Given that k is positive real number.
x + (1 + k)y = 0
(1 - k)x + ky = 1 + k
(1 + k)x + (12 - k)y = -(1 + k)
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Answer:
Given equations are kx+4y=k−4 and 16x+ky=k
Thus, we have
a
1
=k,b
1
=4,c
1
=−(k−4)
a
2
=16,b
2
=k,c
2
=−k
Here condition is
a
2
a
1
=
b
2
b
1
=
c
2
c
1
⇒
16
k
=
k
4
=
(k)
(k−4)
⇒
16
k
=
k
4
Also
k
4
=
k
k−4
⇒ k
2
=64
⇒ 4k=k
2
−4k
⇒ k=±8
⇒ k
2
−8k=0
⇒ k(k−8)=0
k=0 or k=8 but k=0 is not possible other wise equation will be one variable.
⇒ k=8 is correct value for infinite solution.
Step-by-step explanation:
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