The relation R is defined on the set N by R = | vedclass

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(D) Given the relation $R = \{(x, y) | x, y \in N, 2x + y = 41\}$.For $R$ to be reflexive,$(x, x) \in R$ for all $x \in N$. This implies $2x + x = 41$

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None of these

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(D) Given the relation $R = \{(x, y) | x, y \in N, 2x + y = 41\}$.For $R$ to be reflexive,$(x, x) \in R$ for all $x \in N$. This implies $2x + x = 41$,so $3x = 41$,which gives $x = 41/3 
otin N$. Thus,$R$ is not reflexive.For $R$ to be symmetric,if $(x, y) \in R$,then $(y, x) \in R$. Let $(1, 39) \in R$ because $2(1) + 39 = 41$. However,$(39, 1) 
otin R$ because $2(39) + 1 = 79 
eq 41$. Thus,$R$ is not symmetric.For $R$ to be transitive,if $(x, y) \in R$ and $(y, z) \in R$,then $(x, z) \in R$. Let $(1, 39) \in R$ and $(39, z) \in R$. For $(39, z) \in R$,$2(39) + z = 41$,which gives $z = 41 - 78 = -37 
otin N$. Since there is no such $z \in N$,the condition for transitivity fails. Thus,$R$ is not transitive.Therefore,$R$ is none of these.

The relation $R$ is defined on the set $N$ by $R = \{(x, y) | x, y \in N, 2x + y = 41\}$. Then $R$ is:

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