SlideShare a Scribd company logo

3 3 systems of linear equations 1

M
math123a

The document discusses systems of linear equations. It explains that to solve for two unknowns, two equations are needed, and to solve for three unknowns, three equations are required. A system of linear equations consists of two or more linear equations with two or more variables. A solution to the system is a set of numbers for each variable that satisfies all equations. Two examples are provided to demonstrate solving systems of two equations with two unknowns to find the price of items.

EducationTechnologyBusiness
1 of 79
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
Systems of Linear Equations Back to 123a-Home
If 2 hamburgers cost 4$, then one hamburger is 2$. Systems of Linear Equations
If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. Systems of Linear Equations
If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. But if a hamburger and a drink cost $4 then we won’t be able to pin down the price of each item-not enough information. Systems of Linear Equations
If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. But if a hamburger and a drink cost $4 then we won’t be able to pin down the price of each item-not enough information. In general, to find two unknowns, two pieces of information are needed, for three unknowns, three pieces of information, etc… Systems of Linear Equations
If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. But if a hamburger and a drink cost $4 then we won’t be able to pin down the price of each item-not enough information. In general, to find two unknowns, two pieces of information are needed, for three unknowns, three pieces of information, etc… A system of linear equations is a collection two or more linear equations in two or more variables. Systems of Linear Equations
If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. But if a hamburger and a drink cost $4 then we won’t be able to pin down the price of each item-not enough information. In general, to find two unknowns, two pieces of information are needed, for three unknowns, three pieces of information, etc… A system of linear equations is a collection two or more linear equations in two or more variables. A solution for the system is a collection of numbers, one for each variable, that works for all equations in the system. Systems of Linear Equations
If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. But if a hamburger and a drink cost $4 then we won’t be able to pin down the price of each item-not enough information. In general, to find two unknowns, two pieces of information are needed, for three unknowns, three pieces of information, etc… A system of linear equations is a collection two or more linear equations in two or more variables. A solution for the system is a collection of numbers, one for each variable, that works for all equations in the system. For example, 2x + y = 7 x + y = 5 is a system { Systems of Linear Equations
If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. But if a hamburger and a drink cost $4 then we won’t be able to pin down the price of each item-not enough information. In general, to find two unknowns, two pieces of information are needed, for three unknowns, three pieces of information, etc… A system of linear equations is a collection two or more linear equations in two or more variables. A solution for the system is a collection of numbers, one for each variable, that works for all equations in the system. For example, 2x + y = 7 x + y = 5 is a system and that x = 2 and y = 3 is a solution because the pair fits both equations. { Systems of Linear Equations 2(2) + (3) = 7 (2) + (3) = 5 ck:
If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. But if a hamburger and a drink cost $4 then we won’t be able to pin down the price of each item-not enough information. In general, to find two unknowns, two pieces of information are needed, for three unknowns, three pieces of information, etc… A system of linear equations is a collection two or more linear equations in two or more variables. A solution for the system is a collection of numbers, one for each variable, that works for all equations in the system. For example, 2x + y = 7 x + y = 5 is a system and that x = 2 and y = 3 is a solution because the pair fits both equations. This solution is written as (2, 3). { Systems of Linear Equations 2(2) + (3) = 7 (2) + (3) = 5 ck:
Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Example A. Systems of Linear Equations
Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. Example A. Systems of Linear Equations
Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. We can translate the information into the system 2x + y = 7 Example A. E1 Systems of Linear Equations
Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. We can translate the information into the system 2x + y = 7 x + y = 5{ Example A. E1 E2 Systems of Linear Equations
Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. We can translate the information into the system 2x + y = 7 x + y = 5{ The difference between the order is one hamburger and the difference between the cost is $2, so the hamburger is $2. Example A. E1 E2 Systems of Linear Equations
Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. We can translate the information into the system 2x + y = 7 x + y = 5{ The difference between the order is one hamburger and the difference between the cost is $2, so the hamburger is $2. In symbols, subtract the equations, i.e. E1 – E2: Example A. 2x + y = 7 ) x + y = 5 E1 E2 Systems of Linear Equations
Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. We can translate the information into the system 2x + y = 7 x + y = 5{ The difference between the order is one hamburger and the difference between the cost is $2, so the hamburger is $2. In symbols, subtract the equations, i.e. E1 – E2: Example A. 2x + y = 7 ) x + y = 5 x + 0 = 2 so x = 2 E1 E2 Systems of Linear Equations
Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. We can translate the information into the system 2x + y = 7 x + y = 5{ The difference between the order is one hamburger and the difference between the cost is $2, so the hamburger is $2. In symbols, subtract the equations, i.e. E1 – E2: Example A. substituting 2 for x back into E2, we get: 2 + y = 5 or y = 3 2x + y = 7 ) x + y = 5 x + 0 = 2 so x = 2 E1 E2 Systems of Linear Equations
Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. We can translate the information into the system 2x + y = 7 x + y = 5{ The difference between the order is one hamburger and the difference between the cost is $2, so the hamburger is $2. In symbols, subtract the equations, i.e. E1 – E2: Example A. substituting 2 for x back into E2, we get: 2 + y = 5 or y = 3 Hence the solution is (2 , 3) or, $2 for a hamburger, $3 for a salad. 2x + y = 7 ) x + y = 5 x + 0 = 2 so x = 2 E1 E2 Systems of Linear Equations
