A) \[p=2,\,\,q=3\]
B) \[p=15,\,\,q=1\]
C) \[p=15,\,\,q=6\]
D) \[p=5,\,\,q=1\]
Correct Answer: D
Solution :
The given system of equations will have infinite number of solutions, if \[\frac{2}{p+q}=\frac{3}{2p-q}=\frac{7}{21}\] \[\Rightarrow \] \[\frac{2}{p+q}=\frac{3}{2p-q}=\frac{1}{3}\] \[\Rightarrow \] \[\frac{2}{p+q}=\frac{1}{3}\] and \[\frac{3}{2p-q}=\frac{1}{3}\] \[\Rightarrow \]\[p+q=6\]and\[2p-q=9\] \[\Rightarrow \]\[p+q+2p-q=6+9\] \[\Rightarrow \]\[3p=15\Rightarrow p=5\] Putting\[p=5\]in\[p+q\] \[=6\]or\[p-q=0\] We get\[q=1\]
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