In this comparison question, two quadratic equations are given: I. 5x^2 + 28x = -15 (that is, 5x^2 + 28x + 15 = 0) and II. 3y^2 + 11y + 6 = 0. After solving both equations, what is the correct relationship between x and y?

Introduction / Context:
This is a standard quadratic comparison problem often seen in bank and aptitude exams. You are given two quadratic equations in variables x and y. Each equation can have two roots. You must determine the relationship between any possible values of x and y (derived from their respective equations).

Given Data / Assumptions:

• Equation I: 5x^2 + 28x = −15, which can be rewritten as 5x^2 + 28x + 15 = 0.
• Equation II: 3y^2 + 11y + 6 = 0.
• x and y can take any root from their respective equations.
• We must compare all combinations of roots.

Concept / Approach:
First, solve each quadratic equation to find its real roots. Then, consider the possible pairs (x, y), where x is any root of Equation I and y is any root of Equation II. If x is always greater than y, you choose x > y, and similarly for other relations. If sometimes x is greater and sometimes smaller, then the relationship cannot be determined.

Step-by-Step Solution:
Step 1: Rewrite Equation I as 5x^2 + 28x + 15 = 0. Step 2: Factor Equation I: 5x^2 + 28x + 15 = (x + 5)(5x + 3) = 0. Step 3: Roots of Equation I: x = −5 and x = −3 / 5 (−0.6). Step 4: Factor Equation II: 3y^2 + 11y + 6 = (y + 3)(3y + 2) = 0. Step 5: Roots of Equation II: y = −3 and y = −2 / 3 (approximately −0.6667). Step 6: Compare all combinations of roots. Case 1: x = −5, y = −3. Then x < y. Case 2: x = −5, y = −2 / 3. Then x < y again. Case 3: x = −0.6, y = −3. Then x > y. Case 4: x = −0.6, y = −2 / 3 (≈ −0.6667). Then x > y.

Verification / Alternative check:
From the comparisons, we see that sometimes x is less than y (when x = −5) and sometimes x is greater than y (when x = −0.6). Because the problem does not restrict x or y to a particular root, all these combinations are possible. Therefore, there is no single consistent relationship that holds for all allowed values of x and y.

Why Other Options Are Wrong:
x > y is not always true because when x = −5 and y = −3, x is clearly less than y.
x ≥ y and x ≤ y both require a consistent inequality in one direction or equality, which does not occur for all pairs of roots.
x < y is not always true because when x = −0.6 and y = −3, x is greater than y.

Common Pitfalls:
A common mistake is to compare only one pair of roots (for example, the larger root of each equation) and draw a conclusion without checking all possible combinations. Another error is misfactorising the quadratic equations, which leads to incorrect roots and hence incorrect comparisons.

Final Answer:
The correct conclusion is that the relationship between x and y cannot be determined uniquely.