Consider the two equations: I) x^2 − 5x − 24 = 0 II) 2y^2 + 19y + 35 = 0 For all possible real solutions x and y of these equations, what is the correct relationship between x and y?

Introduction / Context:
This question tests understanding of quadratic roots and comparison across multiple possible solutions. Because each equation can produce more than one real root, the pair (x, y) is not unique. The correct choice depends on whether a single inequality holds for every possible pairing of roots.

Given Data / Assumptions:

• Equation I: x^2 − 5x − 24 = 0 • Equation II: 2y^2 + 19y + 35 = 0 • x and y take any real root values from their respective equations

Concept / Approach:
Find the sets of roots for x and y by factoring (or using the quadratic formula). Then compare all combinations of x with y. If one relationship (like x > y) fails for even one valid pair, then that relationship is not universally true. If no single inequality always holds, the relationship cannot be uniquely determined.

Step-by-Step Solution:
1) Solve Equation I: x^2 − 5x − 24 = 0 → (x − 8)(x + 3) = 0 So x ∈ {8, −3} 2) Solve Equation II: 2y^2 + 19y + 35 = 0 Discriminant = 19^2 − 4*2*35 = 361 − 280 = 81 y = (−19 ± 9)/4 → y ∈ {−7, −2.5} 3) Compare possibilities: If x = 8 and y = −7, then x > y If x = −3 and y = −2.5, then x < y 4) Since both x > y and x < y can occur depending on the chosen roots, no single fixed relationship holds for all pairs.

Verification / Alternative check:
List all pairs: (8, −7): x > y (8, −2.5): x > y (−3, −7): x > y (−3, −2.5): x < y Because one pair contradicts the others, a universal inequality is impossible.

Why Other Options Are Wrong:
• x > y for all pairs is false because (−3, −2.5) violates it. • x < y, x ≥ y, or x = y for all pairs also fail due to the mixed comparisons above.

Common Pitfalls:
• Comparing only the largest roots or only one pair. • Mistakes in factoring or computing the discriminant for Equation II.

Final Answer:
The relationship between x and y cannot be uniquely determined