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Let the relations $R_1$ and $R_2$ on the set $\mathrm{X}=\{1,2,3, \ldots, 20\}$ be given by $\mathrm{R}_1=\{(\mathrm{x}, \mathrm{y}): 2 \mathrm{x}-3 \mathrm{y}=2\}$ and $\mathrm{R}_2=\{(\mathrm{x}, \mathrm{y}):-5 \mathrm{x}+4 \mathrm{y}=0\}$. If $\mathrm{M}$ and $\mathrm{N}$ 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 $\mathrm{M}+\mathrm{N}$ equals
$8$
$16$
$12$
$10$
Solution
$ \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\}$
$ \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)\}$
in $\mathrm{R}_1 6$ element needed
in $\mathrm{R}_2 4$ element needed
So, total $6+4=10$ element
