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…
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
12
16
8
10
Solution
$\begin{aligned} & \mathrm{x}=\{1,2,3, \ldots \ldots . .20\} \\ & \mathrm{R}_1=\{(\mathrm{x}, \mathrm{y}): 2 \mathrm{x}-3 \mathrm{y}=2\} \\ & \mathrm{R}_2=\{(\mathrm{x}, \mathrm{y}):-5 \mathrm{x}+4 \mathrm{y}=0\}\end{aligned}$
$\begin{aligned}
& \mathrm{R}_1=\{(4,2),(7,4),(10,6),(13,8),(16,10),(19,12)\} \\
& \mathrm{R}_2=\{(4,5),(8,10),(12,15),(16,20)\}
\end{aligned}$
in $R_1 6$ element needed
in $R_2 4$ element needed
So, total $6+4=10$ element