Question

For what value of K the equations kx+3y-K=0,
(3k-2)x+6y-(k+2) = 3 have more than one solutions .​

Answer 1

given equation are

kx+3y-k=0

(3k-2) x +6y+ (k+2) = 0

now

a1=k, b1= 3,c1 = -k

a2=3k-2 , b2= 6 , c2 = -k+2

then, we know that in case ofbinfinite sol

a1/a2 =b1/be = c1/c2

k/3k-2 = 3/6 = -k/-k+2

By cross multipaction

2k= 3k-2

2k-3k= -2

-k= -2

k=2