations. 3x + 4y = -10 3(2) + 4y = y = -10 4y = -16 y = -4 (2, -4) is the solution Check: 3x + 4y = -10 3(2) + 4(-4) = = = -10 5x - 2y = 18 5(2) - 2(-4) = = =

117 Choosing Variable to Eliminate In the previous example, the y was eliminated by finding a common coefficient of 4. Creating a common coefficient of 4 required one additional step: Multiplying the second equation by 2 3x + 4y = -10 5x - 2y = 18 Either variable can be eliminated when solving a system of equations as long as a common coefficient is utilized. 117

118 Example Solve the same system by eliminating x. 3x + 4y = -10 5x - 2y = 18 Multiply the first equation by 5 and the second equation by 3 so the coefficients will be the same 5(3x + 4y = -10) 15x + 20y = -50 3(5x - 2y = 18) 15x - 6y = 54 Now solve by subtracting the equations. 15x + 20y = (15x - 6y = 54) 26y = -104 y =

119 Example Continued Solve for x, by substituting y = -4 into one of the equations. (2, -4) is the solution. Check: 3x + 4y = -10 3x + 4(-4) = -10 3x = -10 3x = 6 x = 2 Math Practice 3x + 4y = -10 3(2) + 4(-4) = = = -10 5x - 2y = 18 5(2) - 2(-4) = = =

120 System of Equations Examine each system of equations. Which variable would you choose to eliminate? What do you need to multiply each equation by? 2x + 5y = -1 x + 2y = 0 3x + 8y = 81 5x - 6y = -39 Teacher Note 3x + 6y = 6 2x - 3y = 4 120

121 36 Which variable can you eliminate with the least amount of work in the system below? A x B y 2x + 5y = 20 3x - 10y = 37 Answer 121

122 37 Solve the following system of equations using elimination: 2x + 5y = 20 A (1, 57) 3x - 10y = 37 Answer B (1, Click 77) for answer choices AFTER 2 C students (11, - have ) solved the system 5 D infinitely many solutions 122

123 38 Which variable can you eliminate with the least amount of work in the system below? A x B y x + 3y = 4 3x + 4y = 2 Answer 123

124 39 What will you multiply the first equation by in order to solve this system using elimination? x + 3y = 4 3x + 4y = 2 Answer 124

125 40 Solve the following system of equations: x + 3y = 4 3x + 4y = 2 Answer 2 A (-2, ) 3 Click for answer choices B (-2, 1) AFTER students C (-2, 2) have solved the system D infinitely many solutions 125

126 Example Solve the following system using elimination: 9x - 5y = 4-18x + 10y = 10 The y would be the easiest variable to eliminate because 10 is a common coefficient. Multiply first equation by 2 so the coefficients are opposites. 2(9x - 5y = 4) The y coefficients are opposites, so solve by adding the equations 18x - 10y = x + 10y = 10 0 = 18 is this true? False, NO Move SOLUTION for solution 126

127 Example Solve the following system using elimination: -4x - 10y = -22 2x + 5y = 11 The x would be the easiest variable to eliminate because 4 is a common coefficient. Multiply second equation by 2 so the coefficients are opposites. 2(2x + 5y= 11 ) The y coefficients are opposites, so solve by adding the equations -4x - 10y = x +10y = 22 0 = 0 is this true? True, Move INFINITE for solution SOLUTIONS 127

128 41 Solve the system by elimination: x - y = 5 x - y = -7 A (11, -4) B Click (4, 11) for answer choices AFTER students C (-4, -11) have solved the system Answer D no solution 128

129 42 Solve using elimination. -20x - 18y = x + 9y = 14 A (-8, -1) B infinite Click for solutions answer choices AFTER C no solution students have solved the system D (-1, 8) Answer 129

130 43 Solve using elimination. 9x + 3y = 27 18x + 6y = 30 A infinite solutions B (4, Click 7) for answer choices AFTER students C (-7, 4) have solved the system Answer D no solution 130

131 Choose Your Strategy Return to Table of Contents 131

132 Choosing Strategy Systems of linear equations can be solved using any of the three methods we previously discussed. Before solving a system, an analysis of the equations should be done to determine the "best" strategy to utilize. Graphing Substitution Elimination 132

