Question

The value of K fir which kc+3y-k+3=0 and 12x+ky=k,have infinite solutions,is..??​

Answer 1

According to the Question:-

Kx+3y-K+3=0                                                             --------------(1)

&,

=>12x+Ky=K

=>12x+Ky-K=0                                                             --------------(2)

Mate,You can see that both of these equations are in the form of

A1X+ B1Y + C1 = 0 and A2X + B2Y + C2 = 0

Where,

A1=K                      AND,            A2=12

B1=3                                           B2=K

C1=-K+3                                     C2= -K

FOR,Infinite solutions we have:

A1/A2 = B1/B2 = C1/C2                                                             --------------(3)

HENCE,By putting value of A1,B1,C1&A2,B2,C2 in Equation (3)

K/12 = 3/K = -K+3/-K

THEN,

K/12 = 3/K = K-3/K

We Get,

=>K/12=3/K & 3/K=K-3/K

=>K²=36 & K²=6K

=>K=√36 & K=6

=>K=±6 & K=6

∵ Both the values are same so main value of K is ±6