1. Solving Systems of Equations

2. Graphing
There are three methods to solving systems of equations by graphing:
Write both equations in slope – intercept form and graph
Write both equations in slope-intercept form and graph using the calculator
Solve for the x and y intercepts of each equation

3. Graphing Solve the following system of equations by graphing:
-6x +3y = -15
-4x +y = -11

4. Graphing – Method 1 -6x +3y = -15 -4x +y = -11
Write both equations in slope – intercept form and graph. To do this, solve each equation for y
-6x +3y = -15
-4x +y = -11

5. Graphing – Method 1 Writing -6x +3y = -15 in slope intercept form:
+6x x
3y = 6x – 15
y = 2x - 5

6. Graphing – Method 1 Writing -4x + y = -11 in slope intercept form:
+4x x
y = 4x - 11

7. Graphing – Method 1
Graph both equations using their slope and y –intercept by starting at the y-intercept and using their slope to do rise over run.
Equation 1: y = 2x - 5
y intercept is (0, -5) slope is rise 2, run 1
Equation 2: y = 4x – 11
y intercept is (0, -11) slope is rise 4, run 1

8. Graphing – Method 1
The lines intersect at the point (1,3)

9. Graphing – Method 2 -6x +3y = -15 -4x +y = -11
Write both equations in slope-intercept form and graph using the calculator
-6x +3y = -15
-4x +y = -11
The equations were already solved for slope-intercept form in method 1, so:
y = 2x – 5
y = 4x – 11

10. Graphing – Method 2 (TI-84+)
Turn the calculator on
Hit the “Y=” key
Type in the first equation next to Y1
Use the “X,T,O,n” key to type “X”
Hit “Enter”
Type in the second equation next to Y2

11. Graphing – Method 2 (TI-84+)
8) Hit the graph button to see the graph

12. Graphing – Method 2 (TI-84+)
9) If necessary, adjust the graph by changing the zoom
You can zoom in, or out by hitting the zoom button and then selecting option 2 or 3. Once selected, press enter again when you see the graph
Zoom standard goes back to the regular zoom

13. Graphing – Method 2 (TI-84+)
10) When looking at the graph hit the “CALC” button. Do this by hitting the “2ND” key followed by the “TRACE” key
11) Move down to choice five and select “intersect”
12) Press “Enter” and the calculator will return to the graph.

14. Graphing – Method 2 (TI-84+)
13) The calculator will prompt you to select the first curve. Use the arrows to put the blinking cursor on one of the lines
14) Hit “Enter”

15. Graphing – Method 2 (TI-84+)
15) The calculator will prompt you to select the second curve. Use the arrows to put the blinking cursor on the other line (the calculator should have already done this for you)
16) Hit “Enter”

16. Graphing – Method 2 (TI-84+)
17) The calculator will prompt you to guess the location of the intersection. Use the arrow keys to move the flashing curser close to the intersection
18) Hit “Enter”

17. Graphing – Method 2 (TI-84+)
19) The calculator will then tell you the intersection. In this case, “X=3, Y=1”
20) Write your answer as an ordered pair (3,1)

18. Graphing – Method 2 (TI-89)
Turn the calculator on
Hit the “Y=” key by hitting Diamond + F1
Type in the first equation next to Y1
Hit “Enter”
Type in the second equation next to Y2

19. Graphing – Method 2 (TI-89)
7) Hit the graph button to see the graph
- Do this by hitting diamond and then F3

20. Graphing – Method 2 (TI-89)
8) If necessary, adjust the graph by changing the zoom
You can zoom in, or out by hitting the zoom button (F2) and then selecting option 2 or 3. Once selected, press enter again when you see the graph
Zoom standard (option 6) goes back to the regular zoom

21. Graphing – Method 2 (TI-89)
9) When looking at the graph select the “Math” menu. Do this by hitting the “F5” key
10) Move down to choice five and select “intersection”
11) Press “Enter” and the calculator will return to the graph.

