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Given the equation $2x + y = 41$, where $x, y \in N$ (natural numbers). We can express $y$ in terms of $x$ as follows: $y = 41 - 2x$
Since $x$ and $y$ are natural numbers, $x > 0$ and $y > 0$. $y = 41 - 2x > 0$ $2x < 41$ $x < \frac{41}{2}$ $x < 20.5$ Since $x$ is a natural number, the possible values of $x$ are $1, 2, 3, ..., 20$. Therefore, the domain of the relation is $\{1, 2, 3, ..., 20\}$.
For each value of $x$ in the domain, we can find the corresponding value of $y$: When $x = 1$, $y = 41 - 2(1) = 39$ When $x = 2$, $y = 41 - 2(2) = 37$ When $x = 3$, $y = 41 - 2(3) = 35$ ... When $x = 20$, $y = 41 - 2(20) = 1$ The range of the relation is $\{1, 3, 5, ..., 39\}$.
For the relation to be reflexive, $(x, x)$ must be in the relation for all $x$ in the domain. This means $2x + x = 41$, or $3x = 41$, which gives $x = \frac{41}{3}$, which is not a natural number. Therefore, the relation is not reflexive.
For the relation to be symmetric, if $(x, y)$ is in the relation, then $(y, x)$ must also be in the relation. If $(x, y)$ is in the relation, then $2x + y = 41$. If $(y, x)$ is in the relation, then $2y + x = 41$. Let's take an example: $x = 1$, $y = 39$. Then $2(1) + 39 = 41$. Now, check if $2(39) + 1 = 41$. $2(39) + 1 = 78 + 1 = 79 \neq 41$. Therefore, the relation is not symmetric.
For the relation to be transitive, if $(x, y)$ and $(y, z)$ are in the relation, then $(x, z)$ must also be in the relation. If $(x, y)$ is in the relation, then $2x + y = 41$. If $(y, z)$ is in the relation, then $2y + z = 41$. We need to check if $2x + z = 41$. From the first equation, $y = 41 - 2x$. Substituting this into the second equation, $2(41 - 2x) + z = 41$, which simplifies to $82 - 4x + z = 41$, so $z = 41 + 4x - 82 = 4x - 41$. Now, we need to check if $2x + z = 41$. $2x + (4x - 41) = 41$ $6x - 41 = 41$ $6x = 82$ $x = \frac{82}{6} = \frac{41}{3}$, which is not a natural number. Let's take an example: Let $x = 1$, then $y = 41 - 2(1) = 39$. So, $(1, 39)$ is in the relation. Let $y = 39$, then $2(39) + z = 41$, so $78 + z = 41$, which means $z = 41 - 78 = -37$. But $z$ must be a natural number, so this case is not valid. Let's try another approach. Suppose $(x, y)$ and $(y, z)$ are in the relation. Then $2x + y = 41$ and $2y + z = 41$. We want to see if $2x + z = 41$. From the first equation, $y = 41 - 2x$. Substituting into the second equation, $2(41 - 2x) + z = 41$, so $82 - 4x + z = 41$, which gives $z = 4x - 41$. Now we check if $(x, z)$ is in the relation, i.e., if $2x + z = 41$. $2x + (4x - 41) = 6x - 41$. We want to see if $6x - 41 = 41$, which means $6x = 82$, so $x = \frac{41}{3}$, which is not a natural number. Therefore, the relation is not transitive.
Since the relation is not reflexive, symmetric, or transitive, it is not an equivalence relation.
Final Answer: Domain: {1, 2, 3, ..., 20}, Range: {1, 3, 5, ..., 39}. The relation is not reflexive, symmetric, or transitive. Therefore, it is not an equivalence relation.
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