Example B. Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost?
Example B. Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad.
Example B. Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system:
4x + 3y = 18 3x + 2y = 13 Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system: Example B. { E1 E2 Systems of Linear Equations
4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. E1 E2 Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system:
4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate x or y. E1 E2 Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system:
4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate x or y. Suppose we chose to eliminate the y-terms, we find the LCM of 3y and 2y first. It's 6y. E1 E2 Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system:
4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate x or y. Suppose we chose to eliminate the y-terms, we find the LCM of 3y and 2y first. It's 6y. Multiplying E1 by (–2), E2 by 3, E1 E2 Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system:
4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate x or y. Suppose we chose to eliminate the y-terms, we find the LCM of 3y and 2y first. It's 6y. Multiplying E1 by (–2), E2 by 3, E1 E2 – 8x – 6y = –36E1*(–2) : Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system:
4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate x or y. Suppose we chose to eliminate the y-terms, we find the LCM of 3y and 2y first. It's 6y. Multiplying E1 by (–2), E2 by 3, E1 E2 – 8x – 6y = –36 9x + 6y = 39 E1*(–2) : E2*(3): Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system:
4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate x or y. Suppose we chose to eliminate the y-terms, we find the LCM of 3y and 2y first. It's 6y. Multiplying E1 by (–2), E2 by 3, and add the results, the y-terms are eliminated. E1 E2 – 8x – 6y = –36 9x + 6y = 39 E1*(–2) : E2*(3): + ) Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system: We adjust the y’s into opposite signs then add to eliminate them.
4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate x or y. Suppose we chose to eliminate the y-terms, we find the LCM of 3y and 2y first. It's 6y. Multiplying E1 by (–2), E2 by 3, and add the results, the y-terms are eliminated. E1 E2 – 8x – 6y = –36 9x + 6y = 39 E1*(–2) : E2*(3): + ) x = 3 Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system: So a hamburger is 3$. We adjust the y’s into opposite signs then add to eliminate them.
To find y, set 3 for x in E2: 3x + 2y = 13 Systems of Linear Equations
To find y, set 3 for x in E2: and get 3(3) + 2y = 13 3x + 2y = 13 Systems of Linear Equations
To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 3x + 2y = 13 Systems of Linear Equations
To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 3x + 2y = 13 Systems of Linear Equations
To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). 3x + 2y = 13 Systems of Linear Equations
To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). Therefore a hamburger cost $3 and a salad cost $2. 3x + 2y = 13 Systems of Linear Equations
To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). Therefore a hamburger cost $3 and a salad cost $2. 3x + 2y = 13 The above method is called the elimination (addition) method. We summarize the steps below. Systems of Linear Equations
To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). Therefore a hamburger cost $3 and a salad cost $2. 3x + 2y = 13 The above method is called the elimination (addition) method. We summarize the steps below. 1. Line up the x’s & y’s to the left and the numbers to the right. Systems of Linear Equations
To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). Therefore a hamburger cost $3 and a salad cost $2. 3x + 2y = 13 The above method is called the elimination (addition) method. We summarize the steps below. 1. Line up the x’s & y’s to the left and the numbers to the right. 2. Select the variable x or y to eliminate. Systems of Linear Equations
To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). Therefore a hamburger cost $3 and a salad cost $2. 3x + 2y = 13 The above method is called the elimination (addition) method. We summarize the steps below. 1. Line up the x’s & y’s to the left and the numbers to the right. 2. Select the variable x or y to eliminate. 3. Find the LCM of the terms with that variable, and convert these terms to the LCM (by multiplication). Systems of Linear Equations
To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). Therefore a hamburger cost $3 and a salad cost $2. 3x + 2y = 13 The above method is called the elimination (addition) method. We summarize the steps below. 1. Line up the x’s & y’s to the left and the numbers to the right. 2. Select the variable x or y to eliminate. 3. Find the LCM of the terms with that variable, and convert these terms to the LCM (by multiplication). 4. Add or subtract the adjusted equations and solve the resulting equation. Systems of Linear Equations
To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). Therefore a hamburger cost $3 and a salad cost $2. 3x + 2y = 13 The above method is called the elimination (addition) method. We summarize the steps below. 1. Line up the x’s & y’s to the left and the numbers to the right. 2. Select the variable x or y to eliminate. 3. Find the LCM of the terms with that variable, and convert these terms to the LCM (by multiplication). 4. Add or subtract the adjusted equations and solve the resulting equation. 5. Substitute the answer back into any equation to solve for the other variable. Systems of Linear Equations
5x + 4y = 2 2x – 3y = -13 Solve E1 E2 {Example C. Systems of Linear Equations
5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. {Example C. Systems of Linear Equations
5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. {Example C. Systems of Linear Equations
5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 {Example C. 15x + 12y = 63*E1: Systems of Linear Equations
5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 8x – 12y = - 52 3*E1: 4*E2: Systems of Linear Equations
5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: Systems of Linear Equations
5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46 Systems of Linear Equations
5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Systems of Linear Equations
5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 Systems of Linear Equations
5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 Systems of Linear Equations
5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y = 3 Systems of Linear Equations