22. Graphing – Method 2 (TI-89)
12) The calculator will prompt you to select the first curve. Use the arrows to put the blinking cursor on one of the lines
13) Hit “Enter”

23. Graphing – Method 2 (TI-89)
14) The calculator will prompt you to select the second curve. Use the arrows to put the blinking cursor on the other line (the calculator should have already done this for you)
15) Hit “Enter”

24. Graphing – Method 2 (TI-89)
16) The calculator will prompt you to select the lower bound of the intersection. Use the arrow keys to move below or to the left of the intersection
17) Hit “Enter”

25. Graphing – Method 2 (TI-89)
18) The calculator will prompt you to select the upper bound of the intersection. Use the arrow keys to move above or to the right of the intersection
19) Hit Enter

26. Graphing – Method 2 (TI-89)
20) The calculator will then tell you the intersection. In this case, “X=3, Y=1”
21) Write your answer as an ordered pair (3,1)

27. Graphing – Method 3 -6x +3y = -9 -4x +y = -8
Graph by solving for the x and y intercepts of each equation:
-6x +3y = -9
-4x +y = -8

28. Graphing – Method 3 Find the x and y intercepts of the first equation:
-6x +3y = -15
x-intercept, let y=0
-6x +3y = -15
-6x +3(0) = -15
-6x = -15
x = -15/-6 = 5/2
x-int = (5/2,0)
y-intercept, let x=0
-6x +3y = -15
-6(0) + 3y = -15
3y = -15 3
y = -5
y-int = (0,-5)

29. Graphing – Method 3
Find the x and y intercepts of the second equation:
-4x +y = -11
x-intercept, let y=0
-4x +y = -11
-4x +(0) = -11
-4x = -11
x = -11/-4 = 11/4
x-int = (11/4,0)
y-intercept, let x=0
-4x +y = -11
-4(0) + y = -11
y = -11
y-int = (0,-11)

30. Graphing – Method 3
Graph by plotting the x and y intercepts of each line and connecting them to form the line
The solution is the intersection: the point (3,1)

31. Substitution 1) Solve one of the equations for a variable
2) Substitute the solved equation into the OTHER equation in place of the variable you solved for
3) Solve the new equation for the remaining variable.
4) Once you find an answer for the first variable, substitute that answer into one of the original two equations and solve for the second variable.
5) Write your answer as an ordered pair.

32. Substitution
** If both variables cancel out, the lines are either parallel, or they are the same line
If the lines are parallel then there is no solution, and when the equation is solved, it will result in an answer that is NOT true like 0 = 4
If the lines are the same, then there is an infinite number of solutions, and when the equation is solved, it will result in an answer that IS true like 0=0 or 5=5.

33. Substitution – Example 1
Solve the following two equations using substitution:
y = 2x + 1
3x – 2y = -4

34. Substitution – Example 1
Solve one of the equations for a variable
- The first equation is already solved for y:
y = 2x + 1

35. Substitution – Example 1
2) Substitute the solved equation into the OTHER equation in place of the variable you solved for
So, substitute y = 2x + 1 into 3x – 2y = -4
3x – 2(2x + 1) = -4

36. Substitution – Example 1
3) Solve the new equation for the remaining variable.
3x – 2(2x + 1) = -4
3x -2(2x) -2(1) = -4
3x – 4x – 2 = -4
-x – 2 = -4
-x = -2
x = 2

37. Substitution – Example 1
4) Once you find an answer for the first variable, substitute that answer into one of the original two equations and solve for the second variable.
y = 2x + 1 and x = 2, so
y = 2(2) + 1
y = 4 + 1
y = 5

