+965 For a certain value of $k$, the system
x+y+3z=10,6x+4y+z=18−5y−x,kx+7z=8−14y
has no solutions. What is this value of $k$?
+130466 Rearrange as
x + y + 3z = 10
7x + 9y + z = 18
kx + 14y + 7z = 8
Multiply the second equation through by -3 = -21x -27y -3z = -54
Add to the first
-20x - 26y = -44 (1)
Multiply the second equation by -7 = -49x - 63y -7z = -126
Add to the third
(-49 + k)x -49y = -118 (2)
Multiply (1) by -49 and (2) by -26
980x + 1274y = 2156
-26(-49 + k)x + 1274y = 3068
To have no solution
980 = -26(-49 + k)
980 = 1274 - 26k
26k = 1274 - 980
26k = 294
k = 294 / 26 = 147 / 13
