Question

find the value of k for which the following system of equation has infinitely many solutions(K-1)x+3y=7 and(k+1)x+6y=(5k-1)

Answer 1

Your answer is --


Given, equations are

(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 infinite may 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


【 Hope it helps you 】