To solve the system of linear equations using the linear combination (or elimination) method, we need to eliminate one of the variables by combining the two equations. The given system is:
[tex]\[
\begin{array}{l}
6x - 3y = 3 \\
-2x + 6y = 14
\end{array}
\][/tex]
Step 1: Multiply the equations to make the coefficients of one of the variables (either [tex]\(x\)[/tex] or [tex]\(y\)[/tex]) the same (with the same or opposite sign).
We can multiply the first equation by 2 and the second equation by 1 (leaving it unchanged) to make the coefficients of [tex]\(x\)[/tex] compatible for elimination:
[tex]\[
\begin{array}{l}
2(6x - 3y) = 2(3) \\
1(-2x + 6y) = 1(14)
\end{array}
\][/tex]
This simplifies to:
[tex]\[
\begin{array}{l}
12x - 6y = 6 \quad \Rightarrow \quad (1) \\
-2x + 6y = 14 \quad \Rightarrow \quad (2)
\end{array}
\][/tex]
Step 2: Add the two equations to eliminate [tex]\(y\)[/tex]:
[tex]\[
(12x - 6y) + (-2x + 6y) = 6 + 14
\][/tex]
This simplifies to:
[tex]\[
12x - 6y - 2x + 6y = 20
\][/tex]
Combining like terms gives us:
[tex]\[
10x = 20
\][/tex]
Step 3: Solve for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{20}{10}
\][/tex]
[tex]\[
x = 2
\][/tex]
Step 4: Substitute [tex]\(x = 2\)[/tex] back into one of the original equations to find [tex]\(y\)[/tex]. We'll use the first equation:
[tex]\[
6x - 3y = 3
\][/tex]
Substitute [tex]\(x = 2\)[/tex]:
[tex]\[
6(2) - 3y = 3
\][/tex]
[tex]\[
12 - 3y = 3
\][/tex]
Step 5: Solve for [tex]\(y\)[/tex]:
[tex]\[
12 - 3 = 3y
\][/tex]
[tex]\[
9 = 3y
\][/tex]
[tex]\[
y = \frac{9}{3}
\][/tex]
[tex]\[
y = 3
\][/tex]
Solution:
The solution to the system of equations is [tex]\((x, y) = (2, 3)\)[/tex].
Answer:
[tex]\((2, 3)\)[/tex]
