Answer 1

The given system of equation may be written as

x + 2y = 0

2x + ky - 5 = 0

The system of equation is of the form

`a_1x + b_1y + c_1 = 0`

`a_2x + b_2y + c_2 = 0`

Where, `a_1 = 1, b_1 = 2,c_1 = 0`

And `a_2 = 2, b_2 = k, c_2 = -5`

For a unique solution, we must have

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

We have

`a_1/a_2 = 1/2`

`b_1/b_2 = 2/k`

And `c_1/c_2 = 0/(-5)`

Now `a_1/a_2 = b_1/b_2`

`=> 1/2 = 2/k`

`=> k = 4`

Hence, the given system of equations has no solutions, when k = 4