The document discusses systems of linear equations. It defines a system of linear equations as a collection of two or more linear equations with two or more variables. A solution to a system is a set of numbers, one for each variable, that satisfies all equations simultaneously. As an example, the document presents a system representing the costs of hamburgers and fries and solves it to find the cost of each item is $2 for a hamburger and $3 for a fry.

1. Systems of Linear Equations

2. Systems of Linear Equations
In general, we need one piece of information to solve for one
unknown quantity.

3. Systems of Linear Equations
In general, we need one piece of information to solve for one
unknown quantity. If 2 hamburgers cost 4 dollars and x is the
cost of a hamburger,

4. Systems of Linear Equations
In general, we need one piece of information to solve for one
unknown quantity. If 2 hamburgers cost 4 dollars and x is the
cost of a hamburger, then 2x = 4 or x = 2 $.

5. Systems of Linear Equations
In general, we need one piece of information to solve for one
unknown quantity. If 2 hamburgers cost 4 dollars and x is the
cost of a hamburger, then 2x = 4 or x = 2 $.
If there are two unknowns, we need two pieces of information
to solve them,

6. Systems of Linear Equations
In general, we need one piece of information to solve for one
unknown quantity. If 2 hamburgers cost 4 dollars and x is the
cost of a hamburger, then 2x = 4 or x = 2 $.
If there are two unknowns, we need two pieces of information
to solve them, three unknowns needs three pieces of
information, etc…

7. Systems of Linear Equations
In general, we need one piece of information to solve for one
unknown quantity. If 2 hamburgers cost 4 dollars and x is the
cost of a hamburger, then 2x = 4 or x = 2 $.
If there are two unknowns, we need two pieces of information
to solve them, three unknowns needs three pieces of
information, etc… These lead to systems of equations.

8. Systems of Linear Equations
In general, we need one piece of information to solve for one
unknown quantity. If 2 hamburgers cost 4 dollars and x is the
cost of a hamburger, then 2x = 4 or x = 2 $.
If there are two unknowns, we need two pieces of information
to solve them, three unknowns needs three pieces of
information, etc… These lead to systems of equations.
A system of linear equations is a collection two or more linear
equations in two or more variables.

9. Systems of Linear Equations
In general, we need one piece of information to solve for one
unknown quantity. If 2 hamburgers cost 4 dollars and x is the
cost of a hamburger, then 2x = 4 or x = 2 $.
If there are two unknowns, we need two pieces of information
to solve them, three unknowns needs three pieces of
information, etc… These lead to systems of equations.
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.

10. Systems of Linear Equations
In general, we need one piece of information to solve for one
unknown quantity. If 2 hamburgers cost 4 dollars and x is the
cost of a hamburger, then 2x = 4 or x = 2 $.
If there are two unknowns, we need two pieces of information
to solve them, three unknowns needs three pieces of
information, etc… These lead to systems of equations.
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.

11. Systems of Linear Equations
In general, we need one piece of information to solve for one
unknown quantity. If 2 hamburgers cost 4 dollars and x is the
cost of a hamburger, then 2x = 4 or x = 2 $.
If there are two unknowns, we need two pieces of information
to solve them, three unknowns needs three pieces of
information, etc… These lead to systems of equations.
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. x = 2 and y = 3 is a solution because it fits both
equations the ordered pair.

12. Systems of Linear Equations
In general, we need one piece of information to solve for one
unknown quantity. If 2 hamburgers cost 4 dollars and x is the
cost of a hamburger, then 2x = 4 or x = 2 $.
If there are two unknowns, we need two pieces of information
to solve them, three unknowns needs three pieces of
information, etc… These lead to systems of equations.
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. x = 2 and y = 3 is a solution because it fits both
equations the ordered pair. This solution is written as (2, 3).

13. Systems of Linear Equations
Example A.
Suppose two hamburgers and a fry cost $7, and one
hamburger and one fry cost $5.

