Question

2x+3y=2. ' (k+2)x-(2k+1)y=3(2k-1)

Answer 1

There is some correction in this question which is as below

2x-3y-7=0, (k+2)x – (2k+1)y - 3(2k-1)=0

2x – 3y – 7 = 0

(k+2)x – (2k+1)y – 3(2k-1) = 0

We know, the equation is consistent with infinitely many solutions, if

= =

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

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

now ,

= , =

= =

= =

On Cross multiplication

2(2k+1) = 3(k+2) 3(6k-3) = 7(2k+1)

4k + 2 = 3k + 6 18k – 9 = 14k + 7

4k-3k = 6-2 18k – 14k = 7 + 9

k = 4 ans 4k = 16

k= 16/4

= 4

The value of k = 4 when the given equations have infinitely many solutions