5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y = 3 Hence the solution is (-2, 3). Systems of Linear Equations
5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y = 3 Hence the solution is (-2, 3). In all the above examples, we obtain exactly one solution in each case. Systems of Linear Equations
5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y = 3 Hence the solution is (-2, 3). In all the above examples, we obtain exactly one solution in each case. However, there are two other possibilities. Systems of Linear Equations
5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y = 3 Hence the solution is (-2, 3). In all the above examples, we obtain exactly one solution in each case. However, there are two other possibilities. Systems of Linear Equations The system might not have any solution
5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y = 3 Hence the solution is (-2, 3). In all the above examples, we obtain exactly one solution in each case. However, there are two other possibilities. Systems of Linear Equations The system might not have any solution or the system might have infinite many solutions.
Two Special Cases: I. Contradictory (Inconsistent) Systems Systems of Linear Equations
Two Special Cases: I. Contradictory (Inconsistent) Systems Example D. { x + y = 2 x + y = 3 (E1) (E2) Systems of Linear Equations
Two Special Cases: I. Contradictory (Inconsistent) Systems Example D. { x + y = 2 x + y = 3 (E1) (E2) Remove the x-terms by subtracting the equations. Systems of Linear Equations
Two Special Cases: I. Contradictory (Inconsistent) Systems Example D. { x + y = 2 x + y = 3 E1 – E2 : x + y = 2 ) x + y = 3 (E1) (E2) Remove the x-terms by subtracting the equations. Systems of Linear Equations
Two Special Cases: I. Contradictory (Inconsistent) Systems Example D. { x + y = 2 x + y = 3 E1 – E2 : x + y = 2 ) x + y = 3 0 = -1 (E1) (E2) Remove the x-terms by subtracting the equations. Systems of Linear Equations
Two Special Cases: I. Contradictory (Inconsistent) Systems Example D. { x + y = 2 x + y = 3 E1 – E2 : x + y = 2 ) x + y = 3 0 = -1 (E1) (E2) Remove the x-terms by subtracting the equations. This is impossible! Such systems are said to be contradictory or inconsistent. Systems of Linear Equations
Two Special Cases: I. Contradictory (Inconsistent) Systems Example D. { x + y = 2 x + y = 3 E1 – E2 : x + y = 2 ) x + y = 3 0 = -1 (E1) (E2) Remove the x-terms by subtracting the equations. This is impossible! Such systems are said to be contradictory or inconsistent. These system don’t have solution. Systems of Linear Equations
II. Dependent Systems { x + y = 3 2x + 2y = 6 Example E. (E1) (E2) Systems of Linear Equations
II. Dependent Systems { x + y = 3 2x + 2y = 6 Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations
II. Dependent Systems { x + y = 3 2x + 2y = 6 2*E1 2x + 2y = 6 Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations
II. Dependent Systems { x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations
II. Dependent Systems { x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0 = 0 Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations
II. Dependent Systems { x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0 = 0 This means the equations are actually the same and it has infinitely many solutions such as (3, 0), (2, 1), (1, 2) etc… Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations
{ x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0 = 0 This means the equations are actually the same and it has infinitely many solutions such as (3, 0), (2, 1), (1, 2) etc… Such systems are called dependent or redundant systems. Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations II. Dependent Systems
{ x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0 = 0 This means the equations are actually the same and it has infinitely many solutions such as (3, 0), (2, 1), (1, 2) etc… Such systems are called dependent or redundant systems. Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations II. Dependent Systems Finally, as before with linear equations, clear the denominators of fractional equations first.
{ x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0 = 0 This means the equations are actually the same and it has infinitely many solutions such as (3, 0), (2, 1), (1, 2) etc… Such systems are called dependent or redundant systems. Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations II. Dependent Systems Finally, as before with linear equations, clear the denominators of fractional equations first. For example, change { x – y = 3 x – y = –1 3 2 2 3 1 2 1 4
{ x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0 = 0 This means the equations are actually the same and it has infinitely many solutions such as (3, 0), (2, 1), (1, 2) etc… Such systems are called dependent or redundant systems. Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations II. Dependent Systems Finally, as before with linear equations, clear the denominators of fractional equations first. For example, change { x – y = 3 x – y = –1 3 2 2 3 1 2 1 4 ( )*6 ( )*4
{ x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0 = 0 This means the equations are actually the same and it has infinitely many solutions such as (3, 0), (2, 1), (1, 2) etc… Such systems are called dependent or redundant systems. Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations II. Dependent Systems Finally, as before with linear equations, clear the denominators of fractional equations first. For example, change { x – y = 3 x – y = –1 3 2 2 3 1 2 1 4 ( )*6 9x – 4y = 18 ( )*4 { 2x – y = –4 then solve.
Exercise. A. Select the solution of the system. {x + y = 3 2x + y = 4 Systems of Linear Equations {x + y = 3 2x + y = 5 {x + y = 3 2x + y = 6 {x + y = 3 2x + y = 31. a. (0, 3) b. (1, 2) c. (2, 1) d.(3, 0) 2. a. (0, 3) b. (1, 2) c. (2, 1) d.(3, 0) 3. a. (0, 3) b. (1, 2) c. (2, 1) d.(3, 0) 4. a. (0, 3) b. (1, 2) c. (2, 1) d.(3, 0) B. Solve the following systems. 5. {y = 3 2x + y = 3 6. {x = 1 2x + y = 4 7. {y = 1 2x + y = 5 8. {x = 2 2x + y = 6 C. Add or subtract vertically. (These are warm ups) 8x –9x+) 9. 8 –9–) 10. –8y –9y–) 11. –8 9–) 12. –4y –5y+) 13. +4x –5x–) 14.
Systems of Linear Equations 18.{–x + 2y = –12 2x + y = 4 D. Solve the following system by the elimination method. 15.{x + y = 3 2x + y = 4 16. 17.{x + 2y = 3 2x – y = 6 {x + y = 3 2x – y = 6 19. {3x + 4y = 3 x – 2y = 6 20. { x + 3y = 3 2x – 9y = –4 21.{–3x + 2y = –1 2x + 3y = 5 22. {2x + 3y = –1 3x + 4y = 2 23. {4x – 3y = 3 3x – 2y = –4 24.{5x + 3y = 2 2x + 4y = –2 25. {3x + 4y = –10 –5x + 3y = 7 26. {–4x + 9y = 1 5x – 2y = 8 { x – y = 3 x – y = –1 3 2 2 3 1 2 1 4 27. { x + y = 1 x – y = –1 1 2 1 5 3 4 1 6 28. 29.{ x + 3y = 4 2x + 6y = 8 E. Which system is inconsistent and which is dependent? 30.{2x – y = 2 8x – 4y = 6

