A) 0
B) 1
C) 2
D) Infinite
Correct Answer: B
Solution :
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, \[\frac{{{k}^{2}}}{-8+9}=\frac{k}{3-2}=\frac{1}{-3+4}\] \[{{k}^{2}}=1\] and \[k=1\]; \ \[k=1\].
You need to login to perform this action.
You will be redirected in
3 sec
