2016/021) w,x,y,z are positive real numbers that satisfy :

$xyz+xy+xz+yz+x+y+z=1\cdots(1)$
$wxy+wx+wy+xy+w+x+y=9\cdots(2)$
$wxz +wx+wz+xz+w+x+z=9\cdots(3)$
$wyz +wy+wz+yz+w+y+z=5\cdots(4)$

Solution

from (1)

$xyz+xy+xz+yz+x+y+z+1=2$
or $(x+1)(y+1)(z+1)=2\cdots(5)$
similarly from (2) (3) and (4)
$(y + 1)(x + 1)(w + 1) = 10\cdots(6)$
$(z + 1)(x + 1)(w + 1) = 10\cdots(7)$
$(y+1)(y+1)(w+1) = 5\cdots(8)$
From (5) (6) (7) and (8) we get
$(x+1)(y+1)(z+1)(w+1) = 10\cdots(9)$
dividing (9) by (5), (6), (7), (8) we get
$(w+1) = 5, y+1 = z+ 1 = 1, x + 1 = 2 => x = 1, w = 4, y= z= 0$