Solve for x and y from power equations, then compare: I. x^3 × 13 = x^2 × 247 II. y^(1/3) × 14 = 294 ÷ y^(2/3) Find the correct relationship between x and y (x > y, x < y, x = y, or cannot be determined).

Introduction / Context:Two separate equations define numeric values for x and y via powers. After solving both exactly, we compare their sizes and select a single relationship statement.Given Data / Assumptions:

• I: x^3 × 13 = x^2 × 247, x ≠ 0 allows cancellation of x^2.
• II: y^(1/3) × 14 = 294 ÷ y^(2/3), y > 0 for real principal fractional powers.

Concept / Approach:In I, divide by x^2 to isolate x. In II, multiply both sides by y^(2/3) to consolidate powers and solve for y. Then compare numeric values of x and y.Step-by-Step Solution:

Verification / Alternative check:Back-substitute: I: 13*19^3 vs 247*19^2 ⇒ both equal 13*19^3. II: Left = 14*21^(1/3); Right = 294/21^(2/3). Since 21^(1/3) * 21^(2/3) = 21, both sides reduce to 14*21^(1/3) and 14*21^(1/3) respectively.

Why Other Options Are Wrong:

• x > y or x = y: Contradicted numerically.
• Relationship cannot be determined: Both values are found uniquely; comparison is definite.

Common Pitfalls:Forgetting domain assumptions with fractional exponents or failing to cancel x^2 when x ≠ 0.

Final Answer: