Let R be a relation defined as `xRy` if and only if `2x+3y = 20`, where `x, y in N`. How many elements of the form `(x, y)` are there in R?

Options

A. 2
B. 3
C. 4
D. 6

Answer

To solve the problem, we need to determine how many pairs \((x, y)\) of natural numbers satisfy the equation \(2x + 3y = 20\).

### Step-by-step Solution:

1. **Start with the equation**:
\[ 2x + 3y = 20 \]

2. **Rearrange the equation to express \(y\) in terms of \(x\)**:
\[ 3y = 20 - 2x \]
\[ y = \frac{20 - 2x}{3} \]

3. **Determine the conditions for \(y\) to be a natural number**:
- For \(y\) to be a natural number, \(20 - 2x\) must be a positive integer and divisible by 3.
- This means \(20 - 2x > 0\) or \(20 > 2x\), which simplifies to \(x < 10\).

4. **Check integer values of \(x\)**:
- Since \(x\) must be a natural number, we can check values of \(x\) from 1 to 9.

5. **Calculate \(y\) for each \(x\)**:
- For \(x = 1\):
\[ y = \frac{20 - 2(1)}{3} = \frac{18}{3} = 6 \quad \text{(valid)} \]
- For \(x = 2\):
\[ y = \frac{20 - 2(2)}{3} = \frac{16}{3} \quad \text{(not valid)} \]
- For \(x = 3\):
\[ y = \frac{20 - 2(3)}{3} = \frac{14}{3} \quad \text{(not valid)} \]
- For \(x = 4\):
\[ y = \frac{20 - 2(4)}{3} = \frac{12}{3} = 4 \quad \text{(valid)} \]
- For \(x = 5\):
\[ y = \frac{20 - 2(5)}{3} = \frac{10}{3} \quad \text{(not valid)} \]
- For \(x = 6\):
\[ y = \frac{20 - 2(6)}{3} = \frac{8}{3} \quad \text{(not valid)} \]
- For \(x = 7\):
\[ y = \frac{20 - 2(7)}{3} = \frac{6}{3} = 2 \quad \text{(valid)} \]
- For \(x = 8\):
\[ y = \frac{20 - 2(8)}{3} = \frac{4}{3} \quad \text{(not valid)} \]
- For \(x = 9\):
\[ y = \frac{20 - 2(9)}{3} = \frac{2}{3} \quad \text{(not valid)} \]

6. **List valid pairs**:
- The valid pairs \((x, y)\) are:
- \((1, 6)\)
- \((4, 4)\)
- \((7, 2)\)

7. **Count the valid pairs**:
- There are **3 valid pairs**.

### Final Answer:
The number of elements of the form \((x, y)\) in the relation \(R\) is **3**.