For a certain value of $k$, the system \begin{align*} x + y + 3z &= 10, \\ -4x + 2y + 5z &= 7, \\ kx + z &= 3 \end{align*} has no solutions. What is this value of $k$?

Guest May 4, 2019

#1**+1 **

x + y + 3z = 10 multiply through by -2 ⇒ -2x - 2y - 6z = -20 (1)

-4x +2y + 5z = 7 (2)

kx + z = 3 (3)

Add (1) and (2) and we have that -6x - z = -17 multiply through by -1 ⇒ 6x + z = 17 (4)

So....we have that....using (3) qnd (4)

6x + z = 17 and

kx + z = 3

And this will have no solution if k = 6

CPhill May 5, 2019