38. Substitution – Example 1
5) Write your answer as an ordered pair.
x = 2 and y = 5 so the answer is:
(2,5)

39. Substitution – Example 1
To check your work, substitute your answer into all of the original equations:
Substitute in (2, 5) for x, y
y = 2x + 1
5 = 2(2) + 1
5 = 4+1
5 = 5
Both equations balance, so our answer is a solution
3x – 2y = -4
3(2) – 2(5) = -4
6 – 10 = -4
-4 = -4

40. Substitution – Example 2
Solve the following two equations using substitution:
3x - 2y = 5
4x + 4y = 20

41. Substitution – Example 2
Solve one of the equations for a variable
- Solving the second equation for y is the easiest:
4x + 4y = 20
-4x x
4y = 20 – 4x
y = 5 - x

42. Substitution – Example 2
2) Substitute the solved equation into the OTHER equation in place of the variable you solved for
So, substitute y = 5 – x into 3x - 2y = 5
3x – 2(5-x) = 5

43. Substitution – Example 2
3) Solve the new equation for the remaining variable.
3x – 2(5-x) = 5
3x - 2(5) - 2(-x) = 5
3x x = 5
5x – 10 = 5
5x = 15 5
x = 3

44. Substitution – Example 2
4) Once you find an answer for the first variable, substitute that answer into one of the original two equations and solve for the second variable.
- Substitute x = 3 into 3x - 2y = 5
3(3) – 2y = 5
9 – 2y = 5
-2y = -4
y = 2

45. Substitution – Example 2
5) Write your answer as an ordered pair.
x = 3 and y = 2 so the ordered pair is:
(3,2)

46. Substitution – Example 3
Solve the following two equations using substitution:
-2x + 4y = -12
-x + 2y = 2

47. Substitution – Example 3
Solve one of the equations for a variable
- Solving the second equation for x is very easy:
-x + 2y = 2
-2y y
-x = 2 – 2y
x = y

48. Substitution – Example 3
2) Substitute the solved equation into the OTHER equation in place of the variable you solved for
So, substitute x= y into -2x + 4y = -12
-2(-2+2y) +4y = -12

49. Substitution – Example 3
3) Solve the new equation for the remaining variable.
-2(-2+2y) +4y = -12
-2(-2) + -2(2y) + 4y = -12
4 – 4y + 4y = -12
4 = -12
** Both variables are eliminated, and
4 = -12 is not true. Thus the lines are parallel and the answer is “no solution”

50. Substitution – Example 4
Solve the following two equations using substitution:
-2x + y = 3
6x + -3y = -9

51. Substitution – Example 4
Solve one of the equations for a variable
**Solving the first equation for y is very easy:
-2x + y = 3
+2x x
y = 2x + 3

52. Substitution – Example 4
2) Substitute the solved equation into the OTHER equation in place of the variable you solved for
So, substitute y = 2x + 3 into 6x + -3y = -9
6x + -3(2x+3) = -9

53. Substitution – Example 4
3) Solve the new equation for the remaining variable.
6x + -3(2x+3) = -9
6x + -3(2x) + -3(3) = -9
6x – 6x – 9 = -9
-9 = -9
0 = 0
** Both variables are eliminated, and 0=0 is true. Thus the lines are the same and the answer is “Infinite solutions”

54. Elimination
The goal of elimination is to add the equations in a system together and have a variable cancel out and be eliminated

55. Elimination
1) Make sure the equations are in the same form. i.e. both in standard form so that the x’s, y’s, and numbers all line up in columns
2) See if the equations can be added together immediately to eliminate a variable
3) If the equations cannot be added together to eliminate a variable, choose a variable to eliminate and multiply one or more of the equations by an integer in order to have the same coefficient in from of the variable in both equations. (note, one should be positive, one should be negative)

56. Elimination
4) Add the two equations together and make sure one variable was eliminated (leaving you with one variable)
5) Solve the equation for the variable
6) Substitute back into one of the original equations to solve for the second variable.
7) Write your answer as an ordered pair

57. Elimination
** If both variables are eliminated, the lines are either parallel, or they are the same line
If the lines are parallel then there is no solution, and when the equation is solved, it will result in an answer that is NOT true like 0 = 4
If the lines are the same, then there is an infinite number of solutions, and when the equation is solved, it will result in an answer that IS true like 0=0 or 5=5.

