Precalculus, Chapter 8, 8.1, Section 8.1, Problem 70

3x-2y+z=-15
-x+y+2z=-10
x-y-4z=14
A=[[3,-2,1],[-1,1,2],[1,-1,-4]]
b=[[-15],[-10],[14]]
[A|b]=[[3,-2,1,-15],[-1,1,2,-10],[1,-1,-4,14]]
Multiply 2nd Row by 3 and add Row 1
[[3,-2,1,-15],[0,1,7,-45],[1,-1,-4,14]]
Multiply 3rd Row by 3 and subtract it from Row 1
[[3,-2,1,-15],[0,1,7,-45],[0,1,13,-57]]
Subtract Row 2 from Row 3
[[3,-2,1,-15],[0,1,7,-45],[0,0,6,-12]]
Now the equations can be written as,
3x-2y+z=-15 ----equation 1
y+7z=-45 ------ equation 2
6z=-12 ------ equation 3
From equation 3,
z=-12/6=-2
Now substitute back z in equation 2,
y+7(-2)=-45
y-14=-45
y=-45+14
y=-31
Substitute back the value of y and z in equation 1,
3x-2(-31)+(-2)=-15
3x+62-2=-15
3x=-15-60
3x=-75
x=-75/3
x=-25
So the solution is x=-25, y=-31 and z=-2