Question

find the value of k for which the system of linear equation (k-1)+3y= 7 and (k+1)x+6y= 4 have no solution​

Answer 1

Step-by-step explanation:

refer to the attachment

Answer 2

Step-by-step explanation:

k-1)x + 3y = 7 or

(k-1)x + 3y -7 =0 ..(1)

(k+1)x+6y = (5k-1) or

(k+1)x+6y-(5k+1) =0 .....(2)

Such that ,

a1 = (k-1) , b1 = 3 , c1 = -7 and

a2 = (k+1) , b2 = 6 , c2 = -(5k+1)

For no solution ,

a1/a2 = b1/b2 = c1/c2

=> (k-1)/(k+1) = 3/6 = 7/(5k+1)

take ,

(k-1)/(k+1) = 3/6

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

=> 2(k-1) = k+1

=> 2k - 2 = k+1

=> [k = 3]

Hence , value of k is 3