(x+2y)+3z=1, (x+2y)+4u=3, so 1-3z=3-4u, 4u=3z+2.
2y+3z+3z+2=-6, 2y+6z+2=-6, y+3z=-4, y=-3z-4.
We now have u and y in terms of z. So we can substitute their values in the last equation: x-6z-8+3z+2=3, x=3z+8+3-2=3z+9.
Now we have u, x, y in terms of z.
x+3z+4u=-7⇒3z+9+3z+3z+2=-7, 9z=-18, z=-2.
Therefore u=(3z+2)/4=-4/4=-1; x=3z+9=-6+9=3; y=-3z-4=6-4=2.
So (u,x,y,z)=(-1,3,2,-2).