A flock of geese on a pond was being observed continuously. At 1:00pm, 1/5 of the geese flew away. At 2:00pm, 1/8 of the geese that remained flew away. At 3:00pm, 3 times as many geese has had flown away at 1:00pm flew away, leaving 28 geese on the pond. At no other time did any geese arrive or fly away.

1:00 -- 1/5 gone, leaving 4/5
2:00 -- 4/5 * 1/8 = 1/10 flew, leaving 4/5 - 1/10 = 7/10
3:00 -- 3 * 1/5 = 3/5 gone, leaving 1/10
x/10 = 28
so the original flock was 280 geese
Trying to do all that in a single equation is possible, but would be very cumbersome. Good luck with that effort.

Is there a different way to do it?
1:00 is 1/5 and that becomes 4/5
2:00 is 1/8 and that becomes 7/8
3:00 is 3x
The equation would equal to 28

Your terminology is rather opaque.
And what do you mean when you finally say

The equation would equal to 28

what equation? An equation is never "equal" to anything.

Then what would the equation be by using 28, 3x, 7/8, and 4/5

Original flock = X geese.

1:00 PM : x - x/5 = 4x/5 remained.
2:00 PM: 4x/5 - 1/8 * 4x/5 = 4x/5 - 4x/40 = 32x/40 - 4x/40 = 7x/10 remained
3:00 PM: 7x/10 - 3x/5 = 7x/10 - 6x/10 = x/10 remained.

x/10 = 28.
X = 280 geese.