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Find the value of k that has NO possible solutions\(\begin{align*} x + y + 3z &= 10, \\ -4x + 2y + 5z &= 7, \\ kx + z &= 3 \end{align*}\)

 

Please help this my homework need to get it done quick

No need to show your work

Web2.0CalcUser123  Nov 3, 2018
edited by Web2.0CalcUser123  Nov 3, 2018
 #1
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LOL!!!...it would be hard to intuit this if some work weren't done  !!!!

 

Multiply the  first equation through by -2  and we have

 

-2x - 2y - 6z  = -20       add this to the second equation  and we get

 

-6x - z  = -13    multiply this through by  - 1

 

6x + z  =  13      (1)

 

So..notice that if k  = 6 we have

 

6x + z  = 3   (2)

 

But  (1)  and (2)  are parallel lines that never intersect.....so no solution is possible when  k = 6

 

 

cool cool cool

CPhill  Nov 3, 2018
edited by CPhill  Nov 3, 2018

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