Let's work through the given system of equations step-by-step to understand how to determine the solution(s).
We are given the system of equations:
[tex]\[ y = x^2 - 6x + 12 \][/tex]
[tex]\[ y = 2x - 4 \][/tex]
### Step 1: Set the equations equal to each other
Set the expressions for [tex]\( y \)[/tex] from both equations equal to each other:
[tex]\[ x^2 - 6x + 12 = 2x - 4 \][/tex]
### Step 2: Move all terms to one side to form a quadratic equation
Rearrange the terms to form a standard quadratic equation:
[tex]\[ x^2 - 6x + 12 - 2x + 4 = 0 \][/tex]
Combine like terms:
[tex]\[ x^2 - 8x + 16 = 0 \][/tex]
### Step 3: Solve the quadratic equation
We now have the quadratic equation:
[tex]\[ x^2 - 8x + 16 = 0 \][/tex]
To solve this quadratic equation, we can use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -8 \)[/tex], and [tex]\( c = 16 \)[/tex].
1. Calculate the discriminant [tex]\( \Delta = b^2 - 4ac \)[/tex]:
[tex]\[ \Delta = (-8)^2 - 4(1)(16) = 64 - 64 = 0 \][/tex]
Since the discriminant is 0, there is exactly one repeated real root.
2. Calculate the root:
[tex]\[ x = \frac{-(-8) \pm \sqrt{0}}{2(1)} = \frac{8 \pm 0}{2} = \frac{8}{2} = 4 \][/tex]
Thus, the value of [tex]\( x \)[/tex] is [tex]\( x = 4 \)[/tex].
### Step 4: Find the corresponding [tex]\( y \)[/tex] value
Substitute [tex]\( x = 4 \)[/tex] into the linear equation [tex]\( y = 2x - 4 \)[/tex]:
[tex]\[ y = 2(4) - 4 = 8 - 4 = 4 \][/tex]
So, the coordinate pair that satisfies both equations is [tex]\( (4, 4) \)[/tex].
### Conclusion
The system of equations has a single solution. The solution is:
[tex]\[ (4, 4) \][/tex]
Thus, the correct answer is:
[tex]\[ (4, 4) \][/tex]
