how many solutions exist for the following system
3x-6y=12 and 2x-4y=7
but lines have the same slope, but not the same intercepts.
So you have two parallel lines, which of course never meet
So .....
To determine how many solutions exist for the given system of equations, we can use the concept of linear independence.
First, let's rewrite the equations in standard form:
3x - 6y = 12 --> 3x - 6y - 12 = 0
2x - 4y = 7 --> 2x - 4y - 7 = 0
Now, we can represent the system of equations using matrix notation:
[A] [X] = [B]
where [A] is the coefficient matrix, [X] is the variables matrix, and [B] is the constants matrix.
| 3 -6 | | x | | 12 |
[A] [X] = [B] --> | 2 -4 | [X] = | 7 |
To determine the number of solutions, we need to find the rank of the coefficient matrix [A], and the rank of the augmented matrix [A|B].
The rank of a matrix is the maximum number of linearly independent rows in the matrix.
Let's find the rank of [A]:
3 -6
2 -4
Perform row operations to reduce [A] to row-echelon form:
R2 = R2 - (2/3)R1
3 -6
0 0
The rank of [A] is 1 since there is one linearly independent row in the matrix.
Now, let's find the rank of [A|B]:
3 -6 | 12
2 -4 | 7
Perform row operations to reduce [A|B] to row-echelon form:
R2 = R2 - (2/3)R1
3 -6 | 12
0 0 | -1
The rank of [A|B] is 2 since there are two linearly independent rows in the matrix.
Now, comparing the ranks of [A] and [A|B], we can determine the number of solutions:
If the rank of [A] is equal to the rank of [A|B], then there is a unique solution.
If the rank of [A] is less than the rank of [A|B], then there are infinitely many solutions.
If the rank of [A] is greater than the rank of [A|B], then there are no solutions.
In this case, the rank of [A] is 1, and the rank of [A|B] is 2. Since the ranks are not equal, the system of equations does not have any solutions.
Therefore, the answer is that there are no solutions for the given system: 3x - 6y = 12 and 2x - 4y = 7.