Question

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

2x + 3y = 2

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

Answer 1

The given system of the equation may be written as

2x + 3y - 2 = 0

(k + 2)x + (2k + 1)y - 2(k - 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 = -2`

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

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

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

`=> 4k +2  = 3k + 6 and 3k - 3 = 2k +1`

`=> 4k - 3k =  6- 2 and 3k - 3k = 1 +3`

=> k = 4 and k =4

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