Question 651693
<pre>
The system of equations:

{{{system(Ax+By=C,Dx+Ey=F)}}}

is 

(1)  consistent and independent if 

AE &#8800; BD 

(2)  consistent and dependent if 

AE = BD and AF = CD

{{{system(kx+3y=k-2,12x+ky=k)}}}

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 &#8800; BD to be false, so we must
require:

  AE = BD

k(k) = 3(12)
  kČ = 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 &#8800; CD


    AF &#8800; CD
  k(k) &#8800; (k-2)(12)

±6(±6) &#8800; (±6-2)(12)

Taking the + we have

   36 &#8800; (6-2)(12)

   36 &#8800; (4)(12)

   36 &#8800; 48

which is true.

Taking the - we have

   36 &#8800; (-6-2)(12)

   36 &#8800; (-8)(12)

   36 &#8800; -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</pre>