For a certain value of $k$, the system \begin{align*} x + y + 3z &= 10, \\ -4x + 3y + 5z &= 7, \\ kx + z &= 3 \end{align*}has no solutions. What is this value of $k$?
The third equation has no y-term, so let's eliminate the y-term from the combination of the first two equations.
x + y + 3z = 10 ---> x -2 ---> -2x - 2y - 6z = -20
-4x + 3y + 5z = 7 ---> -4x + 3y + 5z = 7
Add down the columns: -6x - z = -13
Since kx + z = 3 ---> z = 3 - kx
Substituting these: -6x - (3 - kx) = -15
(k - 6)x = -18
If k - 6 = 0, there is no value for x that will result in a product of -18
so k = 6 results in no possible solution.
For every other value of k, there will be a value of x that can produce a product of -18.
So the answer is k = 6.
