First equation is x-3y+2z=11 We coundt x and we've got x=11+3y-2z
Third equation is 2x-2y-4z=2 We count also x and we've got: 2x=2+2y+4z /:2 x=1+y+2z
so: 11+3y-2z=1+y+2z 2y-4z=-10 /:2 y-2z=-5 y=2z-5 so x=1+y+2z x=1+2z-5+2z x=4z-4
Now substitute this to second equation: -x+4y+3z=5 -(4z-4) + 4(2z-5) +3z=5 -4z+4+8z-20+3z=5 7z-16=5 7z=21 /:7 z=3 And count rest: x=4z-4 = 4*3-4=8 y=2z-5=2*3-5=1 So the solution is: [tex]\begin{cases} x=8 \\ y=1 \\ z=3 \end{cases}[/tex]
