Question

Find k using infinitely many solutions : ( k-3) x +3y-k =0
Kx+ky-12=0

Answer 1

Answer:

k = 6

Step-by-step explanation:

The Given pair of linear equation is :

(k - 3 ) x + 3y = k

kx + ky = 12

We can write these Equations as :

(k - 3 ) x + 3y - k = 0……….(1)

kx + ky - 12 = 0 ………….(2)

On comparing with General form of a pair of linear equations in two variables x & y is:

a1x + b1y + c1 = 0

and a2x + b2y + c2= 0

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

a2= k , b2= k , c2= - 12

a1/a2= k-3 /k , b1/b2= 3/k , c1/c2= -k/-12= k/12

Given: A pair of linear equations has a infinite solution, if

a1/a2 = b1/b2 = c1/c2

k-3 /k =3/k= k/12

I = II = III

Taking the first two terms

a1/a2 = b1/b2

k-3 /k =3/k

k - 3 = 3

k = 3 + 3

k = 6

Taking the II and III terms

3/k= k/12

k² = 36

k =√36

k = 6

Hence, the value of k is 6 .

Answer 2

Answer:

the given pair of equations have infinitly many solutions .......so

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

kx + ky - 12 = 0----------- kx + ky = 12

a1 = k-3. b1 = 3. c1 = k

a2 = k. b2 = k. c2 = 12

a1/a2 = b1/b2 = c1/c2

k-3/k = 3/k

k(k-3) = 3k

k^2 - 3k = 3k

k^2 = 3k + 3k

k^2 = 6k

k = 6

therefore Value of k = 6

I think this is the answer..........I hope it helps you