A) 0
B) 1
C) 2
D) Infinite
Correct Answer: B
Solution :
[b]: For infinitely many solutions, the two equations must be identical \[\Rightarrow \]\[\frac{k+1}{k}=\frac{8}{k+3}=\frac{4k}{3k-1}\] \[\Rightarrow \]\[(k+1)(k+3)=8k\]and\[8(3k-1)=4k(k+3)\] \[\Rightarrow \]\[{{k}^{2}}-4k+3=0\]and\[{{k}^{2}}-3k+2=0\] By cross multiplication method, \[\frac{{{k}^{2}}}{-8+9}=\frac{k}{3-2}=\frac{1}{-3+4}\] \[\Rightarrow \]\[{{k}^{2}}=1\]and\[k=1\] \[\therefore \]\[k=1\].You need to login to perform this action.
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