More Related Content

PPTX
3 4 systems of linear equations 2
math123a
 
PPT
4.3 system of linear equations 1
math123c
 
PPT
4.4 system of linear equations 2
math123c
 
PPTX
6.1 system of linear equations and matrices
math260
 
PPTX
56 system of linear equations
alg1testreview
 
PPTX
2 1 expressions
math123a
 
PPTX
1 s5 variables and evaluation
math123a
 
PPTX
42 linear equations
alg1testreview
 
3 4 systems of linear equations 2
math123a
 
4.3 system of linear equations 1
math123c
 
4.4 system of linear equations 2
math123c
 
6.1 system of linear equations and matrices
math260
 
56 system of linear equations
alg1testreview
 
2 1 expressions
math123a
 
1 s5 variables and evaluation
math123a
 
42 linear equations
alg1testreview
 

What's hot (20)

PPTX
50 solving equations by factoring
alg1testreview
 
PPTX
2linear equations i x
Tzenma
 
PPTX
2 2linear equations i
math123a
 
PPTX
42 linear equations
alg-ready-review
 
PPTX
41 expressions
alg1testreview
 
PPTX
24 variables and evaluation
alg1testreview
 
PPTX
4.6 radical equations
math123b
 
PPTX
6 system of linear equations i x
Tzenma
 
PPT
81 systems of linear equations 1
math126
 
PPTX
2 6 inequalities
math123a
 
PPTX
49 factoring trinomials the ac method and making lists
alg1testreview
 
PPTX
4 3polynomial expressions
math123a
 
PPTX
1 expressions x
Tzenma
 
PPT
82 systems of linear equations 2
math126
 
PPTX
1 f2 fractions
math123a
 
PPTX
1 4 cancellation
math123b
 
PPTX
41 expressions
alg-ready-review
 
PPTX
2 expressions and linear expressions
elem-alg-sample
 
PDF
Maths 11
Mehtab Rai
 
PPTX
5 2factoring trinomial i
math123a
 
50 solving equations by factoring
alg1testreview
 
2linear equations i x
Tzenma
 
2 2linear equations i
math123a
 
42 linear equations
alg-ready-review
 
41 expressions
alg1testreview
 
24 variables and evaluation
alg1testreview
 
4.6 radical equations
math123b
 
6 system of linear equations i x
Tzenma
 
81 systems of linear equations 1
math126
 
2 6 inequalities
math123a
 
49 factoring trinomials the ac method and making lists
alg1testreview
 
4 3polynomial expressions
math123a
 
1 expressions x
Tzenma
 
82 systems of linear equations 2
math126
 
1 f2 fractions
math123a
 
1 4 cancellation
math123b
 
41 expressions
alg-ready-review
 
2 expressions and linear expressions
elem-alg-sample
 
Maths 11
Mehtab Rai
 
5 2factoring trinomial i
math123a
 
Ad

Viewers also liked (20)

PPTX
3 1 rectangular coordinate system
math123a
 
PPTX
3 2 linear equations and lines
math123a
 
PPTX
3 5linear word problems ii
math123a
 
PPTX
2 5literal equations
math123a
 
PPTX
2 4linear word problems
math123a
 
PDF
Lesson 1: Systems of Linear Equations (slides)
Matthew Leingang
 
DOCX
System of linear equations
Cesar Mendoza
 
PPTX
Solution of system of linear equations by elimination
Regie Panganiban
 
PPTX
38 equations of lines-x
math123a
 
PPTX
2 3linear equations ii
math123a
 
PPTX
1 s4 order of operations
math123a
 
PDF
Lesson 15: Linear Approximation and Differentials
Matthew Leingang
 
PPTX
linear programming
ST ZULAIHA NURHAJARURAHMAH
 
PPTX
Linear Systems - Expansion and Movement Joints
Ali Asgar Raja
 
PPTX
Chapter 3 linear systems
leblance
 
PDF
Math 1300: Section 4- 3 Gauss-Jordan Elimination
Jason Aubrey
 
PDF
Math 1300: Section 4-2 Systems of Linear Equations; Augmented Matrices
Jason Aubrey
 
PPTX
1 s2 addition and subtraction of signed numbers
math123a
 
PPTX
Electric Circuit - Lecture 04
Hassaan Rahman
 
PDF
metode iterasi Gauss seidel
Muhammad Ali Subkhan Candra
 
3 1 rectangular coordinate system
math123a
 
3 2 linear equations and lines
math123a
 
3 5linear word problems ii
math123a
 
2 5literal equations
math123a
 
2 4linear word problems
math123a
 
Lesson 1: Systems of Linear Equations (slides)
Matthew Leingang
 
System of linear equations
Cesar Mendoza
 
Solution of system of linear equations by elimination
Regie Panganiban
 
38 equations of lines-x
math123a
 
2 3linear equations ii
math123a
 
1 s4 order of operations
math123a
 
Lesson 15: Linear Approximation and Differentials
Matthew Leingang
 
linear programming
ST ZULAIHA NURHAJARURAHMAH
 
Linear Systems - Expansion and Movement Joints
Ali Asgar Raja
 
Chapter 3 linear systems
leblance
 
Math 1300: Section 4- 3 Gauss-Jordan Elimination
Jason Aubrey
 
Math 1300: Section 4-2 Systems of Linear Equations; Augmented Matrices
Jason Aubrey
 
1 s2 addition and subtraction of signed numbers
math123a
 
Electric Circuit - Lecture 04
Hassaan Rahman
 
metode iterasi Gauss seidel
Muhammad Ali Subkhan Candra
 
Ad

Similar to 3 3 systems of linear equations 1 (20)

