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

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

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

=> \( \Large x \left(x-5\right) -3 \left(x-5\right) = 0 \)

=> \( \Large \left(x-5\right) \left(x-3\right) = 0 \)

Therefore, x = 5, 3

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

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

= \( \Large y \left(y - 2\right) - 1 \left(y - 2\right) = 0 \)

= y = 2, 1

Therefore, x> y