Correct Answer: x > y

I. \( \Large x^{2} \) - 8x + 15 = 0

=> \( \Large x^{2} \) - 5x - 3x + 15 = 0

=> x ( x - 5 ) - 3 ( x - 5 ) = 0

=> ( x - 3 ) ( x - 5 ) = 0

\( \Large \therefore \) x = 3 or 5

II. \( \Large y^{2} \) - 3y + 2 = 0

=> \( \Large y^{2} \) - 2y - y + 2 = 0

=> y ( y - 2 ) - 1 ( y - 2 ) = 0

=> ( y - 1) ( y - 2 ) = 0

y = 1 or 2

Clearly, x > y