The system x + y = 10 and x - y = k has a solution in Quadrant I (where both x > 0 and y > 0) for a range of values of k. What is the…

Add the equations: $2x = 10 + k$, so $x = \frac{10 + k}{2}$. Subtract them: $2y = 10 - k$, so $y = \frac{10 - k}{2}$. Quadrant I requires both $x > 0$ and $y > 0$. The first gives $k > -10$, and the second gives $k < 10$. So $-10 < k < 10$, and the largest integer in this range is 9. At $k = 10$, $y = 0$, which lies on the axis, not in the quadrant, so 10 fails. A common wrong answer is 10, from including the endpoint, and another is 5, from using the wrong equation.