Question

the value of k for which Kx+3y-k+3=0 and 12x+Ky=K have Infinity solutions is​

Answer 1

Step by step explanation :

If it has infinitely many solutions than:

a1/a2=b1/b2=c1/c2

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

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

k/1=3/k=-k+3/-k

By comparing 2nd and 3rd:

$- 3k = - {k}^{2} + 3k$

${k}^{2} = 6k$

By comparing 1st and 2nd :

${k}^{2} = 3$

By comparing 1st and 3rd :

$- {k}^{2} = - k + 3 \\ {k }^{2} - k + 3 = 0$

By splitting this equation we will get the answer.

Hope this will help you