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To determine which system of equations has the solution [tex]\((1, 4)\)[/tex], we can substitute [tex]\(x = 1\)[/tex] and [tex]\(y = 4\)[/tex] into each equation and verify if both equations are satisfied.
### System 1:
[tex]\[ \begin{cases} y = -3x - 1 \\ y = -x + 5 \end{cases} \][/tex]
Substitute [tex]\(x = 1\)[/tex] and [tex]\(y = 4\)[/tex]:
First equation:
[tex]\[ 4 = -3(1) - 1 \][/tex]
[tex]\[ 4 = -3 - 1 \][/tex]
[tex]\[ 4 = -4 \][/tex] (which is false)
Since the first equation is not satisfied, this system does not have the solution [tex]\((1, 4)\)[/tex].
### System 2:
[tex]\[ \begin{cases} y = 3x + 1 \\ y = -x + 5 \end{cases} \][/tex]
Substitute [tex]\(x = 1\)[/tex] and [tex]\(y = 4\)[/tex]:
First equation:
[tex]\[ 4 = 3(1) + 1 \][/tex]
[tex]\[ 4 = 3 + 1 \][/tex]
[tex]\[ 4 = 4 \][/tex] (which is true)
Second equation:
[tex]\[ 4 = -(1) + 5 \][/tex]
[tex]\[ 4 = -1 + 5 \][/tex]
[tex]\[ 4 = 4 \][/tex] (which is also true)
Both equations are satisfied, so this system does have the solution [tex]\((1, 4)\)[/tex].
### System 3:
[tex]\[ \begin{cases} y = 3x + 1 \\ y = x - 5 \end{cases} \][/tex]
Substitute [tex]\(x = 1\)[/tex] and [tex]\(y = 4\)[/tex]:
First equation:
[tex]\[ 4 = 3(1) + 1 \][/tex]
[tex]\[ 4 = 3 + 1 \][/tex]
[tex]\[ 4 = 4 \][/tex] (which is true)
Second equation:
[tex]\[ 4 = 1 - 5 \][/tex]
[tex]\[ 4 = -4 \][/tex] (which is false)
Since the second equation is not satisfied, this system does not have the solution [tex]\((1, 4)\)[/tex].
### System 4:
[tex]\[ \begin{cases} y = 3x + 1 \\ y = -x - 5 \end{cases} \][/tex]
Substitute [tex]\(x = 1\)[/tex] and [tex]\(y = 4\)[/tex]:
First equation:
[tex]\[ 4 = 3(1) + 1 \][/tex]
[tex]\[ 4 = 3 + 1 \][/tex]
[tex]\[ 4 = 4 \][/tex] (which is true)
Second equation:
[tex]\[ 4 = -(1) - 5 \][/tex]
[tex]\[ 4 = -1 - 5 \][/tex]
[tex]\[ 4 = -6 \][/tex] (which is false)
Since the second equation is not satisfied, this system does not have the solution [tex]\((1, 4)\)[/tex].
### Conclusion:
The system that has the solution [tex]\((1, 4)\)[/tex] is:
[tex]\[ \begin{cases} y = 3x + 1 \\ y = -x + 5 \end{cases} \][/tex]
So, the answer is the second system of equations.
### System 1:
[tex]\[ \begin{cases} y = -3x - 1 \\ y = -x + 5 \end{cases} \][/tex]
Substitute [tex]\(x = 1\)[/tex] and [tex]\(y = 4\)[/tex]:
First equation:
[tex]\[ 4 = -3(1) - 1 \][/tex]
[tex]\[ 4 = -3 - 1 \][/tex]
[tex]\[ 4 = -4 \][/tex] (which is false)
Since the first equation is not satisfied, this system does not have the solution [tex]\((1, 4)\)[/tex].
### System 2:
[tex]\[ \begin{cases} y = 3x + 1 \\ y = -x + 5 \end{cases} \][/tex]
Substitute [tex]\(x = 1\)[/tex] and [tex]\(y = 4\)[/tex]:
First equation:
[tex]\[ 4 = 3(1) + 1 \][/tex]
[tex]\[ 4 = 3 + 1 \][/tex]
[tex]\[ 4 = 4 \][/tex] (which is true)
Second equation:
[tex]\[ 4 = -(1) + 5 \][/tex]
[tex]\[ 4 = -1 + 5 \][/tex]
[tex]\[ 4 = 4 \][/tex] (which is also true)
Both equations are satisfied, so this system does have the solution [tex]\((1, 4)\)[/tex].
### System 3:
[tex]\[ \begin{cases} y = 3x + 1 \\ y = x - 5 \end{cases} \][/tex]
Substitute [tex]\(x = 1\)[/tex] and [tex]\(y = 4\)[/tex]:
First equation:
[tex]\[ 4 = 3(1) + 1 \][/tex]
[tex]\[ 4 = 3 + 1 \][/tex]
[tex]\[ 4 = 4 \][/tex] (which is true)
Second equation:
[tex]\[ 4 = 1 - 5 \][/tex]
[tex]\[ 4 = -4 \][/tex] (which is false)
Since the second equation is not satisfied, this system does not have the solution [tex]\((1, 4)\)[/tex].
### System 4:
[tex]\[ \begin{cases} y = 3x + 1 \\ y = -x - 5 \end{cases} \][/tex]
Substitute [tex]\(x = 1\)[/tex] and [tex]\(y = 4\)[/tex]:
First equation:
[tex]\[ 4 = 3(1) + 1 \][/tex]
[tex]\[ 4 = 3 + 1 \][/tex]
[tex]\[ 4 = 4 \][/tex] (which is true)
Second equation:
[tex]\[ 4 = -(1) - 5 \][/tex]
[tex]\[ 4 = -1 - 5 \][/tex]
[tex]\[ 4 = -6 \][/tex] (which is false)
Since the second equation is not satisfied, this system does not have the solution [tex]\((1, 4)\)[/tex].
### Conclusion:
The system that has the solution [tex]\((1, 4)\)[/tex] is:
[tex]\[ \begin{cases} y = 3x + 1 \\ y = -x + 5 \end{cases} \][/tex]
So, the answer is the second system of equations.
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