14. Systems of Linear Equations
Example A.
Suppose two hamburgers and a fry cost $7, and one
hamburger and one fry cost $5.
Let x = cost of a hamburger, y = cost of a fry.

15. Systems of Linear Equations
Example A.
Suppose two hamburgers and a fry cost $7, and one
hamburger and one fry cost $5.
Let x = cost of a hamburger, y = cost of a fry.
We can translate these into the system:
{2x + y = 7

16. Systems of Linear Equations
Example A.
Suppose two hamburgers and a fry cost $7, and one
hamburger and one fry cost $5.
Let x = cost of a hamburger, y = cost of a fry.
We can translate these into the system:
{2x + y = 7
x+y=5

17. Systems of Linear Equations
Example A.
Suppose two hamburgers and a fry cost $7, and one
hamburger and one fry cost $5.
Let x = cost of a hamburger, y = cost of a fry.
We can translate these into the system:
{2x + y = 7
x+y=5
E1
E2

18. Systems of Linear Equations
Example A.
Suppose two hamburgers and a fry cost $7, and one
hamburger and one fry cost $5.
Let x = cost of a hamburger, y = cost of a fry.
We can translate these into the system:
{ 2x + y = 7
x+y=5
E1
E2
The difference between the order is one hamburger and the
difference between the cost is $2,

19. Systems of Linear Equations
Example A.
Suppose two hamburgers and a fry cost $7, and one
hamburger and one fry cost $5.
Let x = cost of a hamburger, y = cost of a fry.
We can translate these into the system:
{ 2x + y = 7
x+y=5
E1
E2
The difference between the order is one hamburger and the
difference between the cost is $2, so the hamburger is $2.

20. Systems of Linear Equations
Example A.
Suppose two hamburgers and a fry cost $7, and one
hamburger and one fry cost $5.
Let x = cost of a hamburger, y = cost of a fry.
We can translate these into the system:
{ 2x + y = 7
x+y=5
E1
E2
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:

21. Systems of Linear Equations
Example A.
Suppose two hamburgers and a fry cost $7, and one
hamburger and one fry cost $5.
Let x = cost of a hamburger, y = cost of a fry.
We can translate these into the system:
{ 2x + y = 7
x+y=5
E1
E2
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:
2x + y = 7
) x+y=5

22. Systems of Linear Equations
Example A.
Suppose two hamburgers and a fry cost $7, and one
hamburger and one fry cost $5.
Let x = cost of a hamburger, y = cost of a fry.
We can translate these into the system:
{ 2x + y = 7
x+y=5
E1
E2
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:
2x + y = 7
) x+y=5
x+0=2

23. Systems of Linear Equations
Example A.
Suppose two hamburgers and a fry cost $7, and one
hamburger and one fry cost $5.
Let x = cost of a hamburger, y = cost of a fry.
We can translate these into the system:
{ 2x + y = 7
x+y=5
E1
E2
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:
2x + y = 7
) x+y=5
x+0=2
so x=2

24. Systems of Linear Equations
Example A.
Suppose two hamburgers and a fry cost $7, and one
hamburger and one fry cost $5.
Let x = cost of a hamburger, y = cost of a fry.
We can translate these into the system:
{ 2x + y = 7
x+y=5
E1
E2
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:
2x + y = 7 Substitute 2 for x back into E2,
) x+y=5 we get: 2 + y = 5
x+0=2
so x=2

25. Systems of Linear Equations
Example A.
Suppose two hamburgers and a fry cost $7, and one
hamburger and one fry cost $5.
Let x = cost of a hamburger, y = cost of a fry.
We can translate these into the system:
{ 2x + y = 7
x+y=5
E1
E2
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:
2x + y = 7 Substitute 2 for x back into E2,
) x+y=5 we get: 2 + y = 5
or y = 3
x+0=2
so x=2

26. Systems of Linear Equations
Example A.
Suppose two hamburgers and a fry cost $7, and one
hamburger and one fry cost $5.
Let x = cost of a hamburger, y = cost of a fry.
We can translate these into the system:
{ 2x + y = 7
x+y=5
E1
E2
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:
2x + y = 7 Substitute 2 for x back into E2,
) x+y=5 we get: 2 + y = 5
or y = 3
x+0=2
Hence the solution is (2 , 3) or,
so x=2
$2 for a hamburger, $3 for a fry.

