Sure, let's solve for the [tex]\( y \)[/tex] values using the given [tex]\( x \)[/tex] values in the system of equations:
[tex]\[
\begin{array}{c}
y = x^2 - 2x + 3 \\
y = -2x + 12
\end{array}
\][/tex]
### For [tex]\( x = -3 \)[/tex]
1. Substitute [tex]\( x = -3 \)[/tex] into the first equation:
[tex]\[
y = (-3)^2 - 2(-3) + 3 = 9 + 6 + 3 = 18
\][/tex]
So, when [tex]\( x = -3 \)[/tex], the [tex]\( y \)[/tex] value for the first equation is [tex]\( 18 \)[/tex].
2. Substitute [tex]\( x = -3 \)[/tex] into the second equation:
[tex]\[
y = -2(-3) + 12 = 6 + 12 = 18
\][/tex]
So, when [tex]\( x = -3 \)[/tex], the [tex]\( y \)[/tex] value for the second equation is also [tex]\( 18 \)[/tex].
Thus, the point for [tex]\( x = -3 \)[/tex] is [tex]\((-3, 18)\)[/tex].
### For [tex]\( x = 3 \)[/tex]
1. Substitute [tex]\( x = 3 \)[/tex] into the first equation:
[tex]\[
y = 3^2 - 2(3) + 3 = 9 - 6 + 3 = 6
\][/tex]
So, when [tex]\( x = 3 \)[/tex], the [tex]\( y \)[/tex] value for the first equation is [tex]\( 6 \)[/tex].
2. Substitute [tex]\( x = 3 \)[/tex] into the second equation:
[tex]\[
y = -2(3) + 12 = -6 + 12 = 6
\][/tex]
So, when [tex]\( x = 3 \)[/tex], the [tex]\( y \)[/tex] value for the second equation is also [tex]\( 6 \)[/tex].
Thus, the point for [tex]\( x = 3 \)[/tex] is [tex]\((3, 6)\)[/tex].
In conclusion, the solution points are:
[tex]\[
(-3, 18) \quad \text{and} \quad (3, 6)
\][/tex]
