Let's start by examining each system of linear and quadratic equations one by one and compare the solutions.
1. [tex]\(\left\{\begin{array}{l}
y + 12 = x^2 + x \\
x + y = 3 \\
\end{array}\right.\)[/tex]
2. [tex]\(\left\{\begin{array}{l}
y + 5 = x^2 - 3x \\
2x + y = 1 \\
\end{array}\right.\)[/tex]
3. [tex]\(\left\{\begin{array}{l}
y - 17 = x^2 - 9x \\
-x + y = 1 \\
\end{array}\right.\)[/tex]
4. [tex]\(\left\{\begin{array}{l}
y - 15 = x^2 + 4x \\
x - y = 1 \\
\end{array}\right.\)[/tex]
5. [tex]\(\left\{\begin{array}{l}
y - 6 = x^2 - 3x \\
x + 2y = 2 \\
\end{array}\right.\)[/tex]
6. [tex]\(\left\{\begin{array}{l}
y - 15 = -x^2 + 4x \\
x + y = 1 \\
\end{array}\right.\)[/tex]
We note the provided solution sets:
- [tex]\(\{(-2, 3), (7, -6)\}\)[/tex]
- [tex]\(\{(-5, 8), (3, 0)\}\)[/tex]
- [tex]\(\{(-2, 5), (3, -5)\}\)[/tex]
- [tex]\(\{(2, 3), (8, 9)\}\)[/tex]
After evaluating these systems step-by-step, we find that none of the systems match these solution sets perfectly.
Therefore, the matches are:
[tex]\[ (\square, \square, \square, \square) = ([], [], [], []) \][/tex]
Because none of the pairings from the systems of equations seems to match any given solution sets.