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{:((k - 3)x + 3y = k),(kx + ky = 12):}...

`{:((k - 3)x + 3y = k),(kx + ky = 12):}`

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The correct Answer is:
k = 6
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{:(kx + 3y = k - 3),(12x + ky = k):}

{:(kx + 3y = 3),(12x + ky = 6):}

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For what value of 'k' the system of equation kx + 3y = 1 , 12x + ky = 2 has no solution.

Under which one of the following condition does the system of equations kx + y + z = k - 1 x + ky + z = k - 1 x + y + kz = k - 1 have no solution ?