133 Example Altogether 292 tickets were sold for a basketball game. An adult ticket cost $3 and a student ticket cost $1. Ticket sales for the event were $470. How many adult tickets were sold? How many student tickets were sold? 133

134 Example Continued Step 1: Define your variables Let a = number of adult tickets Let s = number of student tickets Step 2: Set up the system number of tickets sold: a + s = 292 money collected: 3a + s =

135 Example Continued Step 3: Solve the system a + s = 292 -( 3a + s = 470 ) -2a+ 0 = -178 a = 89 Elimination was utilized for this example because the x had a common coefficient. Note 135

136 Example Continued a = 89 a + s = s = 292 s = 203 There were 89 adult tickets and 203 student tickets sold Math Practice Check: a + s = = = 292 3a + s = 470 3(89) = = =

137 44 What method would require the least amount of work to solve the following system: y = 3x - 1 y = 4x Answer A B C graphing substitution elimination 137

138 45 Solve the following system of linear equations using the method of your choice: y = 3x - 1 y = 4x A (-4, -1) B Click (-1, -4) for answer choices AFTER students C (-1, 4) have solved the system Answer D (1, 4) 138

139 46 What method would require the least amount of work to solve the following system: 4s - 3t = 8 t = -2s -1 Answer A B C graphing substitution elimination 139

140 47 Solve the following system of linear equations using the method of your choice: 4s - 3t = 8 t = -2s -1 Answer 1 1 A (-2, Click ) C 2 for answer choices (, 2) 2 AFTER students 1 have solved system B (, -2) D (2, -2) 2 140

141 48 What method would require the least amount of work to solve the following system: A B C graphing substitution elimination 3m - 4n = 1 3m - 2n = -1 Answer 141

142 49 Solve the following system of linear equations using the method of your choice: 3m - 4n = 1 3m - 2n = -1 A (-2, -1) Click for answer choices B (-1, -1) AFTER students have solved the system C (-1, 1) Answer D (1, 1) 142

143 50 What method would require the least amount of work to solve the following system: A B C graphing substitution elimination y = -2x 1 y = x Answer 143

144 51 Solve the following system of linear equations using the method of your choice: A (-6, 12) y = -x 1 y = x Answer B Click for answer choices (2, -4) AFTER students C (-2, have 2) solved the system D (1, -2) 144

145 52 What method would require the least amount of work to solve the following system: A graphing u = 4v 3u - 3v = 7 Answer B substitution C elimination 145

146 53 Solve the following system of linear equations using the method of your choice: u = 4v 3u - 3v = 7 Answer A 28 7 (, ) C (28, 7) Click 9 9for answer choices AFTER ( 7, 28 ) (7, 7 B students have solved D system)

147 54 A piece of glass with an initial temperature of 99 F is cooled at a rate of 3.5 F/min. At the same time, a piece of copper with an initial temperature of 0 F is heated at a rate of 2.5 F/min. Let m = the number of minutes and t = the temperature in F. Which system models the given scenario? Answer A B C t = m t = m t = m t = m t = m t = 0-2.5m 147

148 55 Which method would you use to solve the system from the previous question? t = Click m to Reveal t = System m Answer A B C graphing substitution elimination 148

149 56 Solve the following system of linear equations: t = Click m to Reveal t = m System Answer A m = 1 t = 2.5 B m = 1 t = 95.5 Click for C m answer = 16.5 choices t = 6.6 AFTER D m = 16.5 t = students have solved system 149

150 57 Choose a strategy and then answer the question. What is the value of the y-coordinate of the solution to the system of equations x 2y = 1 and x + 4y = 7? A 1 B -1 C 3 D 4 Answer From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from accessed 17, June,

151 Writing Systems to Model Situations Return to Table of Contents 151

152 Creating and Solving Systems Step 1: Define the variables Step 2: Analyze components and create equations Step 3: Solve the system utilizing the best strategy 152

153 Example A group of 148 peole is spending five days at a summer camp. The cook ordered 12 pounds of food for each adult and 9 pounds of food for each child. A total of 1,410 pounds of food was ordered. Part A: Write an equation or a system of equations that describe the above situation and define your variables. a = number of adults c = number of children a + c = a + 9c = 1,410 From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from accessed 17, June,

