k का मान ज्ञात कीजिए ताकि निम्न समीकरण युग्म का कोई हल नहीं हो।
`(3k+1)x+3y-2=0`
`(k^(2)+1)x+(k-2)y-5=0`

5x + 2y = 2k , 2 (k + 1 ) x + ky = (3k + 4)

3x - y - 5 = 0 , 6x - 2y + k = 0 ( k ne 0 )

( k - 1 ) x - y = 5 , (k + 1 ) x + ( 1 - k ) y = ( 3k + 1 )

Find the value of k that the following system of linear equation has no solution (3k+1)x+3y-2=0, (k^2+1)x+(k-2)y-5=0 .

Determine the value of k so that the following linear equations have no solution: (3k+1)x+3y-2=0,\ \ \ \ (k^2+1)x+(k-2)y-5=0

Determine the value of k so that the following linear equations have no solution: (3k+1)x+3y-2=0,quad (k^(2)+1)x+(k-2)y-5=0

{:(2x + 3y = 2),((k + 2)x + (2k + 1)y = 2(k - 1)):}

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