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=[tex] \frac{k+3}{k-12} [/tex] Also y=[tex] \frac{5k}{k-12} [/tex] We can not divide with zero. k-12=0 k=12 For k= 12 equations have no solutions.