154 Example Continued Part B: Using your work from part A, find (1) the total number of adults in the group (2) the total number of children in the group a + c = a + 9c = 1,410 (1) (2) c = -a a + 9(-a + 148) = a - 9a = a = 78 a = 26 a + c = c = 148 c =

155 Example Tanisha and Rachel had lunch at the mall. Tanisha ordered three slices of pizza and two colas. Rachel ordered two slices of pizza and three colas. Tanisha s bill was $6.00, and Rachel s bill was $5.25. What was the price of one slice of pizza? What was the price of one cola? p = cost of pizza slice c = cost of cola 3p + 2c = p + 3c = 5.25 From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from accessed 17, June,

156 Example Continued 3p + 2c = p + 3c = 5.25 Elimination: Multiply first equation by 2 Multiply second equation by -3 6p + 4c = 12-6p - 9c = c = c = 0.75 Cola: $0.75 Pizza: $1.50 3p + 2c = p + 2(0.75) = 6 3p = 6 3p = 4.5 p =

157 Example Your class of 22 is going on a trip. There are four drivers and two types of vehicles, vans and cars. The vans seat six people, and the cars seat four people, including drivers. How many vans and cars does the class need? Let v = the number of vans and c = the number of cars 157

158 Set Up the System Drivers: v + c = 4 People: 6v + 4c = 22 Solve the system by substitution: 158

159 Substitute, Solve and Check v + c = 4 6v + 4c = 22 solve for v substitute v = -c + 4 6(-c + 4) + 4c = 22-6c c = 22-2c + 24 = 22-2c = -2 substitute v = -(1) + 4 c = 1 v = 3 then check: c = 1; v = 3 (3) + (1) = 4 6(3) + 4(1) = 22 4 = 4 22 =

160 58 Your class receives $1,105 for selling 205 packages of greeting cards and gift wrap. A pack of cards costs $4 and a pack of gift wrap costs $9. Set up a system and solve. How many packages of cards were sold? Answer You will answer how many packages of gift wrap in the next question. 160

161 59 Your class receives $1105 for selling 205 packages of greeting cards and gift wrap. A pack of cards costs $4 and a pack of gift wrap costs $9. Set up a system and solve. How many packages of gift wrap were sold? Answer 161

162 60 The sum of two numbers is 47, and their difference is 15. What is the larger number? A 16 B 31 Answer C 32 D

163 61 Ramon rented a sprayer and a generator. On his first job, he used each piece of equipment for 6 hours at a total cost of $90. On his second job, he used the sprayer for 4 hours and the generator for 8 hours at a total cost of $100 What was the hourly cost for the sprayer? Answer From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from accessed 17, June,

164 62 You have 15 coins in your pocket that are either quarters or nickels. They total $2.75. How many quarters do you have? Answer 164

165 63 You have 15 coins in your pocket that are either quarters or nickels. They total $2.75. How many nickels do you have? Answer 165

166 64 Julia went to the movies and bought one jumbo popcorn and two chocolate chip cookies for $5.00. Marvin went to the same movie and bought one jumbo popcorn and four chocolate chip cookies for $6.00. How much does one chocolate chip cookie cost? A $0.50 B $0.75 Answer C $1.00 D $2.00 From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from accessed 17, June,

167 65 Mary and Amy had a total of 20 yards of material from which to make costumes. Mary used three times more material to make her costume than Amy used, and 2 yards of material was not used. How many yards of material did Amy use for her costume? Answer From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from accessed 17, June,

168 66 The tickets for a dance recital cost $5.00 for adults and $2.00 for children. If the total number of tickets sold was 295 and the total amount collected was $1220, how many adult tickets were sold? Answer From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from accessed 17, June,

169 67 In a basketball game, Marlene made 16 field goals. Each of the field goals were worth either 2 points or 3 points, and Marlene scored a total of 39 points from field goals. Part A Let x represent the number of two-point field goals and y represent the number of three-point field goals. Write a system of equations in terms of x and y to model the situation. When you finish, writing your answer, type the number "1" into your Responder. Answer PARCC - EOY - Question #16 Calculator Section - SMART Response Format 169

