How fixed air, gas and fixed weights in rigid airships is computed in this book?
$D$ 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
$$x = 1 - .288 - .055 - .3 = .357$$