For x + y ≤ 20 and x - y ≥ 4, what is the largest integer value of y for which a valid value of x exists?

From $x - y \geq 4$, we get $x \geq y + 4$. For a valid $x$ to exist, the smallest allowed $x$ must also satisfy $x + y \leq 20$. Substituting $x = y + 4$: $(y + 4) + y \leq 20$, so $2y \leq 16$ and $y \leq 8$. At $y = 8$, $x = 12$ works: $12 + 8 = 20 \leq 20$ and $12 - 8 = 4 \geq 4$. At $y = 9$, we would need $x \geq 13$ and $x \leq 11$, which is impossible. So the largest integer is 8. A common wrong answer is 10, from using only $x + y \leq 20$ with $x = 10$.