To solve the system of equations using elimination, we start with the given equations:
- \( 4x - 6y = -48 \)
- \( -2x + 4y = 30 \)
To eliminate one of the variables, we can manipulate the equations. Let's eliminate \( x \) by making the coefficients of \( x \) equal.
First, we can multiply the second equation by 2 to match the coefficient of \( x \) in the first equation:
\[ 2(-2x + 4y) = 2(30) \]
This gives us:
\[ -4x + 8y = 60 \]
Now we have the system:
- \( 4x - 6y = -48 \) (equation 1)
- \( -4x + 8y = 60 \) (equation 2)
Next, we will add the two equations together to eliminate \( x \):
\[ (4x - 6y) + (-4x + 8y) = -48 + 60 \]
This simplifies to:
\[ 0x + 2y = 12 \]
Which simplifies further to:
\[ 2y = 12 \]
Now, divide both sides by 2:
\[ y = 6 \]
Now that we have \( y \), we can substitute \( y \) back into one of the original equations to solve for \( x \). We will use the second equation:
\[ -2x + 4(6) = 30 \]
This simplifies to:
\[ -2x + 24 = 30 \]
Subtract 24 from both sides:
\[ -2x = 30 - 24 \]
\[ -2x = 6 \]
Now divide by -2:
\[ x = -3 \]
The solution to the system of equations is:
\[ \boxed{(-3, 6)} \]