To solve the system of equations:
- \( 2x + 6y = -6 \) (Equation 1)
- \( 4x - 3y = -12 \) (Equation 2)
We can use either the substitution method or the elimination method. Here, I will use the elimination method.
First, let's simplify Equation 1 by dividing all terms by 2:
\[ x + 3y = -3 \quad \text{(Equation 1 simplified)} \]
Now we have:
- \( x + 3y = -3 \) (Equation 3)
- \( 4x - 3y = -12 \) (Equation 2)
Next, let's add Equation 3 and Equation 2 to eliminate \( y \). First, we can multiply Equation 3 by 3 so that the coefficients of \( y \) will cancel out:
\[ 3(x + 3y) = 3(-3) \] \[ 3x + 9y = -9 \quad \text{(Equation 4)} \]
Now we can write the two equations we have:
- \( 3x + 9y = -9 \) (Equation 4)
- \( 4x - 3y = -12 \) (Equation 2)
Now let's eliminate \( y \) by multiplying Equation 2 by 3:
\[ 3(4x - 3y) = 3(-12) \] \[ 12x - 9y = -36 \quad \text{(Equation 5)} \]
Now we have:
- \( 3x + 9y = -9 \) (Equation 4)
- \( 12x - 9y = -36 \) (Equation 5)
Next, we can add Equation 4 and Equation 5:
\[ (3x + 9y) + (12x - 9y) = -9 + (-36) \]
This simplifies to:
\[ 15x = -45 \]
Now, divide both sides by 15:
\[ x = -3 \]
Now that we have \( x \), we can substitute it back into Equation 3 to solve for \( y \):
\[ -3 + 3y = -3 \] \[ 3y = -3 + 3 \] \[ 3y = 0 \] \[ y = 0 \]
Thus, the solution to the system of equations is:
\[ (x, y) = (-3, 0) \]
So, the final answer is:
\[ \boxed{(-3, 0)} \]