To solve the problem, we need to find the relation \( R \) defined on the set \( A = \{1, 2, 3, 4, 5, 6, 7, 8\} \) such that \( R = \{(x, y) : x \in A, y \in A \text{ and } x + 2y = 10\} \). We will then determine the domain and range of \( R \) and its inverse \( R^{-1} \).
### Step 1: Find Ordered Pairs in Relation \( R \)
We start by solving the equation \( x + 2y = 10 \) for \( y \):
\[
2y = 10 - x \implies y = \frac{10 - x}{2}
\]
Next, we will substitute values of \( x \) from the set \( A \) and check if \( y \) is also in \( A \).
1. **For \( x = 1 \)**:
\[
y = \frac{10 - 1}{2} = \frac{9}{2} = 4.5 \quad (\text{not in } A)
\]
2. **For \( x = 2 \)**:
\[
y = \frac{10 - 2}{2} = \frac{8}{2} = 4 \quad (4 \in A) \implies (2, 4)
\]
3. **For \( x = 3 \)**:
\[
y = \frac{10 - 3}{2} = \frac{7}{2} = 3.5 \quad (\text{not in } A)
\]
4. **For \( x = 4 \)**:
\[
y = \frac{10 - 4}{2} = \frac{6}{2} = 3 \quad (3 \in A) \implies (4, 3)
\]
5. **For \( x = 5 \)**:
\[
y = \frac{10 - 5}{2} = \frac{5}{2} = 2.5 \quad (\text{not in } A)
\]
6. **For \( x = 6 \)**:
\[
y = \frac{10 - 6}{2} = \frac{4}{2} = 2 \quad (2 \in A) \implies (6, 2)
\]
7. **For \( x = 7 \)**:
\[
y = \frac{10 - 7}{2} = \frac{3}{2} = 1.5 \quad (\text{not in } A)
\]
8. **For \( x = 8 \)**:
\[
y = \frac{10 - 8}{2} = \frac{2}{2} = 1 \quad (1 \in A) \implies (8, 1)
\]
Now we can compile the ordered pairs:
\[
R = \{(2, 4), (4, 3), (6, 2), (8, 1)\}
\]
### Step 2: Find the Domain and Range of \( R \)
- **Domain of \( R \)**: The set of all first elements (x-values) in the ordered pairs.
\[
\text{Domain} = \{2, 4, 6, 8\}
\]
- **Range of \( R \)**: The set of all second elements (y-values) in the ordered pairs.
\[
\text{Range} = \{4, 3, 2, 1\}
\]
### Step 3: Find the Inverse Relation \( R^{-1} \)
To find \( R^{-1} \), we swap the elements in each ordered pair of \( R \):
\[
R^{-1} = \{(4, 2), (3, 4), (2, 6), (1, 8)\}
\]
### Step 4: Find the Domain and Range of \( R^{-1} \)
- **Domain of \( R^{-1} \)**: The set of all first elements (x-values) in the ordered pairs of \( R^{-1} \).
\[
\text{Domain of } R^{-1} = \{4, 3, 2, 1\}
\]
- **Range of \( R^{-1} \)**: The set of all second elements (y-values) in the ordered pairs of \( R^{-1} \).
\[
\text{Range of } R^{-1} = \{2, 4, 6, 8\}
\]
### Final Answers
- \( R = \{(2, 4), (4, 3), (6, 2), (8, 1)\} \)
- Domain of \( R = \{2, 4, 6, 8\} \)
- Range of \( R = \{4, 3, 2, 1\} \)
- \( R^{-1} = \{(4, 2), (3, 4), (2, 6), (1, 8)\} \)
- Domain of \( R^{-1} = \{4, 3, 2, 1\} \)
- Range of \( R^{-1} = \{2, 4, 6, 8\} \)
