If x1 is any positive number, since x1+ x2= 1, x2= 1- x1. Since x2 must alos be postive, x1 cannot be larger than 1. Then the second equation, 2x1 + x2 + x3 = 2, x3= 2- 2x1- x2= 2- 2x1-(1- x1)= 1- 3x1. Since x3 must be larger than 0, x1 must be less than 1/3.
So the set of all feasible solutions is the set of all triples (x1, 1- x1, 1- 3x1) where x1 is between 0 and 1/3.