To solve the systems of equations, let's address each pair step by step.

### a)
\[
\begin{cases}
2y + x = 0 \\
x + y = 18
\end{cases}
\]

1. From the first equation, solve for \(x\):
\[
x = -2y
\]
2. Substitute \(x\) in the second equation:
\[
-2y + y = 18 \implies -y = 18 \implies y = -18
\]
3. Substitute \(y\) back to find \(x\):
\[
x = -2(-18) = 36
\]
4. So the solution is:
\[
(x, y) = (36, -18)
\]

### b)
\[
\begin{cases}
8x + y = 1 \\
2x + 3y = 9
\end{cases}
\]

1. From the first equation, solve for \(y\):
\[
y = 1 - 8x
\]
2. Substitute \(y\) in the second equation:
\[
2x + 3(1 - 8x) = 9 \implies 2x + 3 - 24x = 9 \implies -22x + 3 = 9
\]
3. Rearranging gives:
\[
-22x = 6 \implies x = -\frac{3}{11}
\]
4. Substitute \(x\) back to find \(y\):
\[
y = 1 - 8(-\frac{3}{11}) = 1 + \frac{24}{11} = \frac{11}{11} + \frac{24}{11} = \frac{35}{11}
\]
5. So the solution is:
\[
\left(x, y\right) = \left(-\frac{3}{11}, \frac{35}{11}\right)
\]

### c)
\[
\begin{cases}
5x + 6y = 1 \\
x + y = -1
\end{cases}
\]

1. From the second equation, solve for \(y\):
\[
y = -1 - x
\]
2. Substitute \(y\) in the first equation:
\[
5x + 6(-1 - x) = 1 \implies 5x - 6 - 6x = 1 \implies -x - 6 = 1
\]
3. Rearranging gives:
\[
-x = 7 \implies x = -7
\]
4. Substitute \(x\) back to find \(y\):
\[
y = -1 - (-7) = 6
\]
5. So the solution is:
\[
(x, y) = (-7, 6)
\]

### Summary of Solutions
- **a)** \( (36, -18) \)
- **b)** \( \left(-\frac{3}{11}, \frac{35}{11}\right) \)
- **c)** \( (-7, 6) \)