+78 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
+129850 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
