Total number of faces in a truncated octahedron=x
Total number of faces in a truncated tetrahedron=y
Then x+y is :

Explanation

To solve the problem, we need to find the total number of faces in a truncated octahedron (denoted as \( x \)) and a truncated tetrahedron (denoted as \( y \)), and then calculate \( x + y \).

### Step-by-Step Solution:

1. **Identify the Faces of a Truncated Octahedron:**
- A truncated octahedron is an Archimedean solid that has two types of faces: hexagonal and square.
- Specifically, it has:
- **8 hexagonal faces**
- **6 square faces**
- Therefore, the total number of faces \( x \) is:
\[ x = 8 + 6 = 14 \]

2. **Identify the Faces of a Truncated Tetrahedron:**
- A truncated tetrahedron is another Archimedean solid that has two types of faces: hexagonal and triangular.
- Specifically, it has:
- **4 hexagonal faces**
- **4 triangular faces**
- Therefore, the total number of faces \( y \) is:
\[ y = 4 + 4 = 8 \]

3. **Calculate \( x + y \):**
- Now that we have both values, we can calculate:
\[ x + y = 14 + 8 = 22 \]

### Final Answer:
Thus, the value of \( x + y \) is \( \boxed{22} \).

---