A) \[6\]
B) \[-4\]
C) \[4\]
D) \[-6\]
Correct Answer: C
Solution :
For infinitely many solutions, |
\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}\Rightarrow \frac{2}{k+2}=\frac{3}{2k+1}=\frac{2}{2(k-1)}\] |
\[\Rightarrow \,\,\,\,\frac{2}{k+2}=\frac{3}{2k+1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{2}{k+2}=\frac{2}{2(k-1)}\] |
\[\Rightarrow \,\,\,\,4k+2=3k+6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,2k-2=k+2\] |
\[\Rightarrow \,\,\,\,k=4\,\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \,k=4\] |
\[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,k=4\] |
So, option [c] is correct. |
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