Let the relations `R_1` and `R_2` on the set `X = {1, 2, 3, …….20}` be given by `R_1 = {(x, y): 2x - 3y = 2}` and `R_2 = {(x, y): -5x + 4y = 0}`. If M and N be the minimum number of elements required to be added in `R_1` and `R_2`, respectively, in order to make the relations symmetric, then `M + N` equals

Options

A. 12
B. 8
C. 10
D. 16

Answer

To solve the problem, we need to analyze the relations \( R_1 \) and \( R_2 \) defined on the set \( X = \{1, 2, 3, \ldots, 20\} \).

### Step 1: Analyze Relation \( R_1 \)

The relation \( R_1 \) is defined by the equation:
\[ 2x - 3y = 2 \]
We can rearrange this to find \( x \) in terms of \( y \):
\[ 2x = 2 + 3y \implies x = 1 + \frac{3}{2}y \]
Since \( x \) and \( y \) must be integers, \( \frac{3}{2}y \) must also be an integer. This means \( y \) must be even. Let \( y = 2k \) where \( k \) is an integer. Then:
\[ x = 1 + 3k \]
Now we will find the values of \( k \) such that both \( x \) and \( y \) are in the set \( X \).

### Step 2: Determine Possible Values for \( y \)

Since \( y \) must be even and in the range from 1 to 20, the possible values of \( y \) are:
\[ 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 \]
Now we can find the corresponding \( x \) values for each \( y \):

- For \( y = 2 \): \( x = 1 + 3(1) = 4 \) → \( (4, 2) \)
- For \( y = 4 \): \( x = 1 + 3(2) = 7 \) → \( (7, 4) \)
- For \( y = 6 \): \( x = 1 + 3(3) = 10 \) → \( (10, 6) \)
- For \( y = 8 \): \( x = 1 + 3(4) = 13 \) → \( (13, 8) \)
- For \( y = 10 \): \( x = 1 + 3(5) = 16 \) → \( (16, 10) \)
- For \( y = 12 \): \( x = 1 + 3(6) = 19 \) → \( (19, 12) \)

Thus, the pairs in \( R_1 \) are:
\[ R_1 = \{(4, 2), (7, 4), (10, 6), (13, 8), (16, 10), (19, 12)\} \]

### Step 3: Make \( R_1 \) Symmetric

To make \( R_1 \) symmetric, we need to add pairs \( (y, x) \) for each \( (x, y) \) in \( R_1 \):
- Add \( (2, 4) \)
- Add \( (4, 7) \)
- Add \( (6, 10) \)
- Add \( (8, 13) \)
- Add \( (10, 16) \)
- Add \( (12, 19) \)

Thus, we need to add 6 pairs to make \( R_1 \) symmetric:
\[ M = 6 \]

### Step 4: Analyze Relation \( R_2 \)

The relation \( R_2 \) is defined by:
\[ -5x + 4y = 0 \implies 4y = 5x \implies y = \frac{5}{4}x \]
For \( y \) to be an integer, \( x \) must be a multiple of 4. The possible values of \( x \) in the set \( X \) are:
\[ 4, 8, 12, 16, 20 \]
Calculating \( y \) for these values:
- For \( x = 4 \): \( y = 5 \)
- For \( x = 8 \): \( y = 10 \)
- For \( x = 12 \): \( y = 15 \)
- For \( x = 16 \): \( y = 20 \)

Thus, the pairs in \( R_2 \) are:
\[ R_2 = \{(4, 5), (8, 10), (12, 15), (16, 20)\} \]

### Step 5: Make \( R_2 \) Symmetric

To make \( R_2 \) symmetric, we need to add pairs \( (y, x) \):
- Add \( (5, 4) \)
- Add \( (10, 8) \)
- Add \( (15, 12) \)
- Add \( (20, 16) \)

Thus, we need to add 4 pairs to make \( R_2 \) symmetric:
\[ N = 4 \]

### Step 6: Calculate \( M + N \)

Finally, we find:
\[ M + N = 6 + 4 = 10 \]

### Final Answer:
\[ \boxed{10} \]