Solution of (x + y - 1)dx + (2x + 2y - 3)dy = 0 is:

$\frac{dy}{dx}=-\frac{x+y-1}{2x+2y-3}$

Put x + y = t;

$1+\frac{dy}{dx}=\frac{dt}{dx};\frac{dt}{dx}-1=-\frac{t-1}{2t-3}$

$\frac{dt}{dx}=\frac{t-2}{2t-3};\int\frac{(2t-3)dt}{t-2}=\int dx$;

2t + log_e (t - 2) = x + c; x + 2y + log_e (x + y - 2) = c