Question

Find the value of k for which each of the following systems of equations have infinitely many solution:
2x-3y=7(k+2)x-(2k+1)y=3(2k-1)

Answer 1

2x-3y=7(k+2)x-(2k+1)y=3(2k-1)

Answer: k = 4

Answer 2

k = 4

Step-by-step explanation:

Given:  

2x -3y - 7 = 0

(k+2)x -(2k+1)y - 3(2k-1) = 0

The system of equations has infinitely many solutions.

a1 = 2, b1 = -3, c1 = -7

a2 = (k+2), b2 = -(2k+1), c2 = - 3(2k-1)

 So  a1/a2 = b1/b2 = c1/c2

2/(k+2) = -3/-(2k+1)

2(2k + 1) = 3(k+2)

4k + 2 = 3k + 6

Therefore k = 4