If the pair of linear equations `(3k+1)x+3y-2 =0` and `(k^2+1)x+(k-2)y-5=0` inconsistent, The value of `k` is

Explanation

The given linear 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 = ( 3k + 1 ), b_1 = 3, c _ 1 = - 2 `
and ` a _2 = ( k ^(2) + 1), b_ 2 = (k - 2 ) , c_ 2 = - 5 `
`therefore ( a_ 1 ) /( a _ 2 ) = ((3k + 1 )) /(( k ^(2) + 1 )) , (b _ 1 ) /(b_ 2) = ( 3 ) /( ( k - 2)) and (c_ 1)/(c _ 2 ) = (-2)/(-5) = ( 2 ) /(5)`
Let the given system of equations have no solution.
Then , ` (a_ 1 )/(a_ 2 ) = ( b_ 1)/(b_ 2 ) ne (c_ 1 )/(c _ 2 ) `
`rArr (( 3k + 1 ))/(( k ^(2) + 1 )) = ( 3) /(( k -2 )) ne ( 2) /( 5)`
` rArr (( 3k + 1 )) /((k ^(2) + 1 )) = ( 3) /(( k - 2 ))" " `... (i) and ` ( 3) /((k - 2 )) ne (2 ) /( 5)" " `...(ii)
Now, ` (( 3k + 1 ))/(( k ^(2)+ 1 )) = ( 3) /((k - 2 )) rArr ( 3k + 1) ( k - 2 ) = 3 ( k ^(2) + 1 ) `
` " " rArr 3k ^(2) - 6k + k - 2 = 3k ^(2) + 3 `
` " " rArr 5k = - 5 rArr k = - 1 ` .
When ` k = - 1 ` , then clearly, ` ( 3)/(( k - 2 )) ne ( 2)/( 5) `.
Hence, the given system of equations has no solution when ` k = - 1 `