To solve the system of equations using the elimination method, we start with the two equations:
- \( 3x - 4y = -4 \) (Equation 1)
- \( x + 3y = -10 \) (Equation 2)
The goal of elimination is to combine these equations to eliminate one of the variables. We can start by multiplying Equation 2 by 3 to align the coefficient of \( x \) in both equations.
Multiply Equation 2 by 3:
\[ 3(x + 3y) = 3(-10) \]
This simplifies to:
\[ 3x + 9y = -30 \quad \text{(Equation 3)} \]
Now we have:
- \( 3x - 4y = -4 \) (Equation 1)
- \( 3x + 9y = -30 \) (Equation 3)
Next, we can subtract Equation 1 from Equation 3 to eliminate \( 3x \):
\[ (3x + 9y) - (3x - 4y) = -30 - (-4) \]
This simplifies to:
\[ 9y + 4y = -30 + 4 \] \[ 13y = -26 \]
Now, divide both sides by 13 to solve for \( y \):
\[ y = -2 \]
Now that we have \( y \), we can substitute it back into either Equation 2 or Equation 1 to solve for \( x \). We'll use Equation 2 for substitution:
\[ x + 3(-2) = -10 \] \[ x - 6 = -10 \]
Now, add 6 to both sides:
\[ x = -10 + 6 \] \[ x = -4 \]
Thus, the solution for the system of equations is:
\[ x = -4, \quad y = -2 \]
The ordered pair representing the solution to the system is:
\[ \boxed{(-4, -2)} \]