Sure! Let's solve the system of equations using the elimination method to find the correct ordered pair.
We are given the following system of equations:
[tex]\[
\begin{array}{l}
6x + 2y = 8 \quad \text{(1)} \\
12x + y = 22 \quad \text{(2)}
\end{array}
\][/tex]
First, we need to eliminate one of the variables. To do this, we can align the coefficients of [tex]\( y \)[/tex] in both equations and subtract or add them.
Since the coefficient of [tex]\( y \)[/tex] in the second equation is 1, and in the first equation it's 2, we can multiply the second equation by 2 to align the coefficients of [tex]\( y \)[/tex]:
[tex]\[
2(12x + y) = 2(22)
\][/tex]
This simplifies to:
[tex]\[
24x + 2y = 44 \quad \text{(3)}
\][/tex]
Now we have the system:
[tex]\[
\begin{array}{l}
6x + 2y = 8 \quad \text{(1)} \\
24x + 2y = 44 \quad \text{(3)}
\end{array}
\][/tex]
Next, we subtract equation (1) from equation (3) to eliminate [tex]\( y \)[/tex]:
[tex]\[
(24x + 2y) - (6x + 2y) = 44 - 8
\][/tex]
Simplifying this, we get:
[tex]\[
24x + 2y - 6x - 2y = 36
\][/tex]
[tex]\[
18x = 36
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{36}{18} = 2
\][/tex]
Now that we have [tex]\( x = 2 \)[/tex], we substitute it back into one of the original equations to find [tex]\( y \)[/tex]. Let's use equation (1):
[tex]\[
6x + 2y = 8
\][/tex]
Substitute [tex]\( x = 2 \)[/tex]:
[tex]\[
6(2) + 2y = 8
\][/tex]
[tex]\[
12 + 2y = 8
\][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[
2y = 8 - 12
\][/tex]
[tex]\[
2y = -4
\][/tex]
[tex]\[
y = \frac{-4}{2} = -2
\][/tex]
So, the solution to the system of equations is [tex]\((x, y) = (2, -2)\)[/tex].
Therefore, the correct ordered pair is:
[tex]\[ \boxed{(2, -2)} \][/tex]
And the corresponding choice from the given options is:
C. [tex]\((2, -2)\)[/tex]
