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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 $k$?

 Apr 6, 2021
 #1
avatar+487 
+1

Hint: one way to define no solutions means that when you simplify everything(without possibly dividing by 0, such as 3x=5x, x=0 is a solution), you are left with two unequal numbers equaling each other, such as $3=5$.

 Apr 6, 2021
 #2
avatar+312 
+2

So, just solve x an z. 

WillBillDillPickle  Apr 6, 2021
 #3
avatar+129849 
+2

Multiply the  first equation by -3

-3x - 3y  - 9z   =   -30

Add this to the  second equation and we get  that

-7x -4z = -22

Mutltiply the  last equation  by  -4

-4kx - 4z  =  -12

 

The  system will have no solutions when

 

-4k  =  -7

k =  -7/-4

k  = 7/4

 

cool cool cool

 Apr 6, 2021
edited by CPhill  Apr 6, 2021
 #4
avatar+487 
+1

Great job @CPhill!

RiemannIntegralzzz  Apr 6, 2021
 #5
avatar+129849 
+1

THX!!!!

 

 

cool cool cool

CPhill  Apr 6, 2021
 #6
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+1

Thanks! Its correct!! :D

 Apr 6, 2021

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