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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 $k$?

 May 4, 2019
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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

 

cool cool cool

 May 5, 2019

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