To solve the system of equations using the elimination method, we start with the two equations:
- \( 4x - 6y = -48 \)
- \( -2x + 4y = 30 \)
We want to eliminate one of the variables. We can do this by making the coefficients of \( x \) or \( y \) the same in both equations.
Let's eliminate \( x \). We will multiply the second equation by \( 2 \) to make the coefficient of \( x \) in the second equation become \( -4 \):
Multiply equation (2) by \( 2 \):
\[ -2x + 4y = 30 \implies -4x + 8y = 60 \]
Now our system of equations is:
- \( 4x - 6y = -48 \)
- \( -4x + 8y = 60 \)
Next, we add these two equations together to eliminate \( x \):
\[ (4x - 6y) + (-4x + 8y) = -48 + 60 \]
The \( 4x \) and \( -4x \) cancel out:
\[ -6y + 8y = 12 \]
Simplifying this gives:
\[ 2y = 12 \]
Now, we can solve for \( y \):
\[ y = \frac{12}{2} = 6 \]
Now that we have \( y \), we can plug this value back into one of the original equations to find \( x \). We can use equation (1):
\[ 4x - 6(6) = -48 \]
This simplifies to:
\[ 4x - 36 = -48 \]
Adding \( 36 \) to both sides gives:
\[ 4x = -48 + 36 \] \[ 4x = -12 \]
Now, divide by \( 4 \):
\[ x = \frac{-12}{4} = -3 \]
Thus, the solution to the system of equations is:
\[ x = -3, \quad y = 6 \]
So, the solution in ordered pair form is:
\[ \boxed{(-3, 6)} \]