Find the values(s) of `k` for which the system of equations `k x-y=2` `6x-2y=3` has (i) a unique solution (ii) no solution. Is there a value of `k` for which the system has infinitely many solutions?

Explanation

The given system of equations is
` k x - y - 2 = 0, 6x - 2y - 3 = 0 `.
This is 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 = - 1, c _ 1 = - 2 and a_ 2 = 6, b_ 2 = - 2, c_ 2 = - 3.`
(i) For a unique solution, we must have ` (a_ 1 ) /(a_ 2) ne(b_ 1 ) /( b_ 2 )`
` therefore (k ) /( 6) ne (-1) /( - 2) rArr ( k ) /( 6) ne (1)/(2) rArr k ne 3 `.
Hence, the given system of equations will have a unique solution when ` k ne 3 `.
(ii) For no solution, we must have ` (a_ 1 ) /(a_ 2 ) = ( b_ 1 ) /(b _ 2 ) ne (c_ 1 )/(c _ 2 )`
` therefore ( k ) /( 6) = (-1)/(-2) ne (-2)/(-3) `
` rArr (k)/( 6) = (1)/(2) ne ( 2 ) /(3 )`
` rArr ( k ) /( 6) = (1)/(2) and (k ) /( 6)ne ( 2) /(3) rArr k = 3 and k =ne 4 `.
Clearly, k = 3 also satsfies the condition `k ne 4`
Hence, the given system of equations will have no solution when k = 3.
(iii) For infinitely many solutions, we must have
`(a_ 1)/(a_ 2 ) = (b_ 1 ) /(b_ 2 ) = (c_ 1 ) /(c_ 2 ) , `
i.e., ` ( k ) /( 6) = (1)/(2) = (1)/(3), ` which is never possible, as `(1)/(2) ne (1)/(3)` .
Hence, there is no real value of k for which the given system of equations has infinitely many solutions.