To solve the system of equations using the linear combination (or elimination) method, we'll follow these steps:
1. Write down the system of equations:
[tex]\[
\begin{cases}
6x - 3y = 3 \\
-2x + 6y = 14
\end{cases}
\][/tex]
2. Find a common coefficient for either [tex]\( x \)[/tex] or [tex]\( y \)[/tex] to facilitate elimination.
Notice that the coefficients of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] in the equations are not directly the same, but they can be manipulated to be. Let's focus on eliminating [tex]\( x \)[/tex]:
Equation 1: [tex]\( 6x - 3y = 3 \)[/tex]
Equation 2: [tex]\( -2x + 6y = 14 \)[/tex]
If we multiply Equation 2 by 3, we get:
[tex]\[
3(-2x + 6y) = 3(14)
\][/tex]
Simplifying, we obtain:
[tex]\[
-6x + 18y = 42
\][/tex]
3. Rewrite the system with the new equation:
[tex]\[
\begin{cases}
6x - 3y = 3 \\
-6x + 18y = 42
\end{cases}
\][/tex]
4. Add the two equations to eliminate [tex]\( x \)[/tex]:
[tex]\[
(6x - 3y) + (-6x + 18y) = 3 + 42
\][/tex]
This simplifies to:
[tex]\[
15y = 45
\][/tex]
5. Solve for [tex]\( y \)[/tex]:
[tex]\[
y = \frac{45}{15}
\][/tex]
[tex]\[
y = 3
\][/tex]
6. Substitute [tex]\( y \)[/tex] back into one of the original equations to solve for [tex]\( x \)[/tex]:
Let's substitute [tex]\( y = 3 \)[/tex] into the first equation:
[tex]\[
6x - 3(3) = 3
\][/tex]
Simplifying,
[tex]\[
6x - 9 = 3
\][/tex]
[tex]\[
6x = 3 + 9
\][/tex]
[tex]\[
6x = 12
\][/tex]
[tex]\[
x = \frac{12}{6}
\][/tex]
[tex]\[
x = 2
\][/tex]
The solution to the system of equations is:
[tex]\((x, y) = (2, 3)\)[/tex]
Therefore, the correct answer is:
[tex]\((2, 3)\)[/tex]
