math
posted by sherri r .
solve the system 32+3z=4 (x2y) 4 (x2yz)= 36 2 (2x+y)+2z=12.

Is it ???
32+3z=4(x2y) then 4x8y3z = 32 (1)
4(x2yz)= 36 then 4x8y+2z = 36 (2)
2(2x+y)+2z=12 then 4x2y+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