Question

infinitely many solution,find k
kx+3y-(k-3)=0
12x+ky-k=0​

Answer 1

Answer:

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