27. Systems of Linear Equations
Example B.
Suppose four hamburgers and three fry cost $18, and three
hamburger and two fry cost $13. How much does each cost?

28. Systems of Linear Equations
Example B.
Suppose four hamburgers and three fry cost $18, and three
hamburger and two fry cost $13. How much does each cost?
Let x = cost of a hamburger, y = cost of a fry.

29. Systems of Linear Equations
Example B.
Suppose four hamburgers and three fry cost $18, and three
hamburger and two fry cost $13. How much does each cost?
Let x = cost of a hamburger, y = cost of a fry.
We translate these into the system:

30. Systems of Linear Equations
Example B.
Suppose four hamburgers and three fry cost $18, and three
hamburger and two fry cost $13. How much does each cost?
Let x = cost of a hamburger, y = cost of a fry.
We translate these into the system:
4x + 3y = 18 E1

31. Systems of Linear Equations
Example B.
Suppose four hamburgers and three fry cost $18, and three
hamburger and two fry cost $13. How much does each cost?
Let x = cost of a hamburger, y = cost of a fry.
We translate these into the system:
{4x + 3y = 18
3x + 2y = 13
E1
E2

32. Systems of Linear Equations
Example B.
Suppose four hamburgers and three fry cost $18, and three
hamburger and two fry cost $13. How much does each cost?
Let x = cost of a hamburger, y = cost of a fry.
We translate these into the system:
{4x + 3y = 18
3x + 2y = 13
E1
E2
Subtracting the equations now will not eliminate the x nor y.

33. Systems of Linear Equations
Example B.
Suppose four hamburgers and three fry cost $18, and three
hamburger and two fry cost $13. How much does each cost?
Let x = cost of a hamburger, y = cost of a fry.
We translate these into the system:
{ 4x + 3y = 18
3x + 2y = 13
E1
E2
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 one of the variable.

34. Systems of Linear Equations
Example B.
Suppose four hamburgers and three fry cost $18, and three
hamburger and two fry cost $13. How much does each cost?
Let x = cost of a hamburger, y = cost of a fry.
We translate these into the system:
{ 4x + 3y = 18
3x + 2y = 13
E1
E2
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 one of the variable.
Suppose we chose to eliminate the y-terms, we find the LCM
of 3y and 2y first.

35. Systems of Linear Equations
Example B.
Suppose four hamburgers and three fry cost $18, and three
hamburger and two fry cost $13. How much does each cost?
Let x = cost of a hamburger, y = cost of a fry.
We translate these into the system:
{ 4x + 3y = 18
3x + 2y = 13
E1
E2
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 one of the variable.
Suppose we chose to eliminate the y-terms, we find the LCM
of 3y and 2y first. It's 6y.

36. Systems of Linear Equations
Example B.
Suppose four hamburgers and three fry cost $18, and three
hamburger and two fry cost $13. How much does each cost?
Let x = cost of a hamburger, y = cost of a fry.
We translate these into the system:
{ 4x + 3y = 18
3x + 2y = 13
E1
E2
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 one of the variable.
Suppose we chose to eliminate the y-terms, we find the LCM
of 3y and 2y first. It's 6y. Then we multiply E1 by 2,

37. Systems of Linear Equations
Example B.
Suppose four hamburgers and three fry cost $18, and three
hamburger and two fry cost $13. How much does each cost?
Let x = cost of a hamburger, y = cost of a fry.
We translate these into the system:
{ 4x + 3y = 18
3x + 2y = 13
E1
E2
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 one of the variable.
Suppose we chose to eliminate the y-terms, we find the LCM
of 3y and 2y first. It's 6y. Then we multiply E1 by 2,
E1*2: 8x + 6y = 36