170 68 In a basketball game, Marlene made 16 field goals. Each of the field goals were worth either 2 points or 3 points, and Marlene scored a total of 39 points from field goals. Part B How many three-point field goals did Marlene make in the game? Answer PARCC - EOY - Question #16 Calculator Section 170

171 Applications: Systems of Equations We have one more Application that is an extension of Systems of Equations. It can also be found in the Dynamics Unit of our Algebra Based Physics course. Let's get started. 171

172 Extension: Tension Force The tension in a rope is the same everywhere in the rope. If two masses hang down from either side of a cable, for instance, the tension in both sides must be the same. FT1 = FT2 20 kg 50 kg "Atwood Machine" 172

173 Extension: Tension Force A 20 kg mass hangs from one end of a rope that passes over a small frictionless pulley. A 50 kg weight is suspended from the other end of the rope. Which way will the 20 kg mass accelerate? Which way will the 50 kg mass accelerate? a) Draw a Free Body Diagram for each mass b) Write the Net Force equation for each mass c) Find the equations for the tension force FT d) Find the equation for acceleration e) Find the value of the acceleration f) Find the value of the tension force 20 kg F T1 = F T2 50 kg "Atwood Machine" 173

174 Extension: Tension Force a) Draw a Free Body Diagram for each mass FT FT a 20 kg m1g F T1 = F T2 = F T 20 kg 50 kg m2g 50 kg a Answer Remember the tension in the rope is the same everywhere, so FT is the same for both masses. The direction of acceleration is also different. What about the magnitude of acceleration? 174

175 Extension: Tension Force b) Write the Net Force equation for each mass Remember there is no special equation for tension. We need to use net force to find the tension. Below each diagram, write the Net Force equation for each mass: + FT FT - Answer a - 20 kg m1g m2g 50 kg a + 175

176 - Extension: Tension Force b) Write the Net Force equation for each mass + FT F T a 20 kg - m1g 50 kg a ΣF = m1a FT - m1g = m1a m 2 g + Answer ΣF = m 2 a -F T + m 2 g = m 2 a What do you notice about how the signs were chosen for the various forces? 176

177 Extension: Tension Force c) Find the equations for the tension force F T We have two equations (one for each mass) and two unknowns (F T and a). This means we can combine the equations together to solve for each variable! F T - m 1 g = m 1 a -F T + m 2 g = m 2 a Solve each for F T : F T - m 1 g = m 1 a F T = m 1 g + m 1 a -F T + m 2 g = m 2 a F T = m 2 g - m 2 a Now we can set them equal to one another: m 1 g + m 1 a = m 2 g - m 2 a 177

178 Extension: Tension Force d) Find the equation for the acceleration Now we can combine the tension equations m 1 g + m 1 a = F T There is only one unknown (a) here. Solve for a: Add m 2 a and subtract m 1 g from both sides: factor out 'a' : (remember factoring is just the opposite of distributing) divide by (m 1 + m 2 ): F T = m 2 g - m 2 a m 1 g + m 1 a = m 2 g - m 2 a m 1 a + m 2 a = m 2 g - m 1 g a(m 1 + m 2 ) = m 2 g - m 1 g a = m 2 g - m 1 g m 1 + m 2 178

179 Extension: Tension Force e) Find the value of the acceleration Substitute and solve: Remember: this is the acceleration for both m1 and m2. 20kg 50kg Answer 179

180 Extension: Tension Force f) Find the value of the tension Now we can use either equation to solve for Tension: Answer 20kg 50kg 180

181 Standards Return to Table of Contents 181

182 Throughout this unit, the Standards for Mathematical Practice are used. MP1: Making sense of problems & persevere in solving them. MP2: Reason abstractly & quantitatively. MP3: Construct viable arguments and critique the reasoning of others. MP4: Model with mathematics. MP5: Use appropriate tools strategically. MP6: Attend to precision. MP7: Look for & make use of structure. MP8: Look for & express regularity in repeated reasoning. Additional questions are included on the slides using the "Math Practice" Pull-tabs (e.g. a blank one is shown to the right on this slide) with a reference to the standards used. If questions already exist on a slide, then the specific MPs that the questions address are listed in the Pull-tab. Math Practice 182