The system of equations:



is 

(1)  consistent and independent if 

AE ≠ BD 

(2)  consistent and dependent if 

AE = BD and AF = CD



Here A=k, B=3, C=k-2, D=12, E=k, F=k

Since we want them to be inconsistent, we must rule out 
both case (1) and case (2).

To rule out case (1) we require AE ≠ BD to be false, so we must
require:

  AE = BD

k(k) = 3(12)
  k² = 36
   k = ±6

But we also must check to see that k = ±6 also rules out 
case (2)

That is we must be sure that AF ≠ CD


    AF ≠ CD
  k(k) ≠ (k-2)(12)

±6(±6) ≠ (±6-2)(12)

Taking the + we have

   36 ≠ (6-2)(12)

   36 ≠ (4)(12)

   36 ≠ 48

which is true.

Taking the - we have

   36 ≠ (-6-2)(12)

   36 ≠ (-8)(12)

   36 ≠ -96

which is also true.

So case (2) is ruled out be taking k = ±6

It was necessary to rule out case (2), however.

Edwin