Let's do it step by step:
Let x = the side of the pyramid's square base (in yards)
Let h = the height of the pyramid (in yards)
Let V = the volume of the pyramid (in cubic yards)
The we know or you can look up or even derive the volume formula for a 3-dimensional square base pyramid:
V = (1/3) h x2. But the problem says that h = (x + 2) yards, and that V = 147 cu. yds
So 147 = (1/3)(x + 2)x2 = [x3 + 2x2]/3 , or [x3 + 2x2] = 3 (147) = 441.
Rather than try to solve this algebraically for x (which isn't easy but can be done), I will approach it numerically since I believe they made x an integer.
To get into the neighborhood of the solution, I would first solve x3 = 441, so x = 7.6 yards approximately.
Now let me do trial and error on both sides of that:
If x = 6 yards, [x3 + 2x2] = 216 + 72 = 288 too low!
If x = 7 yards, [x3 + 2x2] = 343 + 98 = 441 Right on!
If x = 8 yards, [x3 + 2x2] = 512 + 128 = 640 too high!
So the pyramid is 7 yards square at the bottom and 9 yards high in the center.
If I had not hit it exactly, then I would do tenths, x = 6.7, 6.8, 6.9, 7.1, 7.2, 7.3
And then hundreds, each time squeezing the possible solution into a narrower band.