To determine which ordered pair is a solution to the system of linear equations, we need to check each pair to see if it satisfies both equations:
[tex]\[
\begin{array}{l}
x + 4y = 3 \\
y = -4x - 3
\end{array}
\][/tex]
Let's check each pair:
1. For the pair (1, 1):
[tex]\[
\begin{array}{l}
x + 4y = 3 \\
1 + 4(1) = 1 + 4 = 5 \quad \text{(not equal to 3, so this pair does not satisfy the first equation)}
\end{array}
\][/tex]
[tex]\[
\begin{array}{l}
y = -4x - 3 \\
1 = -4(1) - 3 = -4 - 3 = -7 \quad \text{(this pair does not satisfy the second equation either)}
\end{array}
\][/tex]
Thus, [tex]\((1, 1)\)[/tex] is not a solution.
2. For the pair (1, -1):
[tex]\[
\begin{array}{l}
x + 4y = 3 \\
1 + 4(-1) = 1 - 4 = -3 \quad \text{(not equal to 3, so this pair does not satisfy the first equation)}
\end{array}
\][/tex]
[tex]\[
\begin{array}{l}
y = -4x - 3 \\
-1 = -4(1) - 3 = -4 - 3 = -7 \quad \text{(this pair does not satisfy the second equation either)}
\end{array}
\][/tex]
Thus, [tex]\((1, -1)\)[/tex] is not a solution.
3. For the pair (-1, 1):
[tex]\[
\begin{array}{l}
x + 4y = 3 \\
-1 + 4(1) = -1 + 4 = 3 \quad \text{(this pair satisfies the first equation)}
\end{array}
\][/tex]
[tex]\[
\begin{array}{l}
y = -4x - 3 \\
1 = -4(-1) - 3 = 4 - 3 = 1 \quad \text{(this pair satisfies the second equation as well)}
\end{array}
\][/tex]
Thus, [tex]\((-1, 1)\)[/tex] is a solution.
4. For the pair (-1, -1):
[tex]\[
\begin{array}{l}
x + 4y = 3 \\
-1 + 4(-1) = -1 - 4 = -5 \quad \text{(not equal to 3, so this pair does not satisfy the first equation)}
\end{array}
\][/tex]
[tex]\[
\begin{array}{l}
y = -4x - 3 \\
-1 = -4(-1) - 3 = 4 - 3 = 1 \quad \text{(this pair does not satisfy the second equation either)}
\end{array}
\][/tex]
Thus, [tex]\((-1, -1)\)[/tex] is not a solution.
After checking each pair, we find that the only ordered pair that satisfies both equations is [tex]\((-1, 1)\)[/tex].
Thus, the correct answer is:
[tex]\[
\boxed{(-1, 1)}
\][/tex]
