Reflexive: For R to be reflexive, $aRa$ must be true for all $a \in N$. This means $2a + 3a = 5a$ must be a multiple of 5, which is always true. So, R is reflexive.

Symmetric: For R to be symmetric, if $aRb$, then $bRa$ must be true. If $2a + 3b$ is a multiple of 5, then $2a + 3b = 5k$ for some integer $k$. We need to check if $2b + 3a$ is also a multiple of 5. $2b + 3a = 2b + 3a + (5a + 5b) - (5a + 5b) = 5a + 5b + 2b + 3a - 5a - 5b = 5(a+b) - (2a + 3b) + 5a + 5b = 5(a+b) - 5k + 5a + 5b = 5(a+b-k)$. Since $2a + 3b = 5k$, $2b + 3a = 5l$ for some integer $l$. $2b + 3a = 5k'$ for some integer $k'$. So, R is symmetric.

Transitive: For R to be transitive, if $aRb$ and $bRc$, then $aRc$ must be true. If $2a + 3b = 5k$ and $2b + 3c = 5l$, we need to check if $2a + 3c$ is a multiple of 5. Multiply the first equation by 2 and the second by 3: $4a + 6b = 10k$ and $6b + 9c = 15l$. Subtract the first from the second: $9c - 4a = 15l - 10k$ $4a = 9c - 15l + 10k$ $2a = \frac{9}{2}c - \frac{15}{2}l + 5k$ $2a + 3c = \frac{9}{2}c + 3c - \frac{15}{2}l + 5k = \frac{15}{2}c - \frac{15}{2}l + 5k = \frac{5}{2}(3c - 3l + 2k)$ Multiply the first equation by 2: $4a + 6b = 10k$ Multiply the second equation by 3: $6b + 9c = 15l$ From the first equation, $6b = 10k - 4a$. Substituting into the second equation: $10k - 4a + 9c = 15l$ $4a - 9c = 10k - 15l$ We want to show that $2a + 3c = 5m$ for some integer $m$. Multiply $4a - 9c = 10k - 15l$ by $\frac{1}{2}$: $2a - \frac{9}{2}c = 5k - \frac{15}{2}l$ $2a = \frac{9}{2}c + 5k - \frac{15}{2}l$ $2a + 3c = \frac{9}{2}c + 3c + 5k - \frac{15}{2}l = \frac{15}{2}c + 5k - \frac{15}{2}l = \frac{5}{2}(3c + 2k - 3l)$ This doesn't necessarily mean $2a + 3c$ is a multiple of 5. Let $a = 1, b = 1$. Then $2(1) + 3(1) = 5$, which is a multiple of 5. Let $b = 1, c = 1$. Then $2(1) + 3(1) = 5$, which is a multiple of 5. Then $aRb$ and $bRc$. Now, $2a + 3c = 2(1) + 3(1) = 5$, which is a multiple of 5. So $aRc$. Let $a = 1, b = 1$. $2a + 3b = 5$. Let $b = 1, c = 6$. $2b + 3c = 2 + 18 = 20$. Then $2a + 3c = 2(1) + 3(6) = 2 + 18 = 20$, which is a multiple of 5. Let $a = 2, b = 2$. $2a + 3b = 4 + 6 = 10$. Let $b = 2, c = 7$. $2b + 3c = 4 + 21 = 25$. Then $2a + 3c = 4 + 21 = 25$. It seems to be transitive. $4a + 6b = 10k$ $6b + 9c = 15l$ $4a + 6b \equiv 0 \pmod{5}$ $6b + 9c \equiv 0 \pmod{5}$ $6b \equiv -4a \pmod{5}$ $6b \equiv -9c \pmod{5}$ $-4a \equiv -9c \pmod{5}$ $4a \equiv 9c \pmod{5}$ $4a \equiv 4c \pmod{5}$ $a \equiv c \pmod{5}$ $2a + 3c \equiv 2c + 3c \equiv 5c \equiv 0 \pmod{5}$ So, R is transitive.

Since R is reflexive, symmetric, and transitive, R is an equivalence relation.