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Question

The number of values of k, for which the system of equations:
(k+1)x+8y=4k
kx+(k+3)y=3k1
has no solution, is


A
infinite
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B
1
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C
2
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D
3
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Solution

The correct option is A 1
(k+1)x+8y=4k...(1)
kx+(k+3)y=3k1...(2)
k+1k=8k+34k3k1
Now, k+1k=8k+3
k24k+3=0
k=1,3
If k=1
81+3=431
2=2
k=1 is not the correct value
So, if k=3
83+31291
4332
k=3 is the only value for which system of equations has no solution
Hence, number of values of k is 1.

Mathematics

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