We are given the system of equations:
[tex]\[
\begin{cases}
x + y = 1 \\
3x - y = 3
\end{cases}
\][/tex]
We need to determine which of the given points is a solution to this system. Let's evaluate each point in the system one by one.
Option A: (3,0)
Substitute [tex]\( x = 3 \)[/tex] and [tex]\( y = 0 \)[/tex]:
1. [tex]\( x + y = 1 \)[/tex]
[tex]\[ 3 + 0 = 3 \quad \text{(which is not equal to 1)} \][/tex]
2. [tex]\( 3x - y = 3 \)[/tex]
[tex]\[ 3(3) - 0 = 9 \quad \text{(which is not equal to 3)} \][/tex]
So, [tex]\((3,0)\)[/tex] is not a solution.
Option B: (5,-1)
Substitute [tex]\( x = 5 \)[/tex] and [tex]\( y = -1 \)[/tex]:
1. [tex]\( x + y = 1 \)[/tex]
[tex]\[ 5 + (-1) = 4 \quad \text{(which is not equal to 1)} \][/tex]
2. [tex]\( 3x - y = 3 \)[/tex]
[tex]\[ 3(5) - (-1) = 16 \quad \text{(which is not equal to 3)} \][/tex]
So, [tex]\((5,-1)\)[/tex] is not a solution.
Option C: (1,4)
Substitute [tex]\( x = 1 \)[/tex] and [tex]\( y = 4 \)[/tex]:
1. [tex]\( x + y = 1 \)[/tex]
[tex]\[ 1 + 4 = 5 \quad \text{(which is not equal to 1)} \][/tex]
2. [tex]\( 3x - y = 3 \)[/tex]
[tex]\[ 3(1) - 4 = -1 \quad \text{(which is not equal to 3)} \][/tex]
So, [tex]\((1,4)\)[/tex] is not a solution.
Option D: (1,0)
Substitute [tex]\( x = 1 \)[/tex] and [tex]\( y = 0 \)[/tex]:
1. [tex]\( x + y = 1 \)[/tex]
[tex]\[ 1 + 0 = 1 \quad \text{(which is equal to 1)} \][/tex]
2. [tex]\( 3x - y = 3 \)[/tex]
[tex]\[ 3(1) - 0 = 3 \quad \text{(which is equal to 3)} \][/tex]
Both equations are satisfied with this substitution, so [tex]\((1,0)\)[/tex] is indeed a solution to the system of equations.
Thus, the correct answer is:
D. (1,0)
