To determine which point is a solution to the system of linear equations:
[tex]\[
\begin{cases}
y = -x + 3 \\
2x - y = 6
\end{cases}
\][/tex]
we will substitute each point into the system of equations and see which one satisfies both equations.
Checking the point [tex]\((3, 0)\)[/tex]:
1. Substituting [tex]\((3, 0)\)[/tex] into the first equation [tex]\(y = -x + 3\)[/tex]:
[tex]\[
0 = -3 + 3
\][/tex]
[tex]\[
0 = 0 \quad \text{(True)}
\][/tex]
2. Substituting [tex]\((3, 0)\)[/tex] into the second equation [tex]\(2x - y = 6\)[/tex]:
[tex]\[
2(3) - 0 = 6
\][/tex]
[tex]\[
6 = 6 \quad \text{(True)}
\][/tex]
Since the point [tex]\((3, 0)\)[/tex] satisfies both equations, it is a solution to the system.
Checking the point [tex]\((3, -1)\)[/tex]:
1. Substituting [tex]\((3, -1)\)[/tex] into the first equation [tex]\(y = -x + 3\)[/tex]:
[tex]\[
-1 = -3 + 3
\][/tex]
[tex]\[
-1 = 0 \quad \text{(False)}
\][/tex]
Since the point [tex]\((3, -1)\)[/tex] does not satisfy the first equation, it is not a solution.
Checking the point [tex]\((0, 3)\)[/tex]:
1. Substituting [tex]\((0, 3)\)[/tex] into the first equation [tex]\(y = -x + 3\)[/tex]:
[tex]\[
3 = -0 + 3
\][/tex]
[tex]\[
3 = 3 \quad \text{(True)}
\][/tex]
2. Substituting [tex]\((0, 3)\)[/tex] into the second equation [tex]\(2x - y = 6\)[/tex]:
[tex]\[
2(0) - 3 = 6
\][/tex]
[tex]\[
-3 = 6 \quad \text{(False)}
\][/tex]
Since the point [tex]\((0, 3)\)[/tex] does not satisfy the second equation, it is not a solution.
Checking the point [tex]\((-1, 3)\)[/tex]:
1. Substituting [tex]\((-1, 3)\)[/tex] into the first equation [tex]\(y = -x + 3\)[/tex]:
[tex]\[
3 = -(-1) + 3
\][/tex]
[tex]\[
3 = 1 + 3
\][/tex]
[tex]\[
3 = 4 \quad \text{(False)}
\][/tex]
Since the point [tex]\((-1, 3)\)[/tex] does not satisfy the first equation, it is not a solution.
After checking all the points, we find that only the point [tex]\((3, 0)\)[/tex] satisfies both equations. Therefore, the point [tex]\((3, 0)\)[/tex] is the solution to the system of equations.
