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 - (2 - 3kx) = -15
-6x - 2 + 3kx = -15
6x - 2 - 3kx = 15
6x - 3kx = 1
3x(2 - k) = 17
If 2 - k = 0, there is no value for x that will result in a product of 10,
so k = 2 results in no possible solution.
For every other value of k, there will be a value of x that can produce a product of 10.
So the answer is k = 2.
