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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?

 Aug 3, 2019
 #1
avatar+6045 
+2

when k=6 the solutions for x,y,z all have 0 in the denominator which is a nono

 Aug 3, 2019
 #2
avatar+8652 
+2

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?

 

\(2x+2y+6z=20\\ \underline{-4x+2y+5z=7}\\ 6x{\color{blue}{+z}}=13\)           Thanks for the help heureka!

\(\underline{kx+z=3}\\ 6x-kx=10\\ x=\frac{10}{6-k}\\ \color{blue}k=6\)

 

For the value of k = 6, x is indefinite and the functions have no solution.

laugh  !

 Aug 3, 2019
edited by asinus  Aug 3, 2019
 #3
avatar+23342 
+2

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 \(k\)?

\(\begin{array}{|rcll|} \hline \begin{vmatrix} 1&1&3\\ -4 & 2 & 5\\ k & 0 &1 \\ \end{vmatrix} &=& 0 \\\\ k\begin{vmatrix} 1&3\\ 2 & 5\\ \end{vmatrix} + \begin{vmatrix} 1&1\\ -4 & 2\\ \end{vmatrix} &=& 0 \\\\ k*(5-6)+2+4 &=& 0\\ -k &=& -6 \\ \mathbf{k} &=& \mathbf{6} \\ \hline \end{array}\)

 

laugh

 Aug 3, 2019

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