Answer :
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].
[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].
