#### Question

Find the value of *k* for which each of the following system of equations has infinitely many solutions :

2x + (k - 2)y = k

6x + (2k - 1)y - (2k + 5)

#### Solution

The given system of the equation may be written as

2x + (k - 2)y - k = 0

6x + (2k - 1)y - (2k + 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 = 2, b_1 = k - 2, c_1 = -k`

And `a_2 = 6, b_2 = 2k - 1, c_2 = -(2k + 5)`

For a unique solution, we must have

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

`=> 2/6 = (k -2)/(2k - 1) = (-k)/(-2(2k + 5))`

`=> 2/6 = (k -2)/(2k - 1) and (k - 2)/(2k -1) = k/(2k + 5)`

`=> 1/3 = (k -2)/(2k -1) and (k -2)(2k + 5) = k(2k - 1)`

`=> 2k - 1 = 3(k - 2) and 2k^2 + 5k - 4k - 10 = 2k^2 - k`

=> 2k - 3k - 6 and k - 10 = -k

=> 2k - 3k = -6 + 1 and k + k = 10

=> -k = -5 and 2k = 10

`=> k = (-5)/(-1) and k = 10/2`

`=> k = 5 and k = 5`

K = 5 satisfies both the conditions

Hence, the given system of equations will have infinitely many solutions, if k = 5