Question

the value of k for which the equations (3k+1)x+3y=2 and( 2k+1)x+(k-2)y=5 has no solution ,then the value of k is

2

-2

-1


answer please
as soon as possible ​

Answer 1

Ans: -1

Step-by-step explanation:

(3k+1)x+3y-2=0 and (k^2+1)x+ (k-2)y-5=0

This is of the form

a1x+b1y+c1=0, a2x+b2y+c2=0

where a1=3k+1,b1=3,c1=-2 & a2=k^2+1, b2=k-2,c2=-5

For no solution we must have

a1/a2=b1/b2=c1/c2

by solving we get k= -1