We are given the system of equations:
[tex]\[
3x + \frac{1}{2}y = 3
\][/tex]
[tex]\[
6x - y = 2
\][/tex]
### Part 1: Eliminating the [tex]\(x\)[/tex]-terms
To eliminate the [tex]\(x\)[/tex]-terms, we need to manipulate the equations so that the coefficients of [tex]\(x\)[/tex] in both equations are equal in magnitude but opposite in sign.
1. Original equations:
[tex]\[
3x + \frac{1}{2}y = 3 \quad \text{(1)}
\][/tex]
[tex]\[
6x - y = 2 \quad \text{(2)}
\][/tex]
2. Multiply the first equation (1) by [tex]\(-2\)[/tex]:
[tex]\[
-2(3x + \frac{1}{2}y = 3)
\][/tex]
This gives us:
[tex]\[
-6x - y = -6 \quad \text{(3)}
\][/tex]
Now, add the modified equation (3) to equation (2) to eliminate the [tex]\(x\)[/tex]-terms:
[tex]\[
(-6x - y) + (6x - y) = -6 + 2
\][/tex]
Simplifying, we get:
[tex]\[
-2y = -4
\][/tex]
[tex]\[
y = 2
\][/tex]
Thus, [tex]\(\boxed{\text{Multiply the first equation by -2}}\)[/tex] is the correct operation to eliminate the [tex]\(x\)[/tex]-terms.
### Part 2: Eliminating the [tex]\(y\)[/tex]-terms
To eliminate the [tex]\(y\)[/tex]-terms, we can make the coefficients of [tex]\(y\)[/tex] in both equations equal in magnitude but opposite in sign.
1. Original equations:
[tex]\[
3x + \frac{1}{2}y = 3 \quad \text{(1)}
\][/tex]
[tex]\[
6x - y = 2 \quad \text{(2)}
\][/tex]
2. Multiply the first equation (1) by 2:
[tex]\[
2(3x + \frac{1}{2}y = 3)
\][/tex]
This gives us:
[tex]\[
6x + y = 6 \quad \text{(4)}
\][/tex]
Now, add the modified equation (4) to equation (2) to eliminate the [tex]\(y\)[/tex]-terms:
[tex]\[
(6x + y) + (6x - y) = 6 + 2
\][/tex]
Simplifying, we get:
[tex]\[
12x = 8
\][/tex]
Therefore, multiplying the first equation by 2 does not eliminate the [tex]\(y\)[/tex]-terms properly.
### Part 3: Solving for [tex]\(x\)[/tex]
1. We found [tex]\(y = 2\)[/tex] earlier using elimination by eliminating [tex]\(x\)[/tex]-terms.
2. Substitute [tex]\(y = 2\)[/tex] back into the original first equation to solve for [tex]\(x\)[/tex]:
[tex]\[
3x + \frac{1}{2}(2) = 3
\][/tex]
[tex]\[
3x + 1 = 3
\][/tex]
[tex]\[
3x = 2
\][/tex]
[tex]\[
x = \frac{2}{3}
\][/tex]
So, the solution to the system is:
[tex]\[
(x, y) = \left(\frac{2}{3}, 2\right)
\][/tex]
The solution is [tex]\(\boxed{\left(\frac{2}{3}, 2\right)}\)[/tex].
