To solve the given system of equations using elimination, we have:
- \(-6x + 10y = -16\)
- \(6x - 3y = 12\)
First, we can add the two equations together to eliminate \(x\):
\[ -6x + 10y + 6x - 3y = -16 + 12 \]
This simplifies to:
\[ (10y - 3y) = -16 + 12 \]
\[ 7y = -4 \]
Now, solve for \(y\):
\[ y = \frac{-4}{7} \]
Now that we have \(y\), we can substitute this value back into one of the original equations to solve for \(x\). We'll use the second equation:
\[ 6x - 3y = 12 \]
Substituting \(y = \frac{-4}{7}\):
\[ 6x - 3\left(\frac{-4}{7}\right) = 12 \]
\[ 6x + \frac{12}{7} = 12 \]
Now, subtract \(\frac{12}{7}\) from both sides:
\[ 6x = 12 - \frac{12}{7} \]
To perform this subtraction, convert \(12\) into a fraction with a common denominator:
\[ 12 = \frac{84}{7} \]
Now, we can substitute this back in:
\[ 6x = \frac{84}{7} - \frac{12}{7} \]
\[ 6x = \frac{84 - 12}{7} \]
\[ 6x = \frac{72}{7} \]
Now, divide by \(6\) to solve for \(x\):
\[ x = \frac{72}{7} \cdot \frac{1}{6} = \frac{72}{42} = \frac{12}{7} \]
So, we have:
\[ x = \frac{12}{7} \quad \text{and} \quad y = -\frac{4}{7} \]
Finally, the solution to the system of equations is:
\[ \left(\frac{12}{7}, -\frac{4}{7}\right) \]
In terms of the options provided:
\[ \text{The solution is } \left(12/7, -4/7\right) \]