38. Systems of Linear Equations
Example B.
Suppose four hamburgers and three fry cost $18, and three
hamburger and two fry cost $13. How much does each cost?
Let x = cost of a hamburger, y = cost of a fry.
We translate these into the system:
{ 4x + 3y = 18
3x + 2y = 13
E1
E2
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 one of the variable.
Suppose we chose to eliminate the y-terms, we find the LCM
of 3y and 2y first. It's 6y. Then we multiply E1 by 2, E2 by (-3),
E1*2: 8x + 6y = 36
E2*(-3): -9x – 6y = -39

39. Systems of Linear Equations
Example B.
Suppose four hamburgers and three fry cost $18, and three
hamburger and two fry cost $13. How much does each cost?
Let x = cost of a hamburger, y = cost of a fry.
We translate these into the system:
{ 4x + 3y = 18
3x + 2y = 13
E1
E2
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 one of the variable.
Suppose we chose to eliminate the y-terms, we find the LCM
of 3y and 2y first. It's 6y. Then we multiply E1 by 2, E2 by (-3),
and add the results, the y-terms are eliminated.
E1*2: 8x + 6y = 36
E2*(-3): + ) -9x – 6y = -39

40. Systems of Linear Equations
Example B.
Suppose four hamburgers and three fry cost $18, and three
hamburger and two fry cost $13. How much does each cost?
Let x = cost of a hamburger, y = cost of a fry.
We translate these into the system:
{ 4x + 3y = 18
3x + 2y = 13
E1
E2
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 one of the variable.
Suppose we chose to eliminate the y-terms, we find the LCM
of 3y and 2y first. It's 6y. Then we multiply E1 by 2, E2 by (-3),
and add the results, the y-terms are eliminated.
E1*2: 8x + 6y = 36
E2*(-3): + ) -9x – 6y = -39
-x = -3

41. Systems of Linear Equations
Example B.
Suppose four hamburgers and three fry cost $18, and three
hamburger and two fry cost $13. How much does each cost?
Let x = cost of a hamburger, y = cost of a fry.
We translate these into the system:
{ 4x + 3y = 18
3x + 2y = 13
E1
E2
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 one of the variable.
Suppose we chose to eliminate the y-terms, we find the LCM
of 3y and 2y first. It's 6y. Then we multiply E1 by 2, E2 by (-3),
and add the results, the y-terms are eliminated.
E1*2: 8x + 6y = 36
E2*(-3): + ) -9x – 6y = -39
-x = -3  x = 3

42. Systems of Linear Equations
To find y, set 3 for x in E2: 3x + 2y = 13

43. Systems of Linear Equations
To find y, set 3 for x in E2: 3x + 2y = 13
and get 3(3) + 2y = 13

44. Systems of Linear Equations
To find y, set 3 for x in E2: 3x + 2y = 13
and get 3(3) + 2y = 13
9 + 2y = 13

45. Systems of Linear Equations
To find y, set 3 for x in E2: 3x + 2y = 13
and get 3(3) + 2y = 13
9 + 2y = 13
2y = 4
y=2

46. Systems of Linear Equations
To find y, set 3 for x in E2: 3x + 2y = 13
and get 3(3) + 2y = 13
9 + 2y = 13
2y = 4
y=2
Hence the solution is (3, 2).

47. Systems of Linear Equations
To find y, set 3 for x in E2: 3x + 2y = 13
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 fry cost $2.

48. Systems of Linear Equations
To find y, set 3 for x in E2: 3x + 2y = 13
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 fry cost $2.
The above method is called the elimination (addition) method.
We summarize the steps below.

49. Systems of Linear Equations
To find y, set 3 for x in E2: 3x + 2y = 13
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 fry cost $2.
The above method is called the elimination (addition) method.
We summarize the steps below.
1. Select a variable to eliminate.

