Which of the following statements gives the best choice for multiplying to solve this system using the elimination method?

To solve the system of equations using the elimination method, we want to eliminate one of the variables by making their coefficients equal (but opposite in sign) in both equations.

The original equations are:

1. \(3x - 6y = 4\) (which seems to be what you meant for the first equation, correcting the presentation).
2. \(6x + 11y = -2\)

### Let's analyze the given options:

1. **Multiply the first equation by -11 and the second equation by 6, so the y-variables are eliminated:**
- If we multiply the first equation by -11, we would have:
\[-11(3x - 6y) = -11(4) \rightarrow -33x + 66y = -44\]
- If we multiply the second equation by 6, we would have:
\[6(6x + 11y) = 6(-2) \rightarrow 36x + 66y = -12\]
- In this case, the y-terms would not cancel out; they would actually add up.

2. **Multiply the first equation by 2 so the x-variables are eliminated:**
- This leads to:
\[2(3x - 6y) = 2(4) \rightarrow 6x - 12y = 8\]
- The x-terms do not cancel out.

3. **Multiply the first equation by -2 so the x-variables are eliminated:**
- This leads to:
\[-2(3x - 6y) = -2(4) \rightarrow -6x + 12y = -8\]
- So if we add this to the second equation, the x variables will cancel:
\[-6x + 12y + 6x + 11y = -8 - 2\]
\[23y = -10\]
- This looks promising.

4. **Multiply the second equation by 1/2 so the x-variables are eliminated:**
- This leads to:
\[\frac{1}{2}(6x + 11y) = \frac{1}{2}(-2) \rightarrow 3x + \frac{11}{2}y = -1\]
- The x terms would not effectively cancel out with the first equation.

**Conclusion:**
The best choice for multiplying to solve this system using the elimination method is:

**Multiply the first equation by -2 so the x-variables are eliminated.** This will effectively allow us to eliminate the x-variable when adding the modified equations together.