Question

Find the values of k for which the system

2x + ky = 1

3x – 5y = 7

will have (i) a unique solution, and (ii) no solution. Is there a value of k for which the system has infinitely many solutions?

Answer 1

The given system of equation may be written as

2x + ky – 1 = 0

3x – 5y – 7 = 0

It is of the form

`a_1x + b_1y + c_1 = 0`

`a_2x + b_2y + c_2 = 0`

where `a_1 = 2, b_1 = k, c_1 = -1`

And `a_2 = 3, b_2 = -5, c_2 = -7`

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

`a_1/a_2 != b_1/b_2`

`=> 2/3 != k/(-5)`

`=> -10 != 3k`

`=> 3k != - 10`

`=> k != (-10)/3`

So, the given system of equations will have a unique solution if `k = (-10)/3`

2) The given system will have no solution, if

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

We have

`a_1/a_2 = b_1/b_2`

`=> 2/3 = k/(-5)`

`=> -10 = 3k`

=> 3k = -10

`=> k = (-10)/3`

We have

`b_1/b_2 = k/(-5) = (-10)/(3 xx -5) = 2/3`

And `c_1/c_2 = (-1)/(-7) = 1/7`

Clearly `b_1/b_2 != c_1/c_2`

So, the given system of equations will have no solution, if `k = (-10)/3`

For the given system to have infinite number of solutions, we must have

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

We have,

`a_1/a_2 = 2/3, b_1/b_2 = k/(-5)`

And `c_1/c_2 = (-1)/(-7) = 1/7`

Clearly `a_1/a_2 != c_1/c_2`

So, whatever be the value of k, we cannot have

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

Hence, there is no value of k, for which the given system of equations has infinitely many solutions.