To solve the system of equations using elimination, we start with the two equations given:
- \(-3x + 6y = 15\)
- \(x + 3y = -15\)
To use the elimination method, we can multiply the second equation by 3 to eliminate \(x\):
\[ 3(x + 3y) = 3(-15) \]
This simplifies to:
\[ 3x + 9y = -45 \]
Now, we rewrite our system with this new equation:
- \(-3x + 6y = 15\)
- \(3x + 9y = -45\)
Next, we add both equations together:
\[ (-3x + 6y) + (3x + 9y) = 15 - 45 \]
When we combine like terms, we have:
\[ 0x + 15y = -30 \]
This simplifies to:
\[ 15y = -30 \]
Now, we solve for \(y\):
\[ y = \frac{-30}{15} = -2 \]
Next, we substitute \(y = -2\) back into one of the original equations to find \(x\). We can use the second equation:
\[ x + 3(-2) = -15 \]
This simplifies to:
\[ x - 6 = -15 \]
Adding 6 to both sides gives:
\[ x = -15 + 6 = -9 \]
We have found the solution:
\[ x = -9, \quad y = -2 \]
Thus, the solution to the given system of equations is:
\[ \boxed{(-9, -2)} \]
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