Question 1726
There are many different methods involved that you can do to solve this system of equations. By far the easiest way is to enter the value into a matrix and then getting the matrix into reduced-row echelon form to solve the x,y,and z value. You can also do elimination or substitution. Lets look at the second equation first and notice that there are only two variables. We can subtract the y to get x by itself like so: x=4-y. 
Now we look at the 1st equation and the last equation. Since the second equation does not have a z variable in it we want to do an elimination. We can multiply the last equation by -1 to get: -3x-y-z=-8.
We can then take both equations then add them together like so:
 x+2y+z=6
-3x-y-z=-8 
-2x+y=-2 
With our new equation of -2x+y=-2 we can substitute the x=4-y into that to get: {{{-2(4-y)+y=-2}}} which we can simplify to get y by itself which is y=2. 
Now that we have our y value we can back substitute into x=4-y to get our x value of x=2.
We finally substitute the x and y value into any of the equations to get the z value which is z=0. 
Our values are x=2, y=2, and z=0 or (2,2,0).