Question

the value of k for which the system of linear equations (k-3)x+3y=k, kx+ky=12 has infinite number of solution is​

Answer 1

Answer:

k=6

Step-by-step explanation:

for the given equations

(k-3)x+3y=k kx+ky=12

we can write both equations as:

(k-3)x+3y-k=0 kx+ky-12=0

we know that for infinite solutions

a1/a2=b1/b2=c1/c2

on comparing equation with a1x+b1y+c1=0 and a2x+b2y+c2=0

therefore, a1=(k-3) , b1=3 , c1= -k

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

putiing these values in the infinite solution relation:

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

taking a1/a2= b1/b2

we get (k-3)/k=3/k

k(k-3)=3k

k²-3k=3k

k²= 3k + 3k

k²=6k

cancelling k on RHS with square of k on LHS

we get the answer

k=6