for what values of `k` will the following pairs of linear equations have infinitely many solutions `kx+3y-(k-3)=0` and `12x+ky-k=0`

Explanation

Comparing the given equations with
`a_(1)x + b_(1)y + c_(1) = 0`
and `a_(2)x + b_(2)y + c_(2) = 0`
We have
`(a_(1))/(a_(2)) = (k)/(12), (b_(1))/(b_(2)) = (3)/(k)` and `(c_(1))/(c_(2)) = (-(k + 3))/(k)`
Now, for coincident lines
`(a_(1))/(a_(2)) = (b_(1))/(b_(2)) = (c_(1))/(c_(2))`
implies `(k)/(12) = (3)/(k) = - ((k+3))/(k)`
`{:("Taking first two expressions, we get"),((k)/(12) = (3)/(k)" ...(1)"),(implies" "k^(2) = 36),(therefore " "k = pm 6 ):}:|{:("Taking last two expressions, we get"),((3)/(k) = - ((k + 3))/(k)" "...(2)),(implies " "3k = - k^(2) - 3k),(implies " "k^(2)+ 6k = 0 ),(implies " " k(k + 6) = 0),(therefore " k = 0 or k = - 6 "):}`
But k = - 6 is a value which satisfies both the equations (1) and (2).
`therefore` k = - 6