The first two steps in determining the solution set of the system of equations, [tex]$y=x^2-6x+12$[/tex] and [tex]$y=2x-4$[/tex], algebraically are shown in the table.

\begin{tabular}{|c|c|}
\hline
Step & Equation \\
\hline
Step 1 & [tex]$x^2-6x+12=2x-4$[/tex] \\
\hline
Step 2 & [tex][tex]$x^2-8x+16=0$[/tex][/tex] \\
\hline
\end{tabular}

Which represents the solution(s) of this system of equations?

A. (4, 4)
B. (-4, -12)
C. (4, 4) and (-4, 12)
D. (-4, 4) and (4, 12)



Answer :

To solve the system of equations given by [tex]\( y = x^2 - 6x + 12 \)[/tex] and [tex]\( y = 2x - 4 \)[/tex], we need to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations simultaneously.

### Step-by-Step Solution:

1. Set the equations equal to each other:
Since both expressions are equal to [tex]\( y \)[/tex], we can set them equal to each other:
[tex]\[ x^2 - 6x + 12 = 2x - 4 \][/tex]

2. Rearrange to form a quadratic equation:
Move all terms to one side to simplify the equation:
[tex]\[ x^2 - 6x + 12 - 2x + 4 = 0 \][/tex]
Combining like terms, we get:
[tex]\[ x^2 - 8x + 16 = 0 \][/tex]

3. Solve the quadratic equation:
The equation [tex]\( x^2 - 8x + 16 = 0 \)[/tex] can be solved by factoring or using the quadratic formula. This specific equation is a perfect square trinomial, which can be factored as:
[tex]\[ (x - 4)^2 = 0 \][/tex]
Taking the square root of both sides, we have:
[tex]\[ x - 4 = 0 \implies x = 4 \][/tex]

4. Find the corresponding [tex]\( y \)[/tex] value:
Substitute [tex]\( x = 4 \)[/tex] into one of the original equations to find the corresponding [tex]\( y \)[/tex] value. Using [tex]\( y = 2x - 4 \)[/tex]:
[tex]\[ y = 2(4) - 4 = 8 - 4 = 4 \][/tex]

Thus, the point of intersection, and hence the solution to the system of equations, is:
[tex]\[ (x, y) = (4, 4) \][/tex]

### Conclusion:
The solution to the system of equations [tex]\( y = x^2 - 6x + 12 \)[/tex] and [tex]\( y = 2x - 4 \)[/tex] is:
[tex]\[ (4, 4) \][/tex]

Therefore, the correct answer is:
[tex]\[ (4, 4) \][/tex]