Multiple choice

In each of the following questions, two equations are given. You have to solve them and\ I. ((8 + x)^2 = x^2 + 12x)\ II. (7y^2 + 15y + 4 = 6y^2 + 20)

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

Solving Equation I: (8 + x)^2 = x^2 + 12x. Expanding: 64 + 16x + x^2 = x^2 + 12x, which simplifies to 64 + 4x = 0, giving x = -16. Solving Equation II: 7y^2 + 15y + 4 = 6y^2 + 20, which simplifies to y^2 + 15y - 16 = 0. Factoring: (y - 1)(y + 16) = 0, so y = 1 or y = -16. Comparing x = -16 with y values: x = y when y = -16, and x < y when y = 1. Therefore x ≤ y in all cases.

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