For what value of `k` will the following system of linear equations has no solution? `3x+y=1,\ \ \ \ (2k-1)x+(k-1)y=2k+1`

Explanation

The given linear equations are
` 3x + y- 1 = 0 " " `… (i)
` ( 2k - 1 ) x + ( k - 1 ) y - ( 2k +1) = 0 " " `… (ii)
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 = 3, b_ 1 = 1 , c_ 1 = - 1 `
and ` a_ 2= ( 2k - 1 ), b_ 2 = ( k - 1 ) , c_ 2 = - ( 2k + 1)`
` therefore (a_ 1 ) /(a_ 2) = ( 3 ) /(( 2k - 1)), (b_ 1 ) /(b_2 ) = (1)/((k - 1 )), (c_ 1 ) /(c _ 2 ) = (-1)/(-(2k + 1 )) = (1)/((2k + 1 ))`
Let the given system of equations have no solution.
Then, ` (a_ 1 ) /(a_ 2 ) = (b_ 1 )/(b _ 2 ) ne (c _ 1 ) /(c_ 2 ) `
`rArr ( 3)/((2k - 1 )) = (1)/((k- 1 )) ne (1)/((2k + 1 ))`
`rArr ( 3)/((2k - 1 )) = (1)/((k - 1 )) and ( 1 )/(( k - 1 )) ne (1)/((2k + 1 ))`
` rArr 3k - 3 = 2k - 1 and (1)/((k - 1 )) ne (1)/(( 2k + 1))`
` rArr k = 2 and ( 1) /(( k - 1 )) ne (1)/(( 2k + 1 ))`
Clearly, when ` k = 2`, then ` (1 ) /(( k - 1 )) ne ( 1) /(( 2k + 1 )) ` ` [" as "(1)/(( 2- 1 )) ne (1)/(( 4+ 1 )) ] ` .
Hence, the given system of equations will have no solutions, when k = 2.