For the following system to be consistent, what must k not be equal to? $6x - 4y + 4z = 5$
$9x - 6y + kz = -4$
$12x - 8y    = -10$
Originally I just multiplied the first row by $\frac{3}{2}$ and subtracted it from the second, which gives you a value of $6$ for the answer.  However, this is not the correct answer.  Any idea what I am doing wrong?
 A: Hint: 
you need to look at the third equation as well! How is that related to the second equation?
A: Use the first and third equations to find $z$.  Then substitute that into the equation you got from the first two equations.  There is only one value $k$ can be.
A: Subtracting the first equation from the third twice yields $-8z=-20$, so $z=\frac52$. Then multiplying the second equation by two and subtracting the first three times yields
$$-5k-30=-23.$$
This shows that $k=-\frac75$.
A: 
A system of linear equations is called inconsistent if it has no solutions. A
  system which has a solution is called consistent.

Now the augmented matrix 
$
\left[\begin{array}{rrr|r}
6 & -4 & 4 & 5 \\
9 & -6 & k & -4 \\
12 & -8 & 0 & -10
\end{array}\right]$
Replace $R_3 $ by $R_3-2R_1$ and $R_2$ by $ R_2 - \frac{3}{2}R_1$ 
$
\left[\begin{array}{rrr|r}
6 & -4 & 4 & 5 \\
0 & 0 & k-6 & -\frac{23}{2} \\
0 & 0 & -8 & -20
\end{array}\right]$
Replace $R_3$ by
Clearly,  $z=\frac{20}{8}=\frac{5}{2}$ and
$(k-6)z=-\frac{23}{2}\implies k-6=-\frac{23}{5}\implies k=-\frac{23}{5} +6=\frac{7}{5}$
