Let the relations $R_1$ and $R_2$ on the set $X=\{1,2,3, \ldots, 20\}$ be given by $R_1=\{(x, y): 2 x-3 y=2\}$ and $R_2=\{(x, y):-5 x+4 y=0\}$. 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

<p>$$\begin{aligned} & R_1=\{(x, y): 2 x-3 y=2\} \\ & R_2=\{(x, y):-5 x+4 y=0\} \\ & 2 x-3 y=2 \end{aligned}$$</p> <p>So $2 x$ and $3 y$ both has to be even or odd simultaneously and $2 x$ can't be odd so $2 x$ and $3 y$ both will be even</p> <p>$R_1=\{(4,2),(7,4),(10,6),(13,8),(16,10),(19,12)\}$</p> <p>For symmetric we need to add 6 elements as</p> <p>$$\begin{aligned} & (2,4),(4,7),(6,10),(8,13),(10,16),(12,19) \\ & M=6 \end{aligned}$$</p> <p>For $R_2-5 x+4 y=0$</p> <p>$5 x$ and $4 y$ has to be equal $4 y$ is always even so $5 x$ will also be even</p> <p>$R_2=\{(4,5),(8,10),(12,15),(16,20)\}$</p> <p>For symmetric we need to add 4 element as</p> <p>$$\begin{aligned} & (5,4)(10,8)(15,12)(20,16) \\ & N=4 \\ & M+N=6+4=10 \end{aligned}$$</p>