Multiple choice

In each of the following questions, two equations are I and II given. You have to solve them and give answer $I. (2(x + y) = 16) II. (x^2 + y^2 = 32)$

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

  5. If x = y or relationship cannot be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

From equation I: 2(x + y) = 16, we get x + y = 8. Substituting into equation II: x^2 + y^2 = 32. Using (x + y)^2 = x^2 + 2xy + y^2, we get 64 = 32 + 2xy, so xy = 16. This means x and y are roots of t^2 - 8t + 16 = 0, which gives t = 4 (repeated root). Thus x = y = 4, so the relationship is x = y.

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