Solve the system by the method of elimination.
x - 2y + 3z = 11
2x - z = 3
3y + z = -8
The 2nd equation is the simplest
and it already has y eliminated in it.
so let's eliminate y also from the 1st
and 3rd, so we'll have two equations
with y eliminated in them:
To eliminate y from the 1st and 3rd equations,
x - 2y + 3z = 11
3y + z = -8
we multiply the top one by 3 and the bottom one by 2
and add them, and the y's will cancel out:
3x - 6y + 9z = 33
6y + 2z = -16
-------------------
3x + 11z = 17
So we put that with the original second equation:
3x + 11z = 17
2x - z = 3
To eliminate z we multiply the bottom one
through by 11, and add them so the z's will
cancel out:
3x + 11z = 17
22x - 11z = 33
---------------
25x = 50
x = 2
Substitute that in
2x - z = 3
2(2) - z = 3
4 - z = 3
-z = -1
z = 1
Substitute that in
3y + z = -8
3y + 1 = -8
3y = -9
y = -3
(x,y,z) = (2,-3,1)
Edwin
