To solve the system of equations:

1. **First equation:**
\[
\frac{2}{2x - y} + \frac{3}{x - 2y} = \frac{1}{2}
\]

2. **Second equation:**
\[
\frac{2}{2x - y} - \frac{1}{x - 2y} = \frac{1}{18}
\]

**Step 1: Substitutions**

Let:
\[
a = 2x - y
\]
\[
b = x - 2y
\]

Then the equations can be rewritten as:

1. \[
\frac{2}{a} + \frac{3}{b} = \frac{1}{2}
\]

2. \[
\frac{2}{a} - \frac{1}{b} = \frac{1}{18}
\]

**Step 2: Solve for \( \frac{2}{a} \) and \( \frac{1}{b} \)**

From the first equation:
\[
\frac{2}{a} + \frac{3}{b} = \frac{1}{2} \quad \Rightarrow \quad \frac{2}{a} = \frac{1}{2} - \frac{3}{b}
\]

From the second equation:
\[
\frac{2}{a} - \frac{1}{b} = \frac{1}{18} \quad \Rightarrow \quad \frac{2}{a} = \frac{1}{18} + \frac{1}{b}
\]

**Step 3: Set up the equations for \( \frac{2}{a} \)**

We now have two expressions for \( \frac{2}{a} \):
1. \(\frac{2}{a} = \frac{1}{2} - \frac{3}{b}\)
2. \(\frac{2}{a} = \frac{1}{18} + \frac{1}{b}\)

Setting these equal gives:
\[
\frac{1}{2} - \frac{3}{b} = \frac{1}{18} + \frac{1}{b}
\]

**Step 4: Solve for \( b \)**

Multiply through by \( 18b \) to eliminate denominators:
\[
9b - 54 = b + 18
\]
\[
8b = 72 \quad \Rightarrow \quad b = 9
\]

**Step 5: Substitute \( b \) back to find \( a \)**

Now substitute \( b = 9 \) back into one of the equations for \( \frac{2}{a} \):
\[
\frac{2}{a} = \frac{1}{2} - \frac{3}{9}
\]
\[
\frac{2}{a} = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}
\]
Thus,
\[
a = 12
\]

**Step 6: Translate back to \( x \) and \( y \)**

Now we have:
\[
a = 2x - y = 12 \quad (1)
\]
\[
b = x - 2y = 9 \quad (2)
\]

Substituting (1) into (2):
From (1), we get:
\[
y = 2x - 12
\]

Substituting \( y \) into (2):
\[
x - 2(2x - 12) = 9
\]
\[
x - 4x + 24 = 9
\]
\[
-3x + 24 = 9 \quad \Rightarrow \quad -3x = -15 \quad \Rightarrow \quad x = 5
\]

Now substitute back to find \( y \):
\[
y = 2(5) - 12 = 10 - 12 = -2
\]

Thus, the solution is:
\[
\boxed{(5, -2)}
\]