58. Elimination – Example 1
Solve the following two equations using substitution:
2x + 5y = -12
-5x - 5y = -3

59. Elimination – Example 1 Make sure the equations are in the same form:
2x + 5y = -12 and
-5x - 5y = -3 are both in standard form
** Note how the variables line up!
2) See if the equations can be added together immediately to eliminate a variable
Yes they can be! (skip step 3)

60. Elimination – Example 1
4) Add the two equations together and make sure one variable was eliminated (leaving you with one variable)
2x + 5y = -12
+ -5x - 5y = -3
-3x = => -3x = -15

61. Elimination – Example 1 5) Solve the equation for the variable
x = 5

62. Elimination – Example 1
6) Substitute back into one of the original equations to solve for the second variable.
x = 5; 2x + 5y = -12 so,
2(5) + 5y = -12
10 + 5y = -12
5y = -22 5
y = -22/5

63. Elimination – Example 1 7) Write your answer as an ordered pair
x = 5, y = -22/5 so,
(5, -22/5)

64. Elimination – Example 1 2x + 5y = -12 2(5) + 5(-22/5) = -12
To check your work, substitute your answer into all of the original equations:
Substitute in (5, -22/5) for x, y
2x + 5y = -12
2(5) + 5(-22/5) = -12
-12 = -12
Both equations balance,
so our answer is a solution
-5x - 5y = -3
-5(5) – 5(-22/5) = -3
= -3
= -3
-3 = -3

65. Elimination – Example 2
Solve the following two equations using substitution:
3x + 5y = 2
2x - 4y = 16

66. Elimination – Example 2 Make sure the equations are in the same form:
3x + 5y = and
2x - 4y = are both in standard form
** Note how the variables line up!

67. Elimination – Example 2 3x + 5y = 2 2x - 4y = 16
2) See if the equations can be added together immediately to eliminate a variable
No They cannot (multiply equations by a scalar)
3x + 5y = 2
2x - 4y = 16
Lcm of 3 and 2 is 6, so I will make 6x in the first equation by multiplying the whole first equation by 2, and I will make -6 in the second equation by multiplying the whole second equation by -3

68. Elimination – Example 2 (2)3x + (2)5y = 2(2) (-3)2x – (-3)4y = 16(-3)
Rewriting the equations:
6x + 10y = 4
-6x + 12y = -48
* Now the equations can be added to eliminate a variable

69. Elimination – Example 2
4) Add the two equations together and make sure one variable was eliminated (leaving you with one variable)
6x + 10y = 4
x + 12y = -48
y = -44 => 22y = -44

70. Elimination – Example 2 5) Solve the equation for the variable
22y = -44
y = -2

71. Elimination – Example 2
6) Substitute back into one of the original equations to solve for the second variable.
y = -2; 3x + 5y = 2 so,
3x + 5(-2) = 2
3x – 10 = 2
3x = 12
x = 4

72. Elimination – Example 2 7) Write your answer as an ordered pair
x = 4, y = -2 so,
(4, -2)

73. Elimination – Example 2
To check your work, substitute your answer into all of the original equations:
Substitute in (4, -2) for x, y
3x + 5y = 2
3(4) + 5(-2) = 2
= 2
2 = 2
Both equations balance, so our answer is a solution
2x - 4y = 16
2(4) – 4(-2) = 16
8 + 8 = 16
16 = 16

74. Applications of Linear Systems
Write systems of linear equations
Model real world situations with systems of equations

75. Steps for Solving Application Problems
1) Read the problem carefully. Highlight given information
2) Determine the unknowns and define variables.
3) Write a system of equations
4) Solve using any method (graphically, substitution, or elimination)

76. Example 1
A football club publishes a monthly newsletter. The newsletter costs 90 cents for printing and mailing each copy, plus $600 to write. If the newsletter sells for $1.50 per copy, how many copies of the newsletter must the club sell to break even?

77. Step 1 – Read and Highlight
A football club publishes a monthly newsletter. The newsletter costs 90 cents for printing and mailing each copy, plus $600 to write. If the newsletter sells for $1.50 per copy, how many copies of the newsletter must the club sell to break even?

78. Step 2 – Define Variables
Let x = the number of copies printed and sold
Let y = the amount of money spent or earned

79. Step 3 – Write a System of Equations
Equation 1: Cost of printing
y = $0.90x + $600
Equation 2: Money Earned
y = $1.50x
Note: y is the same in both equations because we want income to equal expenses

80. Step 4 - Solve Method of solving Chosen: Substitution
y = $0.90x + $600
y = $1.50x
$0.90x + $600 = $1.50x
- $0.90x $0.90x
$600 = $0.60x
$ $0.60
1000 = x

81. Step 4 (Continued)- Solve
Substitute back in to solve for y:
y = $1.50x and x = 1000 so,
y = $1.50(1000)
y = $1500
The club must sell 1000 copies to break even. After 1000 copies they will have spent and earned $1500

82. Example 2
A store sells two types of ipod touches. The 8gb version sells for $ The 64gb version sells for $ If the store did $ in sales last year after they sold 30 ipods, how many of each type of ipod did they sell?

83. Step 1 – Read and Highlight
A store sells two types of ipod touches. The 8gb version sells for $ The 64gb version sells for $ If the store did $ in sales last year after they sold 30 ipods, how many of each type of ipod did they sell?

84. Step 2 – Define Variables
Let x = the number of 8gb ipods sold
Let y = the number of 64gb ipods sold

85. Step 3 – Write a System of Equations
Equation 1: Total sales
$ = $249.99x + $329.99y
Equation 2: Total number of ipods
30 = x + y

86. Step 4 - Solve Method of solving Chosen: Substitution
$ = $249.99x + $329.99y
30 = x + y
Solve the 2nd equation for y
-x -x
30 – x = y

87. Step 4 (Continued)- Solve
Substitute the 2nd equation into the first equation:
$ = $249.99x + $329.99y
30 – x = y
$ = $249.99x + $329.99(30-x)
$ = $249.99x + $ $329.99x

88. Step 4 (Continued)- Solve
$ = $249.99x + $ $329.99x
$ = $249.99x - $329.99x + $
$ = -$80x + $
-$ $
-$1520 = -$80x
-$ $80
19 = x

89. Step 4 (Continued)- Solve
Substitute back in to solve for y:
30 = x + y and x = 19 so,
30 = 19 + y
11 = y
So the company sold 19 ipods that were 8gb and 11 ipods that were 64gb

90. Linear Inequalities
A Linear Inequality is used to describe a region of a coordinate plane that is bounded by one or more lines
The solutions of a linear inequality is any point that makes the inequality true.

91. Graphing Linear Inequalities
Write the equation in slope intercept form
Replace the inequality symbol with an equal sign
Graph the equation use a dashed line for > or < and a solid line for ≤ or ≥
Pick test points in each section that the graph bounds (i.e. pick points both above and below the line)
Shade the regions in which the test points are true

92. Example 1 y ≤ -3x + 3 Write in slope intercept form
Change the inequality to an = sign
Graph y = -3x + 3

93. Example 1 y ≤ -3x + 3 4) Pick test points above and below the graph
(-2, 1)
-2 ≤ -3(1) + 3
-2 ≤ 0
(2,3)
2 ≤ -3(3) +3
2 ≤ -6

94. Example 1
y ≤ -3x + 3
5) Shade the region where the test points are true

95. Graphing Inequalities
To shade a graph, first determine whether you need to shade above or below the graph using test points
Go to the y= screen
Hit the y= button on the TI-84
Hit Diamond + F1 on the TI-89

