sorry for the confusion. Let's see if I can clarify what I was trying to talk about...
The VO2 equation for running up a slope must be added to that of level ground VO2. Therefore, VO2 for running up a 2% grade at 300m/min = 63.5 + 10.8 = 74.3 ml/kg/min.
Now, for "even paced running" when running up a slope, the 2 VO2 equations must be equal:
let level running velocity = y and slope running velocity = z
(y x 0.2) + 3.5 = [(z x 0.2) + 3.5] + [z x grade x 1.8]
For a grade of 2% and a level ground running speed of 300m/min, the velocity when running up the hill must equal:
0.2y+3.5 = 0.2z+3.5+0.036z
0.2(300) = 0.236z
60 = 0.236z
z = 254.24 m/min
Therefore, to maintain "an even pace" when running up a 2% grade, you must slow down from 300 m/min (level ground) to a velocity of 254.24 m/min while running up the slope.
Hope this makes sense.