For a certain value of k, the system
x + y + 3z = 10,
-4x + 2y + 5z = 7,
kx + z &= 3
has no solutions. What is this value of k?
+128475 x + y + 3z = 10 ⇒ -2x - 2y - 6z = -20 (1)
-4x + 2y + 5z = 7 (2)
kx + z = 3 (3)
Add (1) and (2) and we get that
-6x - z = -13 Multiply through by -1
6x + z = 13 (4)
Note that, in (3) , if k = 6 we have that
6x + z = 3 (5)
But this is impossible because if 6x + z = 13 in (4), it cannot also = 3 in (5)
So....k = 6 gives no solutions
