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