Question

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

2x − 3y = 7

(k + 2)x − (2k + 1)y − 3(2k − 1)

Answer 1

The given system of the equation may be written as

2x − 3y − 7 = 0

(k + 2)x − (2k + 1)y − 3(2k − 1) = 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 = -3, c_1 = -7`

And

`a_2 = k, b_2 = -(2k + 1), c_2 = -3(2k - 1)`

For a unique solution, we must have

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

`=> 2/(k + 2) = 3/(-(2k + 1)) = (-7)/(-3(2k -1))`

`=> 2/(k+1) = (-3)/(-(2k + 1)) and (-3)/(-(2k + 1)) = (-7)/(-3(2k - 1))` 

`=> 2(2k + 1) =3(k+1)` and `3 xx 3 (2k - 1) = 7(2k + 1)`

`=> 4k + 2 = 3k + 6 and 15k - 9 = 14k + 7`

`=> 4k - 3k = 6 - 2 and 15k - 14k = 7 + 9`

`=> k = 4 and 4k = 16 => k = 4`

`=>k = 4 and k = 4`

Hence, the given system of equations will have infinitely many solutions, if k = 4