Answer :
Sure, let's identify which system of equations out of the listed ones has the point [tex]\( A(1, -3) \)[/tex] as their point of intersection.
We will analyze four potential systems of equations and check if [tex]\( A(1, -3) \)[/tex] satisfies both equations in any of these systems.
The systems to consider are:
1. [tex]\(\begin{cases} y = x - 4 \\ y = 5x - 7 \end{cases}\)[/tex]
2. [tex]\(\begin{cases} y = 3x - 10 \\ y = x - 1 \end{cases}\)[/tex]
3. [tex]\(\begin{cases} y = x - 1 \\ y = 5x - 16 \end{cases}\)[/tex]
4. [tex]\(\begin{cases} y = 2x - 7 \\ y = x - 1 \end{cases}\)[/tex]
### Checking System 1:
[tex]\(\begin{cases} y = x - 4 \\ y = 5x - 7 \end{cases}\)[/tex]
For the point [tex]\( A(1, -3) \)[/tex]:
1. Substitute [tex]\( x = 1 \)[/tex] into [tex]\( y = x - 4 \)[/tex]:
[tex]\[ y = 1 - 4 = -3 \][/tex]
So, the first equation is satisfied.
2. Substitute [tex]\( x = 1 \)[/tex] into [tex]\( y = 5x - 7 \)[/tex]:
[tex]\[ y = 5(1) - 7 = 5 - 7 = -2 \][/tex]
So, the second equation is not satisfied.
Hence, [tex]\( A(1, -3) \)[/tex] is not a point of intersection for System 1.
### Checking System 2:
[tex]\(\begin{cases} y = 3x - 10 \\ y = x - 1 \end{cases}\)[/tex]
For the point [tex]\( A(1, -3) \)[/tex]:
1. Substitute [tex]\( x = 1 \)[/tex] into [tex]\( y = 3x - 10 \)[/tex]:
[tex]\[ y = 3(1) - 10 = 3 - 10 = -7 \][/tex]
So, the first equation is not satisfied.
2. Substitute [tex]\( x = 1 \)[/tex] into [tex]\( y = x - 1 \)[/tex]:
[tex]\[ y = 1 - 1 = 0 \][/tex]
So, the second equation is not satisfied.
Hence, [tex]\( A(1, -3) \)[/tex] is not a point of intersection for System 2.
### Checking System 3:
[tex]\(\begin{cases} y = x - 1 \\ y = 5x - 16 \end{cases}\)[/tex]
For the point [tex]\( A(1, -3) \)[/tex]:
1. Substitute [tex]\( x = 1 \)[/tex] into [tex]\( y = x - 1 \)[/tex]:
[tex]\[ y = 1 - 1 = 0 \][/tex]
So, the first equation is not satisfied.
2. Substitute [tex]\( x = 1 \)[/tex] into [tex]\( y = 5x - 16 \)[/tex]:
[tex]\[ y = 5(1) - 16 = 5 - 16 = -11 \][/tex]
So, the second equation is not satisfied.
Hence, [tex]\( A(1, -3) \)[/tex] is not a point of intersection for System 3.
### Checking System 4:
[tex]\(\begin{cases} y = 2x - 7 \\ y = x - 1 \end{cases}\)[/tex]
For the point [tex]\( A(1, -3) \)[/tex]:
1. Substitute [tex]\( x = 1 \)[/tex] into [tex]\( y = 2x - 7 \)[/tex]:
[tex]\[ y = 2(1) - 7 = 2 - 7 = -5 \][/tex]
So, the first equation is not satisfied.
2. Substitute [tex]\( x = 1 \)[/tex] into [tex]\( y = x - 1 \)[/tex]:
[tex]\[ y = 1 - 1 = 0 \][/tex]
So, the second equation is not satisfied.
Hence, [tex]\( A(1, -3) \)[/tex] is not a point of intersection for System 4.
