You have:

x-y+z = 2

So:

x = y-z+2

Now you also have:

2x -3y + z = 6

If you substitute x, it will become:

2(y-z+2) -3y +z = 6

2y – 2z + 4 – 3y + z = 6

(2- 3) y + (-2 + 1) z = 6 – 4

-y – z = 2

y + z = -2

y = -2 – z

Therefore, using the last equation:

x – y -2z = 8

(y – z + 2) – (-2 -z) – 2z = 8

(-2-z) – z + 2 + 2 +z -2z = 8

(-1 -1 + 1 -2)z -2 +2 + 2 = 8

-3z + 2 = 8

z = (8- 2) / -3

z = -2

Finally,

y = -2 – z

y = -2 – (-2)

y = -2 + 2

y = 0

and

x = y – z + 2

x = 0 +2 + 2

x = 4

So there you go (4, 0 , -2)

2x – 3y + z = 6 (First Equation)

x – y + z = 2 (Second Equation)

x – y – 2z = 8 (Third Equation)

Then;

Subtract equation 2 from equation 3

Thus,

(x – y – 2z = 8)

– (x – y + z = 2)

______________

-3z = 6

z = -2 (Answer)

Subtract equation 2 from equation 1

(2x – 3y + z = 6)

– (x – y + z = 2)

______________

x – 2y = 4

x = 4 + 2y (equation 4)

Consider equation 2…then substitute the values of z and x.

x – y + z = 2

4 + 2y – y + (-2) = 2

y = 0 (Answer)

Consider equation 4..then substitute the value of y

x = 4 + 2(0)

x = 4 (Answer)

Good luck!

2 -3 1 : 6

1 -1 1 : 2

1 -1 -2: 8

add row 3 times -1 plus row two

2 -3 1 : 6

1 -1 1 : 2

0 0 3 : -6

divide row three by 3

0 0 1 : -2 therefore z = -2

multipy row 2 by -2 then add to row one….

HOPE THAT HELPS

if no matrices—substitute

x = y – z + 2

x = y + 2z +8…

x-y+z=2

x=2+y-z

now fill in another equation using x

2+y-z-y-2z=8

2-3z=8

-3z=6

z=-2

now do the third using both

2(2+y–2)-3y+-2=6

4+2y+4-3y-2=6

6-y=6

-y=0

y=0

y=0,z=-2 and then we solve for x

x-0+-2=2

x-2=2

x=4