Integral Inequalities with Absolute Values

Let $x, y \in \{-1, -2, -3, ...\}$. Since $x \le -1$, $|x - 1| = 1 - x \ge 2$. Let $A = |x - 1| \in \{2, 3, 4, 5, ...\}$ and $B = |y + 3| \in \{0, 1, 2, ...\}$. We require $2 \le A \times B \le 5$.

• If $A = 2$ ($x = -1$): $B \in \{1, 2\} \implies |y + 3| = 1 \text{ or } 2 \implies y \in \{-4, -2, -5, -1\}$ (4 values).
• If $A = 3$ ($x = -2$): $B = 1 \implies |y + 3| = 1 \implies y \in \{-4, -2\}$ (2 values).
• If $A = 4$ ($x = -3$): $B = 1 \implies y \in \{-4, -2\}$ (2 values).
• If $A = 5$ ($x = -4$): $B = 1 \implies y \in \{-4, -2\}$ (2 values). Total combinations = $4 + 2 + 2 + 2 = 10$.