solve the system -32+3z=4 (x-2y) 4 (x-2y-z)= -36 -2 (2x+y)+2z=-12.
Is it ???
-32+3z=4(x-2y) then 4x-8y-3z = -32 (1)
4(x-2y-z)= -36 then 4x-8y+2z = -36 (2)
-2(2x+y)+2z=-12 then -4x-2y+2z = -12 (3)
(1) - (2) ---> z = 4
(2) + (3) ----> -10y - 2z = -48
-10y - 8 = -48
y = 4
back in (1)
4x - 32 - 12 = -32
x = 3