How fixed air, gas and fixed weights in rigid airships is computed in this book? I hope I am posting this in the right forum. My current problem is part aviation, part maths.
I have been struggling with the following computation ( Burgess, Charles Paine. Airship Design. Honolulu, Hawaii: University Press of the Pacific, 2004. page 14. The pdf is available at http://www.xlta.org/library/burgess.pdf, note that the first edition of this book was written in 1927. The present snippet uses the typographical conventions from the pdf.)

Problem 1:
Find the volume and horsepower of a rigid airship to carry a military load of 15,000 lbs.at 60 knots (101.3 ft. / sec.). for 60 hours, 85% of the total volume being filled with helium lifting .064 lb./ft.**3 (94 % pure) in the standard atmosphere.
Since the hull is specified in the condition of the problem to be 85% full of gas,
the weight of the gas is 85% D multiplied by the difference between the weight of air and
the lift of gas per unit volume, and divided by the weight of air per unit volume. The total
weight of air and gas is therefore given:

$$
Air\; and\; gas = [ .15 + .85 ( .07635 - .064 ) / .07635 ] \times D = .288 D 
$$
While this formula is not an issue, the subsequent formula has been giving me headaches.

From existing data, the fixed weights exclusive of power plant, power cars, and fuel system
equal -3D, and the crew, stores and ballast amount to .055 D. There remains for power plant, fuel, and military load

$$
( 1 - .288 - .30 - .055) D = .357 D
$$
Question is how did the author arrive at this result? Not matter how I approach solving the equation,
I am unable to get the same result. Problem is that the result of this equation is used at a later
stage. I also considered that 1 might be an i, but there was no mention of any constant i in the chapter this formula was taken from. I have no clue what I am missing here.
$D$ is the total displacement (the weight of the air displaced by the airship)
$0.288 \cdot D$ is the weight of helium and air inside the air ship
$0.055 \cdot D$ is the weight of crew, storage, and ballast
$x \cdot D$ is the unknown weight of power plants, power cars, and the fuel system
$0.3 \cdot D$ is anything else (fixed weights)

The total must be one:
$$.288 + .055 + .3 + x = 1$$
or equivalent, the total must be the displacement:
$$D \cdot (.288 + .055 + .3 + x) = 1 \cdot D$$
solve for x and you get:
$$x \cdot D = (1 - .288 - .055 - .3) \cdot D$$
or just
$$x = 1 - .288 - .055 - .3 = .357$$