To solve the system of equations:
\[ \begin{aligned} -2x - 7y &= 30 \quad \text{(1)} \ 7x + 4y &= 18 \quad \text{(2)} \end{aligned} \]
we will use the substitution or elimination method. Here, I will use the elimination method.
First, let's manipulate the equations to eliminate \(y\). We can multiply equation (1) by 4 and equation (2) by 7 to align the coefficients of \(y\):
\[ \begin{aligned} 4(-2x - 7y) &= 4(30) \ -8x - 28y &= 120 \quad \text{(3)} \end{aligned} \]
And for equation (2):
\[ \begin{aligned} 7(7x + 4y) &= 7(18) \ 49x + 28y &= 126 \quad \text{(4)} \end{aligned} \]
Now we have a new system of equations:
\[ \begin{aligned} -8x - 28y &= 120 \quad \text{(3)} \ 49x + 28y &= 126 \quad \text{(4)} \end{aligned} \]
Next, we can add equations (3) and (4) together to eliminate \(y\):
\[ (-8x - 28y) + (49x + 28y) = 120 + 126 \]
This simplifies to:
\[ 41x = 246 \]
Now, divide both sides by 41:
\[ x = \frac{246}{41} = 6 \]
Now that we have \(x\), we can substitute it back into one of the original equations to find \(y\). We can use equation (2):
\[ 7x + 4y = 18 \]
Substituting \(x = 6\):
\[ 7(6) + 4y = 18 \]
This simplifies to:
\[ 42 + 4y = 18 \]
Subtract 42 from both sides:
\[ 4y = 18 - 42 \] \[ 4y = -24 \]
Now divide by 4:
\[ y = \frac{-24}{4} = -6 \]
The solution of the system of equations is:
\[ x = 6, \quad y = -6 \]
Thus, the final answer is:
\[ \boxed{x = 6, \quad y = -6} \]