Given relations on $X= \{1,2,3,\dots,20\}$: $R_1=\{(x,y):2x-3y=2\}, \qquad R_2=\{(x,y):-5x+4y=0\}$

We need the minimum number of ordered pairs to be added so that each relation becomes symmetric.

A relation is symmetric if whenever $(x,y)\in R$, then $(y,x)\in R$ also.

Find all pairs in $R_1$.

Equation: $2x-3y=2$ So, $x=\frac{3y+2}{2}$ For $x$ to be an integer, $y$ must be even.

Let $y=2k$. Then $x=\frac{6k+2}{2}=3k+1$

Since $x,y\in \{1,2,\dots,20\}$: $y=2k\le 20 \Rightarrow k\le 10$ $x=3k+1\le 20 \Rightarrow k\le 6$

Thus $k=1,2,3,4,5,6$.

Corresponding pairs are: $R_1=\{(4,2),(7,4),(10,6),(13,8),(16,10),(19,12)\}$

Check symmetry requirement for $R_1$.

For symmetry, we need the reverse pairs: $(2,4),(4,7),(6,10),(8,13),(10,16),(12,19)$

Check whether any of these are already in $R_1$.

Substitute into $2x-3y=2$:

• For $(2,4)$: $2(2)-3(4)=4-12=-8\ne 2$
• Similarly none of the reversed pairs satisfy the equation.

Also, no pair in $R_1$ is of the form $(a,a)$, since $2a-3a=2 \Rightarrow -a=2$ impossible in $X$.

Hence all 6 reverse pairs are missing.

Therefore, $M=6$

Find all pairs in $R_2$.

Equation: $-5x+4y=0 \Rightarrow 4y=5x \Rightarrow y=\frac{5x}{4}$

For $y$ to be an integer, $x$ must be a multiple of 4.

Let $x=4k$. Then $y=5k$

Since $x,y\in \{1,2,\dots,20\}$: $4k\le 20 \Rightarrow k\le 5$ $5k\le 20 \Rightarrow k\le 4$

Thus $k=1,2,3,4$.

So, $R_2=\{(4,5),(8,10),(12,15),(16,20)\}$

Check symmetry requirement for $R_2$.

Required reverse pairs are: $(5,4),(10,8),(15,12),(20,16)$

Check if any reverse pair already lies in $R_2$:

• For $(5,4)$: $-5(5)+4(4)=-25+16=-9\ne 0$
• Similarly none of the reversed pairs satisfy the equation.

Also, no diagonal pair $(a,a)$ is possible because $-5a+4a=0 \Rightarrow -a=0 \Rightarrow a=0$ but $0\notin X$.

Hence all 4 reverse pairs must be added.

Therefore, $N=4$

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

Hence, the answer agrees with the stored answer.