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
