Comparison of x and y from two quadratics (use mapping below): I. 18x^2 + 18x + 4 = 0 II. 12y^2 + 29y + 14 = 0 Mapping: 1 → x > y, 2 → x < y, 3 → x = y, 4 → Relationship cannot be determined.

Introduction / Context:
When both x and y are defined by quadratics with two real roots, comparisons must consider all combinations. If some combinations yield equality and others yield strict inequality, the overall relationship is indeterminate.

Given Data / Assumptions:

• I: 18x^2 + 18x + 4 = 0
• II: 12y^2 + 29y + 14 = 0

Concept / Approach:
Find the roots of each. Then examine whether a single relation (>, <, or =) holds for every admissible pair (x, y). If not, choose “Relationship cannot be determined.”

Step-by-Step Solution:
I: Divide by 2 → 9x^2 + 9x + 2 = 0 = (3x + 1)(3x + 2)x ∈ {−1/3, −2/3}II: Discriminant = 29^2 − 4*12*14 = 169 ⇒ √D = 13y = (−29 ± 13)/24 ⇒ y ∈ {−2/3, −7/4}Comparisons: x = −2/3 and y = −2/3 give equality; x = −1/3 compared with y = −2/3 gives x > y. Different relations occur.

Verification / Alternative check:
Narrowest gap occurs at the common root −2/3. Since equality is possible and x > y also occurs, no single relation always holds.

Why Other Options Are Wrong:
Choosing x > y or x = y ignores that both cases occur; x < y never occurs for any admissible pair.

Common Pitfalls:
Concluding based on a single pair, missing the shared root case that forces equality.

Final Answer:
Relationship cannot be determined