96. Graphing Inequalities (TI-84+)
3) In the y= screen, type the equation
4) To change the shading, move your curser to highlight the \ on the far left of the equation. Each time you hit enter will change how the calculator graphs

97. Graphing Inequalities (TI-84+)
5) Hit enter two times to shade the graph below the line. The \ will have changed to this:

98. Graphing Inequalities (TI-84+)
6) Hit enter a third time to shade the graph below the line. The \ will have changed to this:

99. Graphing Inequalities (TI-84+)
6) Hit enter a total of 6 times (3 more times) time to graph a dashed line instead of a solid line. The \ will have changed to this:

100. Graphing Inequalities
To shade a graph, first determine whether you need to shade above or below the graph using test points
Go to the y= screen
Hit the y= button on the TI-84
Hit Diamond + F1 on the TI-89

101. Graphing Inequalities (TI-89)
3) In the y= screen, type the equation

102. Graphing Inequalities (TI-89)
4) To change the shading, hit F6 by hitting “2nd” and then “F1”

103. Graphing Inequalities (TI-89)
5) A menu will appear
Select choice 7 (Above) to shade the region above the line
Select choice 8 (Below) to shade the region below the line
Select choice 2 (Dot) to graph a dotted line

104. Example 2 – Test Points
Determine whether or not the point (2, -3) is a solution to a linear inequality y > -3x + 5
Substitute in for y and x
y > -3x + 5
-3 > -3(2) + 5
-3 > -6 +5
-3 > -1 => NO, this point is NOT a solution

105. Example 3 – Test Points
Determine whether or not the point (6, -1) is a solution to a linear inequality y > (2/3)x -2
Substitute in for y and x
y < (2/3)x - 2
-1 < (2/3)(6) - 2
-1 < 4 - 2
-1 < 2 => YES, this point IS a solution

106. Example 4 – Describing Graphs
Given the graph shown to the right, which equation describes the graph?
y = -2x + 1
y < -2x + 1
y > -2x + 1
y ≥ -2x + 1
y ≤ -2x + 1

107. Example 4 – Describing Graphs
y = -2x + 1 => Wrong because a region of the graph is shaded. The equation is just the equation of a line, not a region of a plane.

108. Example 4 – Describing Graphs
b) y < -2x + 1 => Wrong because the equation is < which means the line should be dashed, not solid

109. Example 4 – Describing Graphs
c) y > -2x + 1 => Wrong because the equation is > which means the line should be dashed, not solid

110. Example 4 – Describing Graphs
y ≥ -2x + 1 => Correct. The line is shaded and the test point (2,3) means that ≥ -2(2) + 1 so 3 ≥ thus, 3 ≥ -3, which is true so the shaded region is above the graph

111. Example 4 – Describing Graphs
e) y ≤ -2x + 1 => Wrong because the test point (2,3) means that 3 ≤ -2(2) + 1 so 3 ≤ thus, 3 ≤ -3, which is false.

112. Example 5 –Graphing Graph 4x - 2y ≥ 8 Write in Slope-Intercept form
-4x x
-2y ≥ -4x + 8
y ≤ 2x – 4
** Note: The sign changes because we divided by a negative**

113. Example 5 –Graphing y ≤ 2x – 4 2) Change the inequality to an = sign
3) Graph – the line is solid because it is ≤

114. Example 5 –Graphing Below: (0, -5) Above: (0,0) y ≤ 2x – 4 y ≤ 2x – 4
4) Pick test points above and below the graph.
Below: (0, -5)
y ≤ 2x – 4
-5 ≤ 2(0) -4
-5 ≤ -4 => True
Above: (0,0)
y ≤ 2x – 4
0 ≤ 2(0) – 4
0 ≤ -4 =>False
Thus, we shade the region below the line

115. Example 5 –Graphing