Question

for what value of k will the following pair of linear equations have infinitely many solutions?

(a) (k-3)x+3y=k

(b) Kx+Ky=12

Answer 1

Answer:

Given 

kx+3y−(k−3)=0

Comparing with a1x+b1y+c1=0

∴a1=k, b1=3, c=−(k−3)

12x+ky−k=0

Comparing with a1x+b1y+c1=0

∴a1=12, b1=k, c=−k

Since equation has infinite number of solutions

So, a2a1=b2b1=c2c1

12k=k3=kk−3

12k=k3

Step-by-step explanation:

Hope it helps you