Question

For what value of k will these pairs of simultaneous equations have no solution?

Answer 1

Kx-3y=k
y=4x+1 ( we will put it in 1st equation )
kx - 3 (4x+1)=k
kx - 12 x - 3 =k
kx - 12 x = k+3
x(k-12) = k+3
x=$\frac{k+3}{k-12}$
Also y=$\frac{5k}{k-12}$
We can not divide with zero.
k-12=0
k=12
For k= 12 equations have no solutions.