infinitely many solution,find k
kx+3y-(k-3)=0
12x+ky-k=0
Answers
Answered by
2
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
Similar questions