Let's solve the system of linear equations using the linear combination (or elimination) method.
Given the system of equations:
[tex]\[
\begin{array}{l}
6x - 3y = 3 \quad \text{(1)} \\
-2x + 6y = 14 \quad \text{(2)}
\end{array}
\][/tex]
Step 1: Align the coefficients
To eliminate [tex]\( y \)[/tex], we can align the coefficients of [tex]\( y \)[/tex] by manipulating the equations. We see that the coefficient of [tex]\( y \)[/tex] in equation [tex]\((1)\)[/tex] is [tex]\(-3y\)[/tex] and in [tex]\((2)\)[/tex] it is [tex]\(6y\)[/tex]. One way to align them is by multiplying equation [tex]\((1)\)[/tex] by 2, so the coefficients of [tex]\( y \)[/tex] become opposites:
[tex]\[
2 \cdot (6x - 3y) = 2 \cdot 3 \implies 12x - 6y = 6 \quad \text{(3)}
\][/tex]
Step 2: Add the equations
Now, we add equation [tex]\((3)\)[/tex] and equation [tex]\((2)\)[/tex]:
[tex]\[
\begin{array}{rcl}
12x - 6y & = & 6 \quad \text{(3)} \\
-2x + 6y & = & 14 \quad \text{(2)} \\
\hline
(12x - 2x) + (-6y + 6y) & = & 6 + 14 \\
10x & = & 20
\end{array}
\][/tex]
Step 3: Solve for [tex]\( x \)[/tex]
[tex]\[
10x = 20 \implies x = \frac{20}{10} = 2
\][/tex]
Step 4: Substitute [tex]\( x \)[/tex] back into one of the original equations
Let's substitute [tex]\( x = 2 \)[/tex] into the first equation [tex]\((1)\)[/tex]:
[tex]\[
6(2) - 3y = 3 \implies 12 - 3y = 3
\][/tex]
Step 5: Solve for [tex]\( y \)[/tex]
[tex]\[
12 - 3y = 3 \implies -3y = 3 - 12 \implies -3y = -9 \implies y = \frac{-9}{-3} = 3
\][/tex]
The solution to the system of equations is:
[tex]\[
(x, y) = (2, 3)
\][/tex]
Thus, the correct solution is:
[tex]\((2, 3)\)[/tex]
