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 equation is
` x + 2y - 5 = 0 " " `… (i)
` 3x + k y - 15 = 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 = 1 , b_ 1 = 2, c_ 1 = - 5 and a_ 2 = 3, b_ 2 = k, c_ 2 = - 15`
` therefore (a_ 1 ) /(a_ 2) = (1)/(3), (b_ 1 ) /( b_ 2 ) = ( 2)/(k) and (c_ 1)/(c_ 2 ) = ( - 5)/( - 15) = (1)/(3)`
Let the given system of equations have no solution.
Then, ` (a_ 1)/(a_ 2 ) = (b_ 1 ) /(b_ 2 ) ne (c_ 1 ) /(c_2 ) `
` rArr (1)/(3) = ( 2)/(k) ne (1)/(3) `
` rArr (1 )/(3) = (2)/(k) and (2)/(k) ne (1)/(3)`
`rArr k = 6 and k = - 6 ` , which is impossible.
Hence, there is no value of k for which the given system of equations has no solution.