Let the relations R_1 and R_2 on the set X = | vedclass

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(D) Given the set $X = \{1, 2, 3, \ldots, 20\}$.For $R_1 = \{(x, y) : 2x - 3y = 2\}$,we find the ordered pairs $(x, y)$ such that $x, y \in X$:If $y =

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$10$

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(D) Given the set $X = \{1, 2, 3, \ldots, 20\}$.For $R_1 = \{(x, y) : 2x - 3y = 2\}$,we find the ordered pairs $(x, y)$ such that $x, y \in X$:If $y = 2, x = 4$; if $y = 4, x = 7$; if $y = 6, x = 10$; if $y = 8, x = 13$; if $y = 10, x = 16$; if $y = 12, x = 19$.So,$R_1 = \{(4, 2), (7, 4), (10, 6), (13, 8), (16, 10), (19, 12)\}$.Since there are $6$ elements and none are of the form $(a, a)$,to make $R_1$ symmetric,we need to add the reverse of each pair,which is $6$ elements.Thus,$M = 6$.For $R_2 = \{(x, y) : -5x + 4y = 0\}$,which implies $4y = 5x$ or $y = \frac{5}{4}x$,we find the ordered pairs $(x, y)$ such that $x, y \in X$:If $x = 4, y = 5$; if $x = 8, y = 10$; if $x = 12, y = 15$; if $x = 16, y = 20$.So,$R_2 = \{(4, 5), (8, 10), (12, 15), (16, 20)\}$.Since there are $4$ elements and none are of the form $(a, a)$,to make $R_2$ symmetric,we need to add the reverse of each pair,which is $4$ elements.Thus,$N = 4$.Therefore,$M + N = 6 + 4 = 10$.

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