PPTX
3 linear equations
elem-alg-sample
 
PPTX
7 system of linear equations ii x
Tzenma
 
PPTX
35 Special Cases System of Linear Equations-x.pptx
math260
 
PPTX
Ppt materi spltv pembelajaran 1 kelas x
MartiwiFarisa
 
PPT
Systems of equations and matricies
ST ZULAIHA NURHAJARURAHMAH
 
PDF
Bahan ajar materi spltv kelas x semester 1
MartiwiFarisa
 
PPTX
Alg II 3-3 Systems of Inequalities
jtentinger
 
PPTX
Alg II Unit 3-3-systemsinequalities
jtentinger
 
PPTX
LINEAR EQUATION IN TWO VARIABLES
Manah Chhabra
 
PDF
Mat2 mu0c4 estudiante
FranciscoPineda46
 
PPTX
Categories of systems of linear equation.pptx
nor_dumogho
 
PPT
Systems of equations by graphing by graphing sect 6 1
tty16922
 
PPT
A1, 6 1, solving systems by graphing (rev)
kstraka
 
PPTX
Lecture3
Hazel Joy Chong
 
PPT
6.1 presentation
Randall Micallef
 
PPT
Solving linear systems by the substitution method
butterflyrose0411
 
DOCX
Christmas 2013 8th_grade_semester_exam_study_guide Math Pre-Algebra
Taleese
 
PPT
Applications of Systmes
Jessica Polk
 
PPT
3 Systems Jeopardy - Copy.ppt
ssusere48744
 
PPTX
(7) Lesson 1.6 - Solve Proportional Relationships
wzuri
 
3 linear equations
elem-alg-sample
 
7 system of linear equations ii x
Tzenma
 
35 Special Cases System of Linear Equations-x.pptx
math260
 
Ppt materi spltv pembelajaran 1 kelas x
MartiwiFarisa
 
Systems of equations and matricies
ST ZULAIHA NURHAJARURAHMAH
 
Bahan ajar materi spltv kelas x semester 1
MartiwiFarisa
 
Alg II 3-3 Systems of Inequalities
jtentinger
 
Alg II Unit 3-3-systemsinequalities
jtentinger
 
LINEAR EQUATION IN TWO VARIABLES
Manah Chhabra
 
Mat2 mu0c4 estudiante
FranciscoPineda46
 
Categories of systems of linear equation.pptx
nor_dumogho
 
Systems of equations by graphing by graphing sect 6 1
tty16922
 
A1, 6 1, solving systems by graphing (rev)
kstraka
 
Lecture3
Hazel Joy Chong
 
6.1 presentation
Randall Micallef
 
Solving linear systems by the substitution method
butterflyrose0411
 
Christmas 2013 8th_grade_semester_exam_study_guide Math Pre-Algebra
Taleese
 
Applications of Systmes
Jessica Polk
 
3 Systems Jeopardy - Copy.ppt
ssusere48744
 
(7) Lesson 1.6 - Solve Proportional Relationships
wzuri
 

More from math123a (20)

PPTX
1 numbers and factors eq
math123a
 
PPTX
37 more on slopes-x
math123a
 
PPTX
36 slopes of lines-x
math123a
 
PDF
123a ppt-all-2
math123a
 
PPTX
7 inequalities ii exp
math123a
 
PPTX
115 ans-ii
math123a
 
PPTX
14 2nd degree-equation word problems
math123a
 
PPTX
Soluiton i
math123a
 
PPTX
123a test4-sample
math123a
 
DOC
Sample fin
math123a
 
DOCX
12 4- sample
math123a
 
DOC
F12 2 -ans
math123a
 
DOCX
F12 1-ans-jpg
math123a
 
PPT
Sample1 v2-jpg-form
math123a
 
PPTX
1exponents
math123a
 
PPTX
3 6 introduction to sets-optional
math123a
 
PPTX
1 f5 addition and subtraction of fractions
math123a
 
PPTX
1 f4 lcm and lcd
math123a
 
PPTX
1 f2 fractions
math123a
 
PPTX
1 f7 on cross-multiplication
math123a
 
1 numbers and factors eq
math123a
 
37 more on slopes-x
math123a
 
36 slopes of lines-x
math123a
 
123a ppt-all-2
math123a
 
7 inequalities ii exp
math123a
 
115 ans-ii
math123a
 
14 2nd degree-equation word problems
math123a
 
Soluiton i
math123a
 
123a test4-sample
math123a
 
Sample fin
math123a
 
12 4- sample
math123a
 
F12 2 -ans
math123a
 
F12 1-ans-jpg
math123a
 
Sample1 v2-jpg-form
math123a
 
1exponents
math123a
 
3 6 introduction to sets-optional
math123a
 
1 f5 addition and subtraction of fractions
math123a
 
1 f4 lcm and lcd
math123a
 
1 f2 fractions
math123a
 
1 f7 on cross-multiplication
math123a
 

Recently uploaded (20)

PDF
Microbial disease of the cardiovascular and lymphatic systems
KeyaSharma
 
PDF
Black Hat USA 2025 - Micro ICS Summit - ICS/OT Threat Landscape
Chris Sistrunk
 
PDF
What if we spent less time fighting change, and more time building what’s rig...
Derek Wenmoth
 
PDF
STATICS OF THE RIGID BODIES Hibbelers.pdf
pascualasturiaskaye4
 
PPTX
master seminar digital applications in india
ananyaray35
 
PDF
A GUIDE TO GENETICS FOR UNDERGRADUATE MEDICAL STUDENTS
DrBincy A. S
 
PDF
Chinmaya Tiranga quiz Grand Finale.pdf
Nitin Suresh
 
PDF
Classroom Observation Tools for Teachers
Nicaramirez
 
PDF
Practical Manual AGRO-233 Principles and Practices of Natural Farming
MilindKhedekar2
 
PDF
GENETICS IN BIOLOGY IN SECONDARY LEVEL FORM 3
ElishamichaelSamwel
 
PPTX
Final Presentation General Medicine 03-08-2024.pptx
ZaheerAhmad228692
 
PDF
Module 4: Burden of Disease Tutorial Slides S2 2025
Jonathan Hallett
 
PDF
grade 11-chemistry_fetena_net_5883.pdf teacher guide for all student
banteamlakmebratu738
 
PDF
Yogi Goddess Pres Conference Studio Updates
LDMMia Reiki Lectures
 
PDF
LDMMIA Reiki Yoga Finals Review Spring Summer
LDMMia Reiki Lectures
 
PPTX
Final Presentation General Medicine 03-08-2024.pptx
ZaheerAhmad228692
 
PPTX
UNIT III MENTAL HEALTH NURSING ASSESSMENT
Sonali Gupta
 
PDF
The Lost Whites of Pakistan by Jahanzaib Mughal.pdf
jahanzaibmughal6
 
PPTX
1st Inaugural Professorial Lecture held on 19th February 2020 (Governance and...
kawisojames10
 
PPTX
Cell Types and Its function , kingdom of life
ClaireMangundayao1
 
Microbial disease of the cardiovascular and lymphatic systems
KeyaSharma
 
Black Hat USA 2025 - Micro ICS Summit - ICS/OT Threat Landscape
Chris Sistrunk
 
What if we spent less time fighting change, and more time building what’s rig...
Derek Wenmoth
 
STATICS OF THE RIGID BODIES Hibbelers.pdf
pascualasturiaskaye4
 
master seminar digital applications in india
ananyaray35
 
A GUIDE TO GENETICS FOR UNDERGRADUATE MEDICAL STUDENTS
DrBincy A. S
 
Chinmaya Tiranga quiz Grand Finale.pdf
Nitin Suresh
 
Classroom Observation Tools for Teachers
Nicaramirez
 
Practical Manual AGRO-233 Principles and Practices of Natural Farming
MilindKhedekar2
 
