## Elementary Linear Algebra 7th Edition

This system won't have any solutions for $$k=\frac{8}{3}.$$
We need to make these two equations inconsistent by choosing the right $k$. To do this follow the steps below: Step 1: Multiply the first equation by $3$: \begin{align*} 3x+6y+3kz = &18\\ 3x+6y+8z =& 4. \end{align*} If we choose $3k = 8$ i.e. $k=8/3$ then the left sides of both equations will be equal which would mean that their right sides must be equal. Since obviously $18\neq4$ this is impossible so for this particular value of $k$ the system is inconsistent i.e. it has no solutions.