Multiple choice

In the following question two equations numbered I and II are given. You have to solve both the equations and give answer. $(\text{I.}\quad y^2 + 2y - 3 = 0)$ $(\text{II.}\quad 2x^2 - 7x + 6 = 0)$

  1. $\(x > y\)$
  2. $\(x \ge y\)$
  3. $\(x < y\)$
  4. $\(x \le y\)$

  5. x = y, or relation cannot be established
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A Correct answer
Explanation

Solving equation I gives y² + 2y - 3 = 0, which factors to (y+3)(y-1) = 0, so y = -3 or y = 1. Solving equation II gives 2x² - 7x + 6 = 0, which factors to (2x-3)(x-2) = 0, so x = 1.5 or x = 2. Comparing the values, both x solutions (1.5 and 2) are greater than both y solutions (-3 and 1), therefore x > y is always true.

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