In this aptitude (simplification) question, two quadratic equations in real variables x and y are given. You must solve both equations completely and then compare all possible values of x and y to decide the correct overall relationship. I. 2x^2 + 19x + 42 = 0 II. 4y^2 + 43y + 30 = 0

Introduction / Context:
This is another quadratic comparison question that appears frequently in quantitative aptitude. The candidate must find all real roots for each quadratic equation and then analyse how these sets of roots compare. Such questions are designed to test conceptual understanding of quadratic equations, factorisation skills, and logical reasoning about inequalities rather than pure calculation speed alone.

Given Data / Assumptions:
Equation I: 2x^2 + 19x + 42 = 0.
Equation II: 4y^2 + 43y + 30 = 0.
All real roots of each equation are valid values that x and y can take respectively.

Concept / Approach:
We factorise each quadratic to obtain its two real roots. Then we list the set of x values and the set of y values. To determine a global comparison such as x > y, x < y, x = y, or x >= y, the relationship must hold for every possible pairing of roots. If we can find at least two pairs with conflicting order (for example one pair where x > y and another where x < y), then the relationship cannot be determined uniquely.

Step-by-Step Solution:
Factor equation I: 2x^2 + 19x + 42 = 0. We can factor this as (2x + 7)(x + 6) = 0. Thus x = -7/2 or x = -6. Now factor equation II: 4y^2 + 43y + 30 = 0. Factorisation gives (4y + 5)(y + 6) = 0. Thus y = -5/4 or y = -6. Now compare different combinations. When x = -6 and y = -6 we get x = y. When x = -7/2 (that is -3.5) and y = -5/4 (that is -1.25) we have x < y. Hence we have at least equality and x < y. By taking x = -7/2 and y = -6 we get x > y. So all three types of relations occur depending on the chosen roots.

Verification / Alternative check:
To verify, convert all roots to decimals. For x we have approximately -3.5 and -6. For y we have approximately -1.25 and -6. Now form pairs: (-6, -6) gives equality. (-3.5, -1.25) gives x < y. (-3.5, -6) gives x > y. Because the relative order changes across different valid root pairs, no single inequality or equality can summarise the relationship between x and y across all possibilities.

Why Other Options Are Wrong:
Options x > y, x < y, and x = y all demand a single strict relation for every pair, which is not true. Option x >= y also fails because we have a pair where x < y. Therefore none of these fixed relationships is valid for all combinations. The only correct conclusion is that the relationship cannot be determined in a unique way.

Common Pitfalls:
A frequent error is to choose only one convenient pair of roots and draw a global conclusion, ignoring other possible combinations. Another mistake is incorrect factorisation, which leads to wrong roots and then a wrong comparison. Learners should also be cautious when working with negative numbers, as sign confusion can easily invert the inequality. Always list all roots clearly before comparing them.

Final Answer:
Because x can be less than, equal to, or greater than y depending on the chosen roots, the correct option is Relationship cannot be determined.