Personally, I'd write it as a matrix and row reduce for workflow simplicity:
So :
#((2, 3, -2),(4, -3, 1), (1, 2, -4))((x),(y),(z)) = ((-4),(25),(-12))#
Becomes in augmented form:
#((2, 3, -2),(4, -3, 1), (1, 2, -4)) ((-4),(25),(-12))#
#R2 to R2 - 2 R1; R3 to R3 - 1/2 R1#
#((2, 3, -2),(0, -9, 5), (0, 1/2, -3)) ((-4),(33),(-10))#
#R3 to R3 +1/18 R2#
#((2, 3, -2),(0, -9, 5), (0, 0, -49/18)) ((-4),(33),(-49/6))#
So we have:
#((2, 3, -2),(0, -9, 5), (0, 0, -49/18))((x),(y),(z)) ((-4),(33),(-49/6))#
We back substitute starting at the bottom:
From row 3: # -49/18 z = -49/6 implies z = 3#
From row 2: #- 9 y + 5 (3) = 33 implies y = -2#
From row 1: #2 x + 3(-2) - 2(3) = -4 implies x = 4#
So:
#((x),(y),(z)) = ((4),(-2),(3))#
