I.x2−19x+84=0
II.y2−25y+156=0
(d)
x2 - 19 x + 84 = 0
=> x2 - 12 x - 7 x + 84 = 0
=> x(x - 12) - 7 (x - 12) = 0
=> (x - 7) (x - 12) = 0
So, x = 7 or x = 12
y2 - 25 y + 156 = 0
=> y2 - 12 y - 13 y + 156 = 0
=> y(y - 12) - 13 (y - 12) = 0
=> (y - 13) (y - 12) = 0
So, y = 12 or 13
Thus, we see that x≤y