If `R = {(x,y):x, y in N, x+2y=8}` then range of R is

Options

A. {1, 2, 3, ... 7,8}
B. {1, 2, 3, ... 9, 10}
C. {1,3,5,7 ... 19}
D. {1,3,5,7, ... 15}

Answer

To find the range of the relation \( R = \{(x,y) : x, y \in \mathbb{N}, x + 2y = 8\} \), we will follow these steps:

### Step 1: Understand the equation
The equation given is \( x + 2y = 8 \). Here, \( x \) and \( y \) are natural numbers (i.e., positive integers).

### Step 2: Rearranging the equation
We can express \( y \) in terms of \( x \):
\[ y = \frac{8 - x}{2} \]

### Step 3: Determine valid values for \( x \)
Since \( y \) must be a natural number, \( 8 - x \) must be a positive even number. This means that \( 8 - x \) must be greater than 0 and even.

### Step 4: Find possible values for \( x \)
To ensure \( y \) is a natural number, we can check values of \( x \) from 1 to 8 (since \( x \) must be less than or equal to 8):
- For \( x = 1 \): \( y = \frac{8 - 1}{2} = \frac{7}{2} \) (not a natural number)
- For \( x = 2 \): \( y = \frac{8 - 2}{2} = 3 \) (valid)
- For \( x = 3 \): \( y = \frac{8 - 3}{2} = \frac{5}{2} \) (not a natural number)
- For \( x = 4 \): \( y = \frac{8 - 4}{2} = 2 \) (valid)
- For \( x = 5 \): \( y = \frac{8 - 5}{2} = \frac{3}{2} \) (not a natural number)
- For \( x = 6 \): \( y = \frac{8 - 6}{2} = 1 \) (valid)
- For \( x = 7 \): \( y = \frac{8 - 7}{2} = \frac{1}{2} \) (not a natural number)
- For \( x = 8 \): \( y = \frac{8 - 8}{2} = 0 \) (not a natural number)

### Step 5: Compile valid pairs
From the calculations above, the valid pairs \((x, y)\) are:
- \( (2, 3) \)
- \( (4, 2) \)
- \( (6, 1) \)

### Step 6: Determine the range of \( R \)
The range of \( R \) consists of the second elements \( y \) from the valid pairs:
- From \( (2, 3) \), we get \( y = 3 \)
- From \( (4, 2) \), we get \( y = 2 \)
- From \( (6, 1) \), we get \( y = 1 \)

Thus, the range of \( R \) is:
\[ \{1, 2, 3\} \]

### Final Answer
The range of \( R \) is \( \{1, 2, 3\} \).

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