Let's solve the system of equations given the x-values -3 and 3.
The system consists of the following equations:
1. [tex]\( y = x^2 - 2x + 3 \)[/tex]
2. [tex]\( y = -2x + 12 \)[/tex]
We need to calculate the corresponding y-values for [tex]\( x = -3 \)[/tex] and [tex]\( x = 3 \)[/tex] using both equations.
### For [tex]\( x = -3 \)[/tex]:
#### Equation 1: [tex]\( y = x^2 - 2x + 3 \)[/tex]
[tex]\[ y = (-3)^2 - 2(-3) + 3 \][/tex]
[tex]\[ y = 9 + 6 + 3 \][/tex]
[tex]\[ y = 18 \][/tex]
#### Equation 2: [tex]\( y = -2x + 12 \)[/tex]
[tex]\[ y = -2(-3) + 12 \][/tex]
[tex]\[ y = 6 + 12 \][/tex]
[tex]\[ y = 18 \][/tex]
So, for [tex]\( x = -3 \)[/tex], the [tex]\( y \)[/tex]-values are both 18. Thus, we have the point [tex]\((-3, 18)\)[/tex] for both equations.
### For [tex]\( x = 3 \)[/tex]:
#### Equation 1: [tex]\( y = x^2 - 2x + 3 \)[/tex]
[tex]\[ y = 3^2 - 2(3) + 3 \][/tex]
[tex]\[ y = 9 - 6 + 3 \][/tex]
[tex]\[ y = 6 \][/tex]
#### Equation 2: [tex]\( y = -2x + 12 \)[/tex]
[tex]\[ y = -2(3) + 12 \][/tex]
[tex]\[ y = -6 + 12 \][/tex]
[tex]\[ y = 6 \][/tex]
So, for [tex]\( x = 3 \)[/tex], the [tex]\( y \)[/tex]-values are both 6. Thus, we have the point [tex]\((3, 6)\)[/tex] for both equations.
### Summary
Using the system of equations and the given [tex]\( x \)[/tex]-values:
- For [tex]\( x = -3 \)[/tex]:
- [tex]\( y \)[/tex] from Equation 1: 18
- [tex]\( y \)[/tex] from Equation 2: 18
- Resulting point: [tex]\((-3, 18)\)[/tex]
- For [tex]\( x = 3 \)[/tex]:
- [tex]\( y \)[/tex] from Equation 1: 6
- [tex]\( y \)[/tex] from Equation 2: 6
- Resulting point: [tex]\((3, 6)\)[/tex]
Thus, the final results are:
[tex]\[
((-3, 18, 18), (3, 6, 6))
\][/tex]
