Let's label the equation x-2y=-9 as (1), the equation 3x-y+z=4 as (2), and the equation y-z=3 as (3), and then we have it.

(2)-( 1)* 3 =(3x-y+z)-(x-2y)* 3 = 4-(-9)x3，

Is 5y+z=3 1, marked as (4).

(3)+(4)=(y-z)+(5y+z)=3+3 1，

That is, 6y=34, and y= 17/3 can be obtained.

Substitute y= 17/3 (1),

Then you can get x-2* 17/3=-9 and x=7/3.

Substitute y= 17/3 into (3),

Then 17/3-z=3, and you can get z=8/3.

To sum up: x=7/3, y= 17/3, z=8/3.

The answer to question 2 is as follows:

If equation 2x+3y-z= 12 is labeled as (1), equation 2z+x=47 is labeled as (2), and equation x+y+z=6 is labeled as (3), then there is

(3)* 3-( 1)=(x+y+z)* 3-(2x+3y-z)= 6 * 3- 12，

X+4z=6, marked as (4)

Then (4)-(3)=(x+4z)-(2z+x)=6-47,

That is 2z=-4 1, then z=-20.5.

Substituting z=-20.5 into (4) gives x = 6-4 *(20.5)= 88.

Then substitute x = 88 and z =-20.5 into (3) to get.

y=6-88-(-20.5)=-6 1.5

To sum up, we can get: x=88, y=-6 1.5, z=-20.5.