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?
+35 I think the answer is 12/19
To get no solutions the variable side must cancel out.
Since the last equation only has variables x and z, we have to cancel the y so
Multiply x+y+3z=10 by 8 to get 8x+8y+24z=80
substract the second equation to get 12x+19z=63
then we have to cancel the z
multiply the last equation by 19 to get 19xz+19z=57
(12x+19z=63)-(19xk+19z=57) to get 12x-19xk=6
the left side of the equation must equal zero, so 12x=19xk
19k=12
divide by 19
k=12/19