GENETICS IN BIOLOGY IN SECONDARY LEVEL FORM 3
ElishamichaelSamwel
 
Final Presentation General Medicine 03-08-2024.pptx
ZaheerAhmad228692
 
Module 4: Burden of Disease Tutorial Slides S2 2025
Jonathan Hallett
 
grade 11-chemistry_fetena_net_5883.pdf teacher guide for all student
banteamlakmebratu738
 
Yogi Goddess Pres Conference Studio Updates
LDMMia Reiki Lectures
 
LDMMIA Reiki Yoga Finals Review Spring Summer
LDMMia Reiki Lectures
 
Final Presentation General Medicine 03-08-2024.pptx
ZaheerAhmad228692
 
UNIT III MENTAL HEALTH NURSING ASSESSMENT
Sonali Gupta
 
The Lost Whites of Pakistan by Jahanzaib Mughal.pdf
jahanzaibmughal6
 
1st Inaugural Professorial Lecture held on 19th February 2020 (Governance and...
kawisojames10
 
Cell Types and Its function , kingdom of life
ClaireMangundayao1
 

3 3 systems of linear equations 1

  • 1. Systems of Linear Equations Back to 123a-Home
  • 2. If 2 hamburgers cost 4$, then one hamburger is 2$. Systems of Linear Equations
  • 3. If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. Systems of Linear Equations
  • 4. If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. But if a hamburger and a drink cost $4 then we won’t be able to pin down the price of each item-not enough information. Systems of Linear Equations
  • 5. If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. But if a hamburger and a drink cost $4 then we won’t be able to pin down the price of each item-not enough information. In general, to find two unknowns, two pieces of information are needed, for three unknowns, three pieces of information, etc… Systems of Linear Equations
  • 6. If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. But if a hamburger and a drink cost $4 then we won’t be able to pin down the price of each item-not enough information. In general, to find two unknowns, two pieces of information are needed, for three unknowns, three pieces of information, etc… A system of linear equations is a collection two or more linear equations in two or more variables. Systems of Linear Equations
  • 7. If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. But if a hamburger and a drink cost $4 then we won’t be able to pin down the price of each item-not enough information. In general, to find two unknowns, two pieces of information are needed, for three unknowns, three pieces of information, etc… A system of linear equations is a collection two or more linear equations in two or more variables. A solution for the system is a collection of numbers, one for each variable, that works for all equations in the system. Systems of Linear Equations
  • 8. If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. But if a hamburger and a drink cost $4 then we won’t be able to pin down the price of each item-not enough information. In general, to find two unknowns, two pieces of information are needed, for three unknowns, three pieces of information, etc… A system of linear equations is a collection two or more linear equations in two or more variables. A solution for the system is a collection of numbers, one for each variable, that works for all equations in the system. For example, 2x + y = 7 x + y = 5 is a system { Systems of Linear Equations
  • 9. If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. But if a hamburger and a drink cost $4 then we won’t be able to pin down the price of each item-not enough information. In general, to find two unknowns, two pieces of information are needed, for three unknowns, three pieces of information, etc… A system of linear equations is a collection two or more linear equations in two or more variables. A solution for the system is a collection of numbers, one for each variable, that works for all equations in the system. For example, 2x + y = 7 x + y = 5 is a system and that x = 2 and y = 3 is a solution because the pair fits both equations. { Systems of Linear Equations 2(2) + (3) = 7 (2) + (3) = 5 ck:
  • 10. If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. But if a hamburger and a drink cost $4 then we won’t be able to pin down the price of each item-not enough information. In general, to find two unknowns, two pieces of information are needed, for three unknowns, three pieces of information, etc… A system of linear equations is a collection two or more linear equations in two or more variables. A solution for the system is a collection of numbers, one for each variable, that works for all equations in the system. For example, 2x + y = 7 x + y = 5 is a system and that x = 2 and y = 3 is a solution because the pair fits both equations. This solution is written as (2, 3). { Systems of Linear Equations 2(2) + (3) = 7 (2) + (3) = 5 ck:
  • 11. Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Example A. Systems of Linear Equations
  • 12. Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. Example A. Systems of Linear Equations
  • 13. Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. We can translate the information into the system 2x + y = 7 Example A. E1 Systems of Linear Equations
  • 14. Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. We can translate the information into the system 2x + y = 7 x + y = 5{ Example A. E1 E2 Systems of Linear Equations
  • 15. Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. We can translate the information into the system 2x + y = 7 x + y = 5{ The difference between the order is one hamburger and the difference between the cost is $2, so the hamburger is $2. Example A. E1 E2 Systems of Linear Equations
  • 16. Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. We can translate the information into the system 2x + y = 7 x + y = 5{ The difference between the order is one hamburger and the difference between the cost is $2, so the hamburger is $2. In symbols, subtract the equations, i.e. E1 – E2: Example A. 2x + y = 7 ) x + y = 5 E1 E2 Systems of Linear Equations
  • 17. Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. We can translate the information into the system 2x + y = 7 x + y = 5{ The difference between the order is one hamburger and the difference between the cost is $2, so the hamburger is $2. In symbols, subtract the equations, i.e. E1 – E2: Example A. 2x + y = 7 ) x + y = 5 x + 0 = 2 so x = 2 E1 E2 Systems of Linear Equations
  • 18. Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. We can translate the information into the system 2x + y = 7 x + y = 5{ The difference between the order is one hamburger and the difference between the cost is $2, so the hamburger is $2. In symbols, subtract the equations, i.e. E1 – E2: Example A. substituting 2 for x back into E2, we get: 2 + y = 5 or y = 3 2x + y = 7 ) x + y = 5 x + 0 = 2 so x = 2 E1 E2 Systems of Linear Equations
  • 19. Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. We can translate the information into the system 2x + y = 7 x + y = 5{ The difference between the order is one hamburger and the difference between the cost is $2, so the hamburger is $2. In symbols, subtract the equations, i.e. E1 – E2: Example A. substituting 2 for x back into E2, we get: 2 + y = 5 or y = 3 Hence the solution is (2 , 3) or, $2 for a hamburger, $3 for a salad. 2x + y = 7 ) x + y = 5 x + 0 = 2 so x = 2 E1 E2 Systems of Linear Equations
