x + y + z =12 x + 2y - z = 12 x + 3y + 3z =24 cari persamaan linear 3 variabel

jwb =
(eliminasi)
x + y + z = 12 (dikali 3)
x + 3y + 3z = 24
-----
3x + 3y + 3z = 36
x + 3y + 3z = 24
------------------------- -
2x = 12
x = 6

(eliminasi)
x + y + z = 12 (kali 2)
x + 2y - z = 12
-----
2x + 2y + 2z = 24
x + 2y - z = 12
------------------------- -
x + 3z = 12

(substitusi)
x + 3z = 12 (x = 6)
6 + 3z = 12
3z = 12 - 6
3z = 6
z = 2

(substitusi)
x + y + z = 12
6 + y + 2 = 12
y + 8 = 12
y = 12 - 8
y = 4