For which value (s) of k will the pair of equations
`kx + 3y = k - 3`,
`12x + ky = k`
has no solution ?

Explanation

The given equations are
` kx + 3y + ( 2- k ) = 0 and 12 x + ky - k = 0 `
These equations are of the form
` a_ 1 x + b_1 y + c_ 1 = 0 and a_ 2 x + b_ 2 y + c_ 2 = 0 `,
where ` a_ 1 = k , b_ 1 = 3, c_ 1 = (2 - k ) and a_ 2 = 12, b_ 2 = k, c_ 2 = - k `
` therefore (a_ 1 )/(a _ 2 ) = (k)/( 12) , (b_ 1 ) /(b _ 2 ) = ( 3)/(k) and (c_ 1 ) /(c_ 2 ) = ((2 - k ))/(- k ) = ((k - 2 ))/( k )`
Let the given sytem of equations have no solution.
Thenm ` (a_ 1 ) /(a_ 2 ) = (b_ 1 ) /( b _ 2 ) ne (c_ 1 ) /(c_ 2 )`
` rArr ( k ) /(12) = ( 3 ) /(k ) ne (( k - 2 ))/( k ) `
` rArr (k)/( 12) = (3 ) /(k ) and ( 3 ) /( k ) ne ((k - 2 ))/( k ) `
` rArr k ^(2) = 36 and k ^(2) - 2k ne 3k `
` rArr k ^(2) = 36 and k ^(2) - 5k ne 0 `
`rArr ( k = 6 or k = - 6) and k ( k - 5) ne 0 `
`rArr ( k = 6 or k = - 6) and k ( k - 5) ne 0 `
Case 1. When k = 6
In this case, `k ( k - 5)= 6( 6- 5) = 6xx 1 = 6 ne 0 ` .
Case 2. When `k = - 6`
In this case, `k (k -l 5) = (-6) ( - 6- 5) = (-6) xx (-11) = 66 ne 0`
Thus, in each case, the given system has no solution.
Hence, `k = 6 or k = -6`