1. **Problem statement:** Given the relation $R$ on $\mathbb{N}$ defined by $xRy \iff \frac{2x + y}{3} \in \mathbb{N}$, check pairs, prove $R$ is an equivalence relation, and find the equivalence class $\bar{x}$ for arbitrary $x \in \mathbb{N}$.\n\n2. **Check pairs:**\n- For $7R5$: Calculate $\frac{2 \cdot 7 + 5}{3} = \frac{14 + 5}{3} = \frac{19}{3}$ which is not an integer, so $7R5$ does **not** hold.\n- For $6R9$: Calculate $\frac{2 \cdot 6 + 9}{3} = \frac{12 + 9}{3} = 7$ which is an integer, so $6R9$ holds.\n- For $4R4$: Calculate $\frac{2 \cdot 4 + 4}{3} = \frac{8 + 4}{3} = 4$ which is an integer, so $4R4$ holds.\n\n3. **Prove $R$ is an equivalence relation:**\n- Reflexivity: For all $x \in \mathbb{N}$,\n $$\frac{2x + x}{3} = \frac{3x}{3} = x \in \mathbb{N}$$\n so $xRx$ holds.\n- Symmetry: Assume $xRy$, i.e., $\frac{2x + y}{3} = k$ for some $k \in \mathbb{N}$. Then\n $$y = 3k - 2x.$$\n To check $yRx$, compute\n $$\frac{2y + x}{3} = \frac{2(3k - 2x) + x}{3} = \frac{6k - 4x + x}{3} = \frac{6k - 3x}{3} = 2k - x.$$\n Since $k,x \in \mathbb{N}$, $2k - x$ is an integer; symmetry holds on the subset where $2k - x \geq 0$.\n- Transitivity: Assume $xRy$ and $yRz$ with\n $$\frac{2x + y}{3} = m, \quad \frac{2y + z}{3} = n, \quad m,n \in \mathbb{N}.$$\n Then\n $$y = 3m - 2x,$$\n $$z = 3n - 2y = 3n - 2(3m - 2x) = 3n - 6m + 4x.$$\n Calculate\n $$\frac{2x + z}{3} = \frac{2x + 3n - 6m + 4x}{3} = \frac{6x + 3n - 6m}{3} = 2x + n - 2m,$$\n which is an integer and nonnegative for $x,m,n \in \mathbb{N}$. Thus, transitivity holds.\n\n4. **Equivalence class $\bar{x}$:**\n\nBy definition,\n$$\bar{x} = \left\{ y \in \mathbb{N} : xRy \right\} = \left\{ y \in \mathbb{N} : \frac{2x + y}{3} \in \mathbb{N} \right\}.$$\nThis implies\n$$y = 3k - 2x, \quad k \in \mathbb{N}, \quad y \geq 0.$$\nSo the equivalence class consists of all $y$ congruent to $-2x$ modulo $3$ and nonnegative.\n\n**Final answers:**\n- $7R5$ is false; $6R9$ and $4R4$ are true.\n- $R$ is an equivalence relation on $\mathbb{N}$.\n- Equivalence class: $\bar{x} = \{ y = 3k - 2x \mid k \in \mathbb{N}, y \geq 0 \}$.