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))#