50. Systems of Linear Equations
To find y, set 3 for x in E2: 3x + 2y = 13
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 fry cost $2.
The above method is called the elimination (addition) method.
We summarize the steps below.
1. Select a variable to eliminate.
2. Find the LCM of the terms with that variable, and convert
the corresponding term to the LCM by multiply each equation
by a number.

51. Systems of Linear Equations
To find y, set 3 for x in E2: 3x + 2y = 13
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 fry cost $2.
The above method is called the elimination (addition) method.
We summarize the steps below.
1. Select a variable to eliminate.
2. Find the LCM of the terms with that variable, and convert
the corresponding term to the LCM by multiply each equation
by a number.
3. Add or subtract the adjusted equations to eliminate the
selected variable and solve the resulting equation.

52. Systems of Linear Equations
To find y, set 3 for x in E2: 3x + 2y = 13
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 fry cost $2.
The above method is called the elimination (addition) method.
We summarize the steps below.
1. Select a variable to eliminate.
2. Find the LCM of the terms with that variable, and convert
the corresponding term to the LCM by multiply each equation
by a number.
3. Add or subtract the adjusted equations to eliminate the
selected variable and solve the resulting equation.
4. Substitute the answer back into any equation to solve for the
other variable.

53. Systems of Linear Equations
5x + 4y = 2 E1
Example C. Solve { 2x – 3y = -13 E2

54. Systems of Linear Equations
5x + 4y = 2 E1
Example C. Solve { 2x – 3y = -13 E2
Eliminate the y.

55. Systems of Linear Equations
5x + 4y = 2 E1
Example C. Solve { 2x – 3y = -13 E2
Eliminate the y. The LCM of the y-terms is 12y.

56. Systems of Linear Equations
5x + 4y = 2 E1
Example C. Solve { 2x – 3y = -13 E2
Eliminate the y. The LCM of the y-terms is 12y.
Multiply E1 by 3
3*E1: 15x + 12y = 6

57. Systems of Linear Equations
5x + 4y = 2 E1
Example C. Solve { 2x – 3y = -13 E2
Eliminate the y. The LCM of the y-terms is 12y.
Multiply E1 by 3 and E2 by 4.
3*E1: 15x + 12y = 6
4*E2: 8x – 12y = - 52

58. Systems of Linear Equations
5x + 4y = 2 E1
Example C. Solve { 2x – 3y = -13 E2
Eliminate the y. The LCM of the y-terms is 12y.
Multiply E1 by 3 and E2 by 4.
3*E1: 15x + 12y = 6
4*E2: ) 8x – 12y = - 52
Add:

59. Systems of Linear Equations
5x + 4y = 2 E1
Example C. Solve { 2x – 3y = -13 E2
Eliminate the y. The LCM of the y-terms is 12y.
Multiply E1 by 3 and E2 by 4.
3*E1: 15x + 12y = 6
4*E2: ) 8x – 12y = - 52
Add: 23x + 0 = - 46

60. Systems of Linear Equations
5x + 4y = 2 E1
Example C. Solve { 2x – 3y = -13 E2
Eliminate the y. The LCM of the y-terms is 12y.
Multiply E1 by 3 and E2 by 4.
3*E1: 15x + 12y = 6
4*E2: ) 8x – 12y = - 52
Add: 23x + 0 = - 46  x = - 2

61. Systems of Linear Equations
5x + 4y = 2 E1
Example C. Solve { 2x – 3y = -13 E2
Eliminate the y. The LCM of the y-terms is 12y.
Multiply E1 by 3 and E2 by 4.
3*E1: 15x + 12y = 6
4*E2: ) 8x – 12y = - 52
Add: 23x + 0 = - 46  x = - 2
Set -2 for x in E1 and get 5(-2) + 4y = 2

62. Systems of Linear Equations
5x + 4y = 2 E1
Example C. Solve { 2x – 3y = -13 E2
Eliminate the y. The LCM of the y-terms is 12y.
Multiply E1 by 3 and E2 by 4.
3*E1: 15x + 12y = 6
4*E2: ) 8x – 12y = - 52
Add: 23x + 0 = - 46  x = - 2
Set -2 for x in E1 and get 5(-2) + 4y = 2
-10 + 4y = 2

