For what value of `k` , will the system of equations `x+2y=5,\ \ \ \ 3x+k y-15=0` has (i) a unique solution? (ii) no solution

Explanation

The given system of equations is
x + 2y - 5 = 0 ….(1)
3x + ky - 15 = 0 ….(2)
Here, `a_(1) = 1, b_(1) = 2, c_(1) = - 5` [from (1)]
`a_(2) = 3, b_(2) = k, c_(2) = - 15` [from (2)]
If the equations have no solution then
`(a_(1))/(a_(2)) = (b_(1))/(b_(2)) ne (c_(1))/(c_(2))`
implies `(1)/(3) = (2)/(k) ne (-5)/(-15)`
implies `(1)/(3) = (2)/(k)` and `(2)/(k) ne (-5)/(-15)`
implies k = 6 and `k ne 6` which is impossible.
Hence, there is no such value of k for which the given system has no solution.