Question 160756
From the first equation:
z=1-x-2y

Put this into the second equation:

2x+3y+2(1-x-2y)=0
2x+3y+2-2x-4y=0
-y=-2
y=2


If y=2, put that in the third equation:
-x-3(2)+3(1-x-2(2))=1
-x-6+3-3x-12=1
-4x=16
x=-4

Now put y=2 and x=-4 into our z from step 1.
z=1-(-4)-2(2)=1

Giving solution set  {-4,2,1}. This also can be solved through techniques of linear algebra (Cramer's Rule, RREF, etc). If you need it solved in a way such as that, send me an E-mail to enabla@gmail.com  .