Let's match the systems of equations with their solution sets based on the given information.
### System 1
[tex]\[
\begin{array}{c}
y + 12 = x^2 + x \\
x + y = 3
\end{array}
\][/tex]
Solution set: [tex]\([(-5, 8), (3, 0)]\)[/tex]
### System 2
[tex]\[
\begin{array}{c}
y + 5 = x^2 - 3x \\
2x + y = 1
\end{array}
\][/tex]
Solution set: [tex]\([(-2, 5), (3, -5)]\)[/tex]
### System 3
[tex]\[
\begin{array}{c}
y - 17 = x^2 - 9x \\
-x + y = 1
\end{array}
\][/tex]
Solution set: [tex]\([(2, 3), (8, 9)]\)[/tex]
### System 4
[tex]\[
\begin{array}{c}
y - 15 = x^2 + 4x \\
x - y = 1
\end{array}
\][/tex]
Solution set: [tex]\([(-3/2 - \sqrt{55}i/2, -5/2 - \sqrt{55}i/2), (-3/2 + \sqrt{55}i/2, -5/2 + \sqrt{55}i/2)]\)[/tex]
### System 5
[tex]\[
\begin{array}{c}
y - 6 = x^2 - 3x \\
x + 2y = 2
\end{array}
\][/tex]
Solution set: [tex]\([(5/4 - \sqrt{55}i/4, 3/8 + \sqrt{55}i/8), (5/4 + \sqrt{55}i/4, 3/8 - \sqrt{55}i/8)]\)[/tex]
### System 6
[tex]\[
\begin{array}{c}
y - 15 = -x^2 + 4x \\
x + y = 1
\end{array}
\][/tex]
Solution set: [tex]\([(-2, 3), (7, -6)]\)[/tex]
So, the correct pairs are:
1.
[tex]\[
\begin{array}{c}
y + 12 = x^2 + x \\
x + y = 3
\end{array}
\][/tex]
Solution: [tex]\([(-5, 8), (3, 0)]\)[/tex]
2.
[tex]\[
\begin{array}{c}
y + 5 = x^2 - 3x \\
2x + y = 1
\end{array}
\][/tex]
Solution: [tex]\([(-2, 5), (3, -5)]\)[/tex]
3.
[tex]\[
\begin{array}{c}
y - 17 = x^2 - 9x \\
-x + y = 1
\end{array}
\][/tex]
Solution: [tex]\([(2, 3), (8, 9)]\)[/tex]
4.
[tex]\[
\begin{array}{c}
y - 15 = x^2 + 4x \\
x - y = 1
\end{array}
\][/tex]
Solution: [tex]\([(-3/2 - \sqrt{55}i/2, -5/2 - \sqrt{55}i/2), (-3/2 + \sqrt{55}i/2, -5/2 + \sqrt{55}i/2)]\)[/tex]
5.
[tex]\[
\begin{array}{c}
y - 6 = x^2 - 3x \\
x + 2y = 2
\end{array}
\][/tex]
Solution: [tex]\([(5/4 - \sqrt{55}i/4, 3/8 + \sqrt{55}i/8), (5/4 + \sqrt{55}i/4, 3/8 - \sqrt{55}i/8)]\)[/tex]
6.
[tex]\[
\begin{array}{c}
y - 15 = -x^2 + 4x \\
x + y = 1
\end{array}
\][/tex]
Solution: [tex]\([(-2, 3), (7, -6)]\)[/tex]
