Integer Pairs Satisfying System of Inequalities

We have $y \ge 3$ and $x \ge y+1$. Also $x + y \le 13$.

• For $y = 3$: $x > 3 \implies x \ge 4$. $x + 3 \le 13 \implies x \le 10$. Possible $x$: $4, 5, 6, 7, 8, 9, 10$ (7 values).
• For $y = 4$: $x > 4 \implies x \ge 5$. $x + 4 \le 13 \implies x \le 9$. Possible $x$: $5, 6, 7, 8, 9$ (5 values).
• For $y = 5$: $x > 5 \implies x \ge 6$. $x + 5 \le 13 \implies x \le 8$. Possible $x$: $6, 7, 8$ (3 values).
• For $y = 6$: $x > 6 \implies x \ge 7$. $x + 6 \le 13 \implies x \le 7$. Possible $x$: $7$ (1 value).
• For $y \ge 7$: $x \ge 8 \implies x+y \ge 15 > 13$, no solution. Total number of pairs = $7 + 5 + 3 + 1 = 16$. Wait, let's re-count carefully: for $y=3$, $x \in [4, 10]$ gives 7 pairs. For $y=4$, $x \in [5, 9]$ gives 5 pairs. For $y=5$, $x \in [6, 8]$ gives 3 pairs. For $y=6$, $x=7$ gives 1 pair. Total = 16.