  • 20. Example B. Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost?
  • 21. Example B. Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad.
  • 22. Example B. Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system:
  • 23. 4x + 3y = 18 3x + 2y = 13 Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system: Example B. { E1 E2 Systems of Linear Equations
  • 24. 4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. E1 E2 Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system:
  • 25. 4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate x or y. E1 E2 Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system:
  • 26. 4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate x or y. Suppose we chose to eliminate the y-terms, we find the LCM of 3y and 2y first. It's 6y. E1 E2 Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system:
  • 27. 4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate x or y. Suppose we chose to eliminate the y-terms, we find the LCM of 3y and 2y first. It's 6y. Multiplying E1 by (–2), E2 by 3, E1 E2 Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system:
  • 28. 4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate x or y. Suppose we chose to eliminate the y-terms, we find the LCM of 3y and 2y first. It's 6y. Multiplying E1 by (–2), E2 by 3, E1 E2 – 8x – 6y = –36E1*(–2) : Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system:
  • 29. 4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate x or y. Suppose we chose to eliminate the y-terms, we find the LCM of 3y and 2y first. It's 6y. Multiplying E1 by (–2), E2 by 3, E1 E2 – 8x – 6y = –36 9x + 6y = 39 E1*(–2) : E2*(3): Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system:
  • 30. 4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate x or y. Suppose we chose to eliminate the y-terms, we find the LCM of 3y and 2y first. It's 6y. Multiplying E1 by (–2), E2 by 3, and add the results, the y-terms are eliminated. E1 E2 – 8x – 6y = –36 9x + 6y = 39 E1*(–2) : E2*(3): + ) Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system: We adjust the y’s into opposite signs then add to eliminate them.
  • 31. 4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate x or y. Suppose we chose to eliminate the y-terms, we find the LCM of 3y and 2y first. It's 6y. Multiplying E1 by (–2), E2 by 3, and add the results, the y-terms are eliminated. E1 E2 – 8x – 6y = –36 9x + 6y = 39 E1*(–2) : E2*(3): + ) x = 3 Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system: So a hamburger is 3$. We adjust the y’s into opposite signs then add to eliminate them.
  • 32. To find y, set 3 for x in E2: 3x + 2y = 13 Systems of Linear Equations
  • 33. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 3x + 2y = 13 Systems of Linear Equations
  • 34. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 3x + 2y = 13 Systems of Linear Equations
  • 35. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 3x + 2y = 13 Systems of Linear Equations
  • 36. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). 3x + 2y = 13 Systems of Linear Equations
  • 37. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). Therefore a hamburger cost $3 and a salad cost $2. 3x + 2y = 13 Systems of Linear Equations
  • 38. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). Therefore a hamburger cost $3 and a salad cost $2. 3x + 2y = 13 The above method is called the elimination (addition) method. We summarize the steps below. Systems of Linear Equations
  • 39. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). Therefore a hamburger cost $3 and a salad cost $2. 3x + 2y = 13 The above method is called the elimination (addition) method. We summarize the steps below. 1. Line up the x’s & y’s to the left and the numbers to the right. Systems of Linear Equations
  • 40. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). Therefore a hamburger cost $3 and a salad cost $2. 3x + 2y = 13 The above method is called the elimination (addition) method. We summarize the steps below. 1. Line up the x’s & y’s to the left and the numbers to the right. 2. Select the variable x or y to eliminate. Systems of Linear Equations
  • 41. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). Therefore a hamburger cost $3 and a salad cost $2. 3x + 2y = 13 The above method is called the elimination (addition) method. We summarize the steps below. 1. Line up the x’s & y’s to the left and the numbers to the right. 2. Select the variable x or y to eliminate. 3. Find the LCM of the terms with that variable, and convert these terms to the LCM (by multiplication). Systems of Linear Equations
  • 42. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). Therefore a hamburger cost $3 and a salad cost $2. 3x + 2y = 13 The above method is called the elimination (addition) method. We summarize the steps below. 1. Line up the x’s & y’s to the left and the numbers to the right. 2. Select the variable x or y to eliminate. 3. Find the LCM of the terms with that variable, and convert these terms to the LCM (by multiplication). 4. Add or subtract the adjusted equations and solve the resulting equation. Systems of Linear Equations
  • 43. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). Therefore a hamburger cost $3 and a salad cost $2. 3x + 2y = 13 The above method is called the elimination (addition) method. We summarize the steps below. 1. Line up the x’s & y’s to the left and the numbers to the right. 2. Select the variable x or y to eliminate. 3. Find the LCM of the terms with that variable, and convert these terms to the LCM (by multiplication). 4. Add or subtract the adjusted equations and solve the resulting equation. 5. Substitute the answer back into any equation to solve for the other variable. Systems of Linear Equations
  • 44. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 {Example C. Systems of Linear Equations
  • 45. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. {Example C. Systems of Linear Equations
  • 46. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. {Example C. Systems of Linear Equations
  • 47. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 {Example C. 15x + 12y = 63*E1: Systems of Linear Equations
  • 48. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 8x – 12y = - 52 3*E1: 4*E2: Systems of Linear Equations
  • 49. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: Systems of Linear Equations
  • 50. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46 Systems of Linear Equations
  • 51. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Systems of Linear Equations
  • 52. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 Systems of Linear Equations
  • 53. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 Systems of Linear Equations
  • 54. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y = 3 Systems of Linear Equations
