+0

0
104
1
+69

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*}

No need to show your work

Nov 3, 2018
edited by Web2.0CalcUser123  Nov 3, 2018

#1
+94545
+1

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

Nov 3, 2018
edited by CPhill  Nov 3, 2018