Based on our analysis, we find that none of the four systems of equations have the point [tex]\( A(1, -3) \)[/tex] as their point of intersection. Therefore, the answer is:
[tex]\[ \boxed{\text{None}} \][/tex]
We will analyze four potential systems of equations and check if [tex]\( A(1, -3) \)[/tex] satisfies both equations in any of these systems.
The systems to consider are:
1. [tex]\(\begin{cases} y = x - 4 \\ y = 5x - 7 \end{cases}\)[/tex]
2. [tex]\(\begin{cases} y = 3x - 10 \\ y = x - 1 \end{cases}\)[/tex]
3. [tex]\(\begin{cases} y = x - 1 \\ y = 5x - 16 \end{cases}\)[/tex]
4. [tex]\(\begin{cases} y = 2x - 7 \\ y = x - 1 \end{cases}\)[/tex]
### Checking System 1:
[tex]\(\begin{cases} y = x - 4 \\ y = 5x - 7 \end{cases}\)[/tex]
For the point [tex]\( A(1, -3) \)[/tex]:
1. Substitute [tex]\( x = 1 \)[/tex] into [tex]\( y = x - 4 \)[/tex]:
[tex]\[ y = 1 - 4 = -3 \][/tex]
So, the first equation is satisfied.
2. Substitute [tex]\( x = 1 \)[/tex] into [tex]\( y = 5x - 7 \)[/tex]:
[tex]\[ y = 5(1) - 7 = 5 - 7 = -2 \][/tex]
So, the second equation is not satisfied.
Hence, [tex]\( A(1, -3) \)[/tex] is not a point of intersection for System 1.
### Checking System 2:
[tex]\(\begin{cases} y = 3x - 10 \\ y = x - 1 \end{cases}\)[/tex]
For the point [tex]\( A(1, -3) \)[/tex]:
1. Substitute [tex]\( x = 1 \)[/tex] into [tex]\( y = 3x - 10 \)[/tex]:
[tex]\[ y = 3(1) - 10 = 3 - 10 = -7 \][/tex]
So, the first equation is not satisfied.
2. Substitute [tex]\( x = 1 \)[/tex] into [tex]\( y = x - 1 \)[/tex]:
[tex]\[ y = 1 - 1 = 0 \][/tex]
So, the second equation is not satisfied.
Hence, [tex]\( A(1, -3) \)[/tex] is not a point of intersection for System 2.
### Checking System 3:
[tex]\(\begin{cases} y = x - 1 \\ y = 5x - 16 \end{cases}\)[/tex]
For the point [tex]\( A(1, -3) \)[/tex]:
1. Substitute [tex]\( x = 1 \)[/tex] into [tex]\( y = x - 1 \)[/tex]:
[tex]\[ y = 1 - 1 = 0 \][/tex]
So, the first equation is not satisfied.
2. Substitute [tex]\( x = 1 \)[/tex] into [tex]\( y = 5x - 16 \)[/tex]:
[tex]\[ y = 5(1) - 16 = 5 - 16 = -11 \][/tex]
So, the second equation is not satisfied.
Hence, [tex]\( A(1, -3) \)[/tex] is not a point of intersection for System 3.
### Checking System 4:
[tex]\(\begin{cases} y = 2x - 7 \\ y = x - 1 \end{cases}\)[/tex]
For the point [tex]\( A(1, -3) \)[/tex]:
1. Substitute [tex]\( x = 1 \)[/tex] into [tex]\( y = 2x - 7 \)[/tex]:
[tex]\[ y = 2(1) - 7 = 2 - 7 = -5 \][/tex]
So, the first equation is not satisfied.
2. Substitute [tex]\( x = 1 \)[/tex] into [tex]\( y = x - 1 \)[/tex]:
[tex]\[ y = 1 - 1 = 0 \][/tex]
So, the second equation is not satisfied.
Hence, [tex]\( A(1, -3) \)[/tex] is not a point of intersection for System 4.
Based on our analysis, we find that none of the four systems of equations have the point [tex]\( A(1, -3) \)[/tex] as their point of intersection. Therefore, the answer is:
[tex]\[ \boxed{\text{None}} \][/tex]
