To solve the system of equations using elimination, we have the following two equations:
- \( 4x - 6y = -48 \) (Equation 1)
- \( -2x + 4y = 30 \) (Equation 2)
First, we can modify Equation 2 to align the coefficients of \(x\) when we eliminate the variable \(x\). We can multiply Equation 2 by 2 to make the coefficient of \(x\) match the coefficient of \(x\) in Equation 1.
Multiplying Equation 2 by 2:
\[ 2(-2x + 4y) = 2(30) \]
This gives us:
\[ -4x + 8y = 60 \quad \text{(Equation 3)} \]
Now we have:
- \( 4x - 6y = -48 \) (Equation 1)
- \( -4x + 8y = 60 \) (Equation 3)
Next, we can add Equation 1 and Equation 3 together to eliminate \(x\):
\[ (4x - 6y) + (-4x + 8y) = -48 + 60 \]
This simplifies to:
\[ 0x + 2y = 12 \]
Thus, we have:
\[ 2y = 12 \]
Dividing both sides by 2 gives:
\[ y = 6 \]
Now that we have the value of \(y\), we can substitute it back into one of the original equations to solve for \(x\). We can use Equation 1:
\[ 4x - 6(6) = -48 \]
Calculating this, we have:
\[ 4x - 36 = -48 \]
Adding 36 to both sides:
\[ 4x = -12 \]
Dividing both sides by 4 gives:
\[ x = -3 \]
Thus, the solution to the system of equations is:
\[ (x, y) = (-3, 6) \]
The answer is:
\[ \boxed{(-3, 6)} \]