Let R_1 = \ (a,b) ∈ ℤ × ℤ : ab ≥ 0\ and R_2 = \ (a,b) ∈ ℤ × ℤ : 3a...

For relation $R_1$:
Reflexivity: For any $a \in \mathbb{Z}$, $a \cdot a = a^2 \ge 0$. So, $R_1$ is reflexive.
Symmetry: If $ab \ge 0$, then $ba \ge 0$. So, $R_1$ is symmetric.
Transitivity: Let $a = 1$, $b = 0$, and $c = -1$.
We have $ab = 1 \cdot 0 = 0 \ge 0$ and $bc = 0 \cdot (-1) = 0 \ge 0$.
However, $ac = 1 \cdot (-1) = -1 < 0$.
Thus, $(a,b) \in R_1$ and $(b,c) \in R_1$ does not imply $(a,c) \in R_1$.
Hence, $R_1$ is not transitive and therefore not an equivalence relation.

For relation $R_2$:
Reflexivity: For any $a \in \mathbb{Z}$, $3a + 4a = 7a$, which is a multiple of $7$. So, $R_2$ is reflexive.
Symmetry: Let $3a + 4b = 7k$ for some integer $k$.
Then $3b + 4a = 7(a+b) - (3a + 4b) = 7(a+b) - 7k = 7(a+b-k)$, which is also a multiple of $7$. So, $R_2$ is symmetric.
Transitivity: Let $3a + 4b = 7k$ and $3b + 4c = 7m$.
Adding the two equations gives $(3a + 4b) + (3b + 4c) = 7k + 7m$
$\Rightarrow 3a + 7b + 4c = 7(k+m)$
$\Rightarrow 3a + 4c = 7(k+m-b)$, which is a multiple of $7$.
So, $R_2$ is transitive.
Hence, $R_2$ is an equivalence relation.

Therefore, $R_2$ is an equivalence relation but $R_1$ is not.

Answer: $R_2$ is an equivalence relation but $R_1$ is not