In the following question, two equations numbered I and II are given. You have to solve both the equations and give answer: I. x^2 - 26x + 168 = 0 II. y^2 - 32y + 252 = 0

Explanation

Equation I: x^2 - 26x + 168 = 0 => x^2 - 12x - 14x + 168 = 0 => x(x - 12) - 14(x - 12) = 0 => (x - 12)(x - 14) = 0 => x = 12 or x = 14. Equation II: y^2 - 32y + 252 = 0 => y^2 - 18y - 14y + 252 = 0 => y(y - 18) - 14(y - 18) = 0 => (y - 14)(y - 18) = 0 => y = 14 or y = 18. Value of x / Value of y / Relation: 12, 14 -> x x x = y; 14, 18 -> x < y. Therefore, x <= y.