To solve the system of equations:
- \( 4x - 6y = -48 \) (Equation 1)
- \( -2x + 4y = 30 \) (Equation 2)
we can use the elimination method. We need to make the coefficients of \(x\) or \(y\) equal in both equations.
Let's manipulate Equation 2 to help eliminate \(x\). First, we can multiply Equation 2 by 2 to make the coefficient of \(x\) equivalent to that in Equation 1:
\[ 2(-2x + 4y) = 2(30) \]
This gives us:
\[ -4x + 8y = 60 \quad \text{(Equation 3)} \]
Now we have the modified system of equations:
- \( 4x - 6y = -48 \) (Equation 1)
- \( -4x + 8y = 60 \) (Equation 3)
Next, we can add both equations to eliminate \(x\):
\[ (4x - 6y) + (-4x + 8y) = -48 + 60 \]
This simplifies to:
\[ 0x + 2y = 12 \]
So:
\[ 2y = 12 \]
Now, divide both sides by 2:
\[ y = 6 \]
Next, we substitute \(y = 6\) back into one of the original equations to find \(x\). Let's use Equation 1:
\[ 4x - 6(6) = -48 \]
This simplifies to:
\[ 4x - 36 = -48 \]
Now, solve for \(x\):
\[ 4x = -48 + 36 \] \[ 4x = -12 \] \[ x = -3 \]
The solution to the system of equations is:
\[ (x, y) = (-3, 6) \]
Therefore, the answer is:
O (-3, 6)