Question

For what value of 'K' will the following pair of linear equations have infinitely many solutions? Kx+3y-(K-3)=0 12x+Ky-K=0​

Answer 1

Answer:

k=6

Step-by-step explanation:

Given  

kx+3y−(k−3)=0

Comparing with a  

1

​

x+b  

1

​

y+c1=0

∴a  

1

​

=k, b  

1

​

=3, c=−(k−3)

12x+ky−k=0

Comparing with a  

1

​

x+b  

1

​

y+c1=0

∴a  

1

​

=12, b  

1

​

=k, c=−k

Since equation has infinite number of solutions

So,  

a  

2

​

 

a  

1

​

 

​

=  

b  

2

​

 

b  

1

​

 

​

=  

c  

2

​

 

c  

1

​

 

​

 

12

k

​

=  

k

3

​

=  

k

k−3

​

 

12

k

​

=  

k

3

​

 

k  

2

=12×3

k  

2

=36

k=±6

k

3

​

=  

k

k−3

​

 

3k=k(k−3)

3k=k  

2

−3k

k  

2

−3k−3k=0

k  

2

−6k=0

k(k−6)=0

k=0,6

Therefore, k=6 satisfies both equations

Hence, k=6