Linear Equations in three variables

Hello everyone ! this is sarabjeet.
Today in class first we had quiz ,(that was hard ) and then Mr.Piatek taught us how to solve " systems linear equations in three variables " by substitution method.We had a presentation about it first and then we did some examples.
for example :
x+y-z=2
x-2y+z=-1
3x+y-2z=4

step 1-solve one of the equations for one variable.
x+y-z=2
x=-y+z+2 or x=2+-y+x

step 2-substitute this equation into the other 2 equations .
x-2y+z=-1
(2-y+z)-2y+z=-1
2-y+z-2y+z=-1
-3y+2z=-3
3x+y-2z=4
3(2-y+z)+y-2z=4
6-3y+3z+y-2z=4
-2y-z=4-6
-2y-z=-2

step 3-solve the system that is made with 2 new equations.
-3y+2z=-3
-2y-z=-2
z=-2+2y
-3y+2(-2+2y)=-3
-3y-4+4y=-3
y=-3+4
y=1

z=-2+2y
z=-2+2(1)
z=-2+2
z=0

x=2-y+z
x=2-1+0
x=1

(1,1,0)

step 4- check your answer.
x+y-z=2
1+1-0=2
2=2
x-2y+z=-1
1-2(1)+0=-1
1-2=-1
-1=-1
3x+y-2z=4
3(1)+1-2(0)=4
3+1=4
2=2

so the answer is (1,1,0)
And I think we have quiz tomorrow too,so be ready !