63. Systems of Linear Equations
5x + 4y = 2 E1
Example C. Solve { 2x – 3y = -13 E2
Eliminate the y. The LCM of the y-terms is 12y.
Multiply E1 by 3 and E2 by 4.
3*E1: 15x + 12y = 6
4*E2: ) 8x – 12y = - 52
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

64. Systems of Linear Equations
5x + 4y = 2 E1
Example C. Solve { 2x – 3y = -13 E2
Eliminate the y. The LCM of the y-terms is 12y.
Multiply E1 by 3 and E2 by 4.
3*E1: 15x + 12y = 6
4*E2: ) 8x – 12y = - 52
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).

65. Systems of Linear Equations
5x + 4y = 2 E1
Example C. Solve { 2x – 3y = -13 E2
Eliminate the y. The LCM of the y-terms is 12y.
Multiply E1 by 3 and E2 by 4.
3*E1: 15x + 12y = 6
4*E2: ) 8x – 12y = - 52
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.

66. Systems of Linear Equations
5x + 4y = 2 E1
Example C. Solve { 2x – 3y = -13 E2
Eliminate the y. The LCM of the y-terms is 12y.
Multiply E1 by 3 and E2 by 4.
3*E1: 15x + 12y = 6
4*E2: ) 8x – 12y = - 52
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; there is
no solution or there are infinitely many solutions.

67. Systems of Linear Equations
Two Special Cases:
I. Contradictory (Inconsistent) Systems

68. Systems of Linear Equations
Two Special Cases:
I. Contradictory (Inconsistent) Systems
Example D.
{ x+y= 2
x+y=3
(E1)
(E2)

69. 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.

70. 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.
E1 – E2 : x + y = 2
) x+y=
3

71. 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.
E1 – E2 : x + y = 2
) x+y=
3
0 = -1

72. 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.
E1 – E2 : x + y = 2
) x+y=
3
0 = -1
This is impossible! Such systems are said to be contradictory
or inconsistent.

73. 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.
E1 – E2 : x + y = 2
) x+y=
3
0 = -1
This is impossible! Such systems are said to be contradictory
or inconsistent.
These system don’t have solution.

74. Systems of Linear Equations
II. Dependent Systems
{
Example E. x + y = 3
2x + 2y = 6
(E1)
(E2)

75. Systems of Linear Equations
II. Dependent Systems
{
Example E. x + y = 3
2x + 2y = 6
(E1)
(E2)
Remove the x-terms by multiplying E1 by 2 then subtract E2.

76. Systems of Linear Equations
II. Dependent Systems
{
Example E. x + y = 3
2x + 2y = 6
(E1)
(E2)
Remove the x-terms by multiplying E1 by 2 then subtract E2.
2*E1: 2x + 2y = 6

77. Systems of Linear Equations
II. Dependent Systems
{
Example E. x + y = 3
2x + 2y = 6
(E1)
(E2)
Remove the x-terms by multiplying E1 by 2 then subtract E2.
2*E1 – E2 : 2x + 2y = 6
) 2x + 2y = 6

78. Systems of Linear Equations
II. Dependent Systems
{
Example E. x + y = 3
2x + 2y = 6
(E1)
(E2)
Remove the x-terms by multiplying E1 by 2 then subtract E2.
2*E1 – E2 : 2x + 2y = 6
) 2x + 2y = 6
0= 0

79. Systems of Linear Equations
II. Dependent Systems
{
Example E. x + y = 3
2x + 2y = 6
(E1)
(E2)
Remove the x-terms by multiplying E1 by 2 then subtract E2.
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…

80. Systems of Linear Equations
II. Dependent Systems
{
Example E. x + y = 3
2x + 2y = 6
(E1)
(E2)
Remove the x-terms by multiplying E1 by 2 then subtract E2.
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.