  • 55. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y = 3 Hence the solution is (-2, 3). Systems of Linear Equations
  • 56. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y = 3 Hence the solution is (-2, 3). In all the above examples, we obtain exactly one solution in each case. Systems of Linear Equations
  • 57. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y = 3 Hence the solution is (-2, 3). In all the above examples, we obtain exactly one solution in each case. However, there are two other possibilities. Systems of Linear Equations
  • 58. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y = 3 Hence the solution is (-2, 3). In all the above examples, we obtain exactly one solution in each case. However, there are two other possibilities. Systems of Linear Equations The system might not have any solution
  • 59. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y = 3 Hence the solution is (-2, 3). In all the above examples, we obtain exactly one solution in each case. However, there are two other possibilities. Systems of Linear Equations The system might not have any solution or the system might have infinite many solutions.
  • 60. Two Special Cases: I. Contradictory (Inconsistent) Systems Systems of Linear Equations
  • 61. Two Special Cases: I. Contradictory (Inconsistent) Systems Example D. { x + y = 2 x + y = 3 (E1) (E2) Systems of Linear Equations
  • 62. Two Special Cases: I. Contradictory (Inconsistent) Systems Example D. { x + y = 2 x + y = 3 (E1) (E2) Remove the x-terms by subtracting the equations. Systems of Linear Equations
  • 63. Two Special Cases: I. Contradictory (Inconsistent) Systems Example D. { x + y = 2 x + y = 3 E1 – E2 : x + y = 2 ) x + y = 3 (E1) (E2) Remove the x-terms by subtracting the equations. Systems of Linear Equations
  • 64. Two Special Cases: I. Contradictory (Inconsistent) Systems Example D. { x + y = 2 x + y = 3 E1 – E2 : x + y = 2 ) x + y = 3 0 = -1 (E1) (E2) Remove the x-terms by subtracting the equations. Systems of Linear Equations
  • 65. Two Special Cases: I. Contradictory (Inconsistent) Systems Example D. { x + y = 2 x + y = 3 E1 – E2 : x + y = 2 ) x + y = 3 0 = -1 (E1) (E2) Remove the x-terms by subtracting the equations. This is impossible! Such systems are said to be contradictory or inconsistent. Systems of Linear Equations
  • 66. Two Special Cases: I. Contradictory (Inconsistent) Systems Example D. { x + y = 2 x + y = 3 E1 – E2 : x + y = 2 ) x + y = 3 0 = -1 (E1) (E2) Remove the x-terms by subtracting the equations. This is impossible! Such systems are said to be contradictory or inconsistent. These system don’t have solution. Systems of Linear Equations
  • 67. II. Dependent Systems { x + y = 3 2x + 2y = 6 Example E. (E1) (E2) Systems of Linear Equations
  • 68. II. Dependent Systems { x + y = 3 2x + 2y = 6 Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations
  • 69. II. Dependent Systems { x + y = 3 2x + 2y = 6 2*E1 2x + 2y = 6 Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations
  • 70. II. Dependent Systems { x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations
  • 71. II. Dependent Systems { x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0 = 0 Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations
  • 72. II. Dependent Systems { x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0 = 0 This means the equations are actually the same and it has infinitely many solutions such as (3, 0), (2, 1), (1, 2) etc… Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations
  • 73. { x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0 = 0 This means the equations are actually the same and it has infinitely many solutions such as (3, 0), (2, 1), (1, 2) etc… Such systems are called dependent or redundant systems. Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations II. Dependent Systems
  • 74. { x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0 = 0 This means the equations are actually the same and it has infinitely many solutions such as (3, 0), (2, 1), (1, 2) etc… Such systems are called dependent or redundant systems. Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations II. Dependent Systems Finally, as before with linear equations, clear the denominators of fractional equations first.
  • 75. { x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0 = 0 This means the equations are actually the same and it has infinitely many solutions such as (3, 0), (2, 1), (1, 2) etc… Such systems are called dependent or redundant systems. Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations II. Dependent Systems Finally, as before with linear equations, clear the denominators of fractional equations first. For example, change { x – y = 3 x – y = –1 3 2 2 3 1 2 1 4
  • 76. { x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0 = 0 This means the equations are actually the same and it has infinitely many solutions such as (3, 0), (2, 1), (1, 2) etc… Such systems are called dependent or redundant systems. Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations II. Dependent Systems Finally, as before with linear equations, clear the denominators of fractional equations first. For example, change { x – y = 3 x – y = –1 3 2 2 3 1 2 1 4 ( )*6 ( )*4
  • 77. { x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0 = 0 This means the equations are actually the same and it has infinitely many solutions such as (3, 0), (2, 1), (1, 2) etc… Such systems are called dependent or redundant systems. Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations II. Dependent Systems Finally, as before with linear equations, clear the denominators of fractional equations first. For example, change { x – y = 3 x – y = –1 3 2 2 3 1 2 1 4 ( )*6 9x – 4y = 18 ( )*4 { 2x – y = –4 then solve.
  • 78. Exercise. A. Select the solution of the system. {x + y = 3 2x + y = 4 Systems of Linear Equations {x + y = 3 2x + y = 5 {x + y = 3 2x + y = 6 {x + y = 3 2x + y = 31. a. (0, 3) b. (1, 2) c. (2, 1) d.(3, 0) 2. a. (0, 3) b. (1, 2) c. (2, 1) d.(3, 0) 3. a. (0, 3) b. (1, 2) c. (2, 1) d.(3, 0) 4. a. (0, 3) b. (1, 2) c. (2, 1) d.(3, 0) B. Solve the following systems. 5. {y = 3 2x + y = 3 6. {x = 1 2x + y = 4 7. {y = 1 2x + y = 5 8. {x = 2 2x + y = 6 C. Add or subtract vertically. (These are warm ups) 8x –9x+) 9. 8 –9–) 10. –8y –9y–) 11. –8 9–) 12. –4y –5y+) 13. +4x –5x–) 14.
  • 79. Systems of Linear Equations 18.{–x + 2y = –12 2x + y = 4 D. Solve the following system by the elimination method. 15.{x + y = 3 2x + y = 4 16. 17.{x + 2y = 3 2x – y = 6 {x + y = 3 2x – y = 6 19. {3x + 4y = 3 x – 2y = 6 20. { x + 3y = 3 2x – 9y = –4 21.{–3x + 2y = –1 2x + 3y = 5 22. {2x + 3y = –1 3x + 4y = 2 23. {4x – 3y = 3 3x – 2y = –4 24.{5x + 3y = 2 2x + 4y = –2 25. {3x + 4y = –10 –5x + 3y = 7 26. {–4x + 9y = 1 5x – 2y = 8 { x – y = 3 x – y = –1 3 2 2 3 1 2 1 4 27. { x + y = 1 x – y = –1 1 2 1 5 3 4 1 6 28. 29.{ x + 3y = 4 2x + 6y = 8 E. Which system is inconsistent and which is dependent? 30.{2x – y = 2 8x – 4y = 6