To solve the system of equations using the elimination method, we must eliminate one of the variables to find the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]. Let's follow these steps:
The given system of equations is:
[tex]\[
\begin{array}{r}
4x + y = 8 \quad \text{(1)} \\
3x - 2y = 6 \quad \text{(2)}
\end{array}
\][/tex]
Step 1: Make the coefficients of [tex]\( y \)[/tex] the same
To eliminate [tex]\( y \)[/tex], we need to make the coefficients of [tex]\( y \)[/tex] in both equations equal. The least common multiple of the coefficients [tex]\(1\)[/tex] and [tex]\(-2\)[/tex] is [tex]\(2\)[/tex]. Thus, we will multiply equation (1) by 2 to make the coefficient of [tex]\( y \)[/tex] equal to 2 in absolute value.
[tex]\[
2 \times (4x + y) = 2 \times 8
\][/tex]
[tex]\[
8x + 2y = 16 \quad \text{(3)}
\][/tex]
Now the system of equations looks like this:
[tex]\[
\begin{array}{r}
8x + 2y = 16 \quad \text{(3)} \\
3x - 2y = 6 \quad \text{(2)}
\end{array}
\][/tex]
Step 2: Add the equations
Next, add equations (3) and (2) together to eliminate [tex]\( y \)[/tex].
[tex]\[
(8x + 2y) + (3x - 2y) = 16 + 6
\][/tex]
The [tex]\( 2y \)[/tex] and [tex]\( -2y \)[/tex] cancel out each other:
[tex]\[
8x + 3x = 22
\][/tex]
[tex]\[
11x = 22
\][/tex]
Step 3: Solve for [tex]\( x \)[/tex]
Divide both sides by 11:
[tex]\[
x = \frac{22}{11}
\][/tex]
[tex]\[
x = 2
\][/tex]
Step 4: Substitute [tex]\( x \)[/tex] back to find [tex]\( y \)[/tex]
Now substitute [tex]\( x = 2 \)[/tex] back into one of the original equations to find [tex]\( y \)[/tex]. We will use equation (1):
[tex]\[
4x + y = 8
\][/tex]
Substitute [tex]\( x = 2 \)[/tex]:
[tex]\[
4(2) + y = 8
\][/tex]
[tex]\[
8 + y = 8
\][/tex]
Subtract 8 from both sides:
[tex]\[
y = 0
\][/tex]
Step 5: Verify the solution
We have found [tex]\( x = 2 \)[/tex] and [tex]\( y = 0 \)[/tex]. We should verify these values in the second equation to ensure they satisfy the original system.
Substitute [tex]\( x = 2 \)[/tex] and [tex]\( y = 0 \)[/tex] into equation (2):
[tex]\[
3(2) - 2(0) = 6
\][/tex]
[tex]\[
6 = 6
\][/tex]
Since both equations are satisfied, the solution to the system of equations is:
[tex]\[
x = 2, \quad y = 0
\][/tex]
Therefore, the correct answer is:
[tex]\[
\boxed{D. \, x = 2, \, y = 0}
\][/tex]
