To find the ordered pair solutions for the given system of equations, let's solve for [tex]\( x \)[/tex] and use those values to find the corresponding [tex]\( y \)[/tex] values.
Given the equations:
1. [tex]\( y = x^2 - 2x + 3 \)[/tex]
2. [tex]\( y = -2x + 12 \)[/tex]
First, set the right-hand sides of the equations equal to each other:
[tex]\[ x^2 - 2x + 3 = -2x + 12 \][/tex]
Next, combine like terms to form a standard quadratic equation:
[tex]\[ x^2 - 2x + 2x + 3 - 12 = 0 \][/tex]
This simplifies to:
[tex]\[ x^2 - 9 = 0 \][/tex]
We solve the quadratic equation [tex]\( x^2 - 9 = 0 \)[/tex]. This can be factored as:
[tex]\[ (x - 3)(x + 3) = 0 \][/tex]
Setting each factor to zero gives the solutions for [tex]\( x \)[/tex]:
[tex]\[ x - 3 = 0 \][/tex]
[tex]\[ x = 3 \][/tex]
[tex]\[ x + 3 = 0 \][/tex]
[tex]\[ x = -3 \][/tex]
So, the solutions for [tex]\( x \)[/tex] are [tex]\( x = -3 \)[/tex] and [tex]\( x = 3 \)[/tex].
Next, we substitute these [tex]\( x \)[/tex] values back into one of the original equations to find the corresponding [tex]\( y \)[/tex] values. Let's use the equation [tex]\( y = -2x + 12 \)[/tex] for convenience:
For [tex]\( x = -3 \)[/tex]:
[tex]\[ y = -2(-3) + 12 \][/tex]
[tex]\[ y = 6 + 12 \][/tex]
[tex]\[ y = 18 \][/tex]
So, one ordered pair is [tex]\((-3, 18)\)[/tex].
For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = -2(3) + 12 \][/tex]
[tex]\[ y = -6 + 12 \][/tex]
[tex]\[ y = 6 \][/tex]
So, the other ordered pair is [tex]\((3, 6)\)[/tex].
Therefore, the ordered pair solutions to the system of equations are:
[tex]\[ (-3, 18) \text{ and } (3, 6) \][/tex]
