Answer 1

The given system of equation may be written as

kx + 2y - 5 = 0

3x + y - 1 = 0

It is of the form

`a_1x + b_1y + c_1 = 0`

`a_2x + b_2y + c_2 = 0`

Where `a_1 = k, b_1 = 2, c_1 = -5`

And `a_2 = 3, b_2 = 1, c_2 = -1`

Where, `a_1 = k, b_1 = 2, c_1 = -5`

And `a_2 = 3, b_2 = 1, c_2 = -1`

1) The given system will have a unique solution, if

`a_1/a_2 != b_1/b_2`

`=> k/3 != 2/1`

`=> k != 6`

So, the given system of equations will have a unique solution, if `k != 6`

2) The given system will have no solution, if

`a_1/a_2 - b_1/b_2 != c_1/c_2`

we have

`b_1/b_2 = 2/1 and c_1/c_2 = (-5)/(-1) = 5/1`

Clearly `b_1/b_2 != c_1/c_2`

So, the given system of equations will have no solution, if

`a_1/a_2 = b_1/b_2`

`=> k/3 = 2/1`

=> k = 6

Hence, the given system of equations will have no solution if k = 6