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.
+876 $x + y + 3z = 10$
$-4x + 3y + 5z = 7$
$kx + z = 3$
Rearranging the 3rd equation, we have $z = 3 - kz$
Let's multiply the 1st equation by $3$: $3x + 3y + 9z = 30$
Subtract the 2nd equation from the new 1st: $7x + 4z = 27$
Plug $3-kz$ for $z:$ $7x + 4(3 - kz) = 27$
Simplifying, we have: $(7-4k)x = 15$
Note: If $7 - 4k = (- \infty, 0) \cup (0, \infty)$ then there will always be a solution for $x,$ and following, $y$ and $z$
Thus, $7-4k$ has to equal $0$ for there to be no solutions for $x,$ which will have no solutions also for $y$ and $z$
We have $7 - 4k = 0$
$4k = 7$
$k = \boxed{\frac{7}{4}}$
