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 x−2y=1,x+ky+4=0 and 2x+y=3 to be consistent, ∣∣
∣∣1−211k−4213∣∣
∣∣=0 ⇒(3k+4)+2(11)+(1−2k)=0 ⇒k=−27