To solve the given system of equations by elimination, follow these steps:
Given system of equations:
[tex]\[
\begin{cases}
3a + 6b = 12 \quad \text{(Equation 1)} \\
-3a + 6b = -12 \quad \text{(Equation 2)}
\end{cases}
\][/tex]
### Step 1: Eliminate variable [tex]\(a\)[/tex]
Add Equation 1 and Equation 2 to eliminate [tex]\(a\)[/tex]:
[tex]\[
(3a + 6b) + (-3a + 6b) = 12 + (-12)
\][/tex]
Combine like terms:
[tex]\[
3a - 3a + 6b + 6b = 0
\][/tex]
Simplify:
[tex]\[
0a + 12b = 0 \quad \Rightarrow \quad 12b = 0 \quad \text{(Resulting Equation 1)}
\][/tex]
### Step 2: Eliminate variable [tex]\(b\)[/tex]
Subtract Equation 2 from Equation 1 to eliminate [tex]\(b\)[/tex]:
[tex]\[
(3a + 6b) - (-3a + 6b) = 12 - (-12)
\][/tex]
Combine like terms:
[tex]\[
3a + 3a + 6b - 6b = 24
\][/tex]
Simplify:
[tex]\[
6a + 0b = 24 \quad \Rightarrow \quad 6a = 24 \quad \Rightarrow \quad a = 4 \quad \text{(However, we will keep it as \(6a = 24\) for consistency)}
\][/tex]
### Conclusion:
The possible resulting equations from eliminating one of the variables can be:
1. [tex]\(12b = 0\)[/tex]
2. [tex]\(6a = 0\)[/tex]
From the given options, the correct resulting equations are:
[tex]\[
\boxed{12b = 0 \quad \text{and} \quad 6a = 0}
\][/tex]
