Question

The Value Of K For Which Kx+3y-k+3=0 And 12x+ky=k, Have Infinite Solution Is:

Answer 1

It is the right one dude hope it helps

Answer 2

Given Equation is kx + 3y - (k - 3) = 0 

On comparing with a1x + b1y + c1 = 0, we get

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


Given Equation is 12x + ky - k = 0

On comparing with a2x + b2y + c2 = 0

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


Given that the equation has infinite solutions:

= > (a1/a2) = (b1/b2) = (c1/c2)

(i)

(a1/a2) = (b1/b2)

= > (k/12) = (3/k)

= > k^2 = 36

= > k = 6,-6.


(ii) 

(b1/b2) = (c1/c2)

= > (3/k) = -(k - 3)/-k

= > 3k = k(k - 3)

= > 3k = k^2 - 3k

= > k^2 - 3k - 3k = 0

= > k^2 - 6k = 0

= > k^2 = 6k

= > k = 6.


The value of k satisfies both equations.


Therefore the value of k = 6.



Hope this helps!