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Question

If the system of equations x−2y=1,x+ky+4=0 and 2x+y=3 is consistent, then the value of k is:

A
3
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B
9
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C
27
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D
27
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Solution

The correct option is D 27
For the system of equations x2y=1,x+ky+4=0 and 2x+y=3 to be consistent, ∣ ∣1211k4213∣ ∣=0
(3k+4)+2(11)+(12k)=0
k=27

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