56In the given question, two equations numbered I and II are given. Solve both…

Concept: For quantity-comparison items built from two quadratic equations, first solve each quadratic independently by factorisation (or the quadratic formula) to get both roots for each variable. Since a quadratic generally has two roots, a single relation between x and y (say x > y) holds only if EVERY pairing of a root of x with a root of y gives that SAME relation; if different pairings give different relations, the relationship cannot be uniquely determined.

1. Factorise equation I: x2 - 14x + 33 = 0 → (x - 11)(x - 3) = 0 → x = 11 or x = 3.

2. Factorise equation II: y2 - 20y + 99 = 0 → (y - 11)(y - 9) = 0 → y = 11 or y = 9.

3. Form all four (x, y) pairings from the two root-sets: (11, 11), (11, 9), (3, 11), (3, 9).

4. Compare x and y in each pairing: (11, 11) → x = y; (11, 9) → x > y; (3, 11) → x < y; (3, 9) → x < y.

5. Since the pairings give x = y, x > y, and x < y across different cases, no single comparison sign holds for every valid pairing — the relationship between x and y cannot be uniquely determined.

Cross-check: Verify the factorisations by expanding back — 11 × 3 = 33 and 11 + 3 = 14 match equation I; 11 × 9 = 99 and 11 + 9 = 20 match equation II. With the roots confirmed and multiple relations arising across the valid pairings, the conclusion — x = y or the relation cannot be determined — is confirmed.