To determine which point is a solution to the system of linear equations:
[tex]\[
\begin{cases}
y = -x + 4 \\
x - 3y = 12
\end{cases}
\][/tex]
we need to test each given point [tex]\((x, y)\)[/tex] to see if it satisfies both equations.
#### Point [tex]\((0, 3)\)[/tex]:
1. Substitute [tex]\(x = 0\)[/tex] and [tex]\(y = 3\)[/tex] into the first equation:
[tex]\[
y = -x + 4 \rightarrow 3 = -(0) + 4 \rightarrow 3 = 4 \quad (\text{False})
\][/tex]
Since this point does not satisfy the first equation, it cannot be a solution.
#### Point [tex]\((1, 2)\)[/tex]:
1. Substitute [tex]\(x = 1\)[/tex] and [tex]\(y = 2\)[/tex] into the first equation:
[tex]\[
y = -x + 4 \rightarrow 2 = -(1) + 4 \rightarrow 2 = 3 \quad (\text{False})
\][/tex]
Since this point does not satisfy the first equation, it cannot be a solution.
#### Point [tex]\((6, -2)\)[/tex]:
1. Substitute [tex]\(x = 6\)[/tex] and [tex]\(y = -2\)[/tex] into the first equation:
[tex]\[
y = -x + 4 \rightarrow -2 = -(6) + 4 \rightarrow -2 = -6 + 4 \rightarrow -2 = -2 \quad (\text{True})
\][/tex]
2. Substitute [tex]\(x = 6\)[/tex] and [tex]\(y = -2\)[/tex] into the second equation:
[tex]\[
x - 3y = 12 \rightarrow 6 - 3(-2) = 12 \rightarrow 6 + 6 = 12 \rightarrow 12 = 12 \quad (\text{True})
\][/tex]
Since this point satisfies both equations, [tex]\((6, -2)\)[/tex] is a solution to the system.
#### Point [tex]\((4, -4)\)[/tex]:
1. Substitute [tex]\(x = 4\)[/tex] and [tex]\(y = -4\)[/tex] into the first equation:
[tex]\[
y = -x + 4 \rightarrow -4 = -(4) + 4 \rightarrow -4 = -4 + 4 \rightarrow -4 = 0 \quad (\text{False})
\][/tex]
Since this point does not satisfy the first equation, it cannot be a solution.
After testing all points, we determine that the point [tex]\((6, -2)\)[/tex] satisfies both equations. Therefore, the solution to the system of equations is:
[tex]\[
(6, -2)
\][/tex]
