For the following system to be consistent, we must have $k \neq\;?$ I am helping with some math homework from WebWorK and am stumped.
The problem is asking:

For the following system to be consistent,
  $$
\begin{align}
4x + 6y +7z &= 5\\
-2x -3y +kz &= 6\\
8x +12y +29z &=26
\end{align}
$$
     we must have $k\ne $ ?

We attempted to do the following, please point out any errors.
$$
\begin{vmatrix}
4 & 6 & 7 & 5\\ 
-2 & -3 & k & 6\\
8 & 12 & 29 & 26
\end{vmatrix}
$$
$r_1*2 \to r_1$
$$
\begin{vmatrix}
8 & 12 & 14 & 10\\ 
-2 & -3 & k & 6\\
8 & 12 & 29 & 26
\end{vmatrix}
$$
$r_3 - r_1 \to r_3$
$$
\begin{vmatrix}
8 & 12 & 14 & 10\\ 
-2 & -3 & k & 6\\
0 & 0 & 15 & 16
\end{vmatrix}
$$
We note that there is only a single solution to $z$
$r_2*4 \to r_2$
$$
\begin{vmatrix}
8 & 12 & 14 & 10\\ 
-8 & -12 & 4k & 24\\
0 & 0 & 15 & 16
\end{vmatrix}
$$
$r_1 + r_2 \to r_2$
$$
\begin{vmatrix}
8 & 12 & 14 & 10\\ 
0 & 0 & 14+4k & 34\\
0 & 0 & 15 & 16
\end{vmatrix}
$$
Uh Oh. $k$ is only a function of $z$
Specifically,
$$
(16/15)(14+4k)=34
$$
But that means (correct me if I am wrong) that there is only a single value for $k$ in which the system is consistent. So asking $k\ne$? doesn't make any sense.
Am I making any sense?
 A: This is an interesting problem, indeed.
You would (and many would) think than any $k$ would work, except one for which the determinant is zero.
Alas, the determinant of $A=\left(\begin{matrix}
4 & 6 & 7\\
-2 & -3 & k\\
8 & 12 & 29
\end{matrix}\right)$ is zero, whatever the value of $k$.
This is obvious by a closer look at the first two columns: they are both multiple of $\left(\begin{matrix}
2\\
-1\\
4
\end{matrix}\right)$.
So, what does it mean? The system has no solution, except for one value of $k$ that makes the system consistent, and that's the value you found.
From your computations, you must have $z=\frac{16}{15}$ and $k=\frac14\left(34\cdot\frac{15}{16}-14\right)=\frac{143}{32}$. Then there are infinitely many solutions for $x,y$, given by the first equation $4x+6y+7z=5$ or $2x+3y=-\frac{37}{30}$: for any $x$, you have a working $y$. The general solution is thus:
$$\left(\begin{matrix}
x\\
-\frac{37}{90}-\frac23x\\
\frac{16}{15}
\end{matrix}\right)$$
By replacing in the equation, you can check that your calculations are right (I did).
Conclusion, you are right, the question does not make much sense. Or it may be on purpose, so that students don't see the trick too early. But this can't be the reason if you must answer something in a web application and there is no way to answer correctly.
