For a certain value of k, the system
x + y + 3z = 10
-4x + 8y + 5z = 7
kx + z = 3
has no solutions. What is this value of k?
+34 x + y + 3z = 10 → y = 10 - x - 3z
(we want to get y by itself0
-4x + 2y +5z = 7 → -4x + 2(10 - x - 3z) + 5z = 7 → -4x + 20 -2x - 6z + 5z = 7 → -6x - z = -13 →
6x + z = 13
(substitute y into the second equation)
Notice we have this system now
k*x + z = 3
6x + z = 13
And notice that, if k =6, we would have this system:
6x + z = 3 and
6x + z = 13 but this is impossible, because two things added together can't possibly have two different outcomes
when k = 6, the system has no solution.
