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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. [tex]$(4, 4)$[/tex]

B. [tex]$(-4, -12)$[/tex]

C. [tex][tex]$(4, 4)$[/tex][/tex] and [tex]$(-4, 12)$[/tex]

D. [tex]$(-4, 4)$[/tex] and [tex][tex]$(4, 12)$[/tex][/tex]



Answer :

GinnyAnswer
To solve the system of equations
[tex]\[ y = x^2 - 6x + 12 \][/tex]
and
[tex]\[ y = 2x - 4, \][/tex]
we follow these steps:

1. Set the equations equal to each other to eliminate [tex]\( y \)[/tex]:
[tex]\[ x^2 - 6x + 12 = 2x - 4 \][/tex]

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

3. Factor or solve the quadratic equation:
[tex]\[ x^2 - 8x + 16 = 0 \][/tex]
This can be factored as:
[tex]\[ (x - 4)^2 = 0 \][/tex]
Therefore, we have a double root:
[tex]\[ x = 4 \][/tex]

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

Hence, the solution to the system of equations is:
[tex]\[ (4, 4) \][/tex]

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

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