A) \[x=\frac{2}{5},y=1\]
B) \[x=\frac{1}{2},y=\frac{1}{3}\]
C) \[x=\frac{1}{5},y=1\]
D) \[x=\frac{2}{3},y=-3\]
Correct Answer: C
Solution :
| [c] The given system of equations is |
| \[\frac{2}{x}+\frac{3}{y}=13\] and \[\frac{3}{x}+\frac{6}{y}=21\] |
| Putting \[\frac{1}{x}=p\] and \[\frac{1}{y}=q,\] we get |
| \[2p+3q=13\] ...(1) |
| and \[3p+6q=21\Rightarrow p+2q=7\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p=7-2q\] ...(2) |
| Substituting the value of p in eq. (1), we get |
| \[2(7-2q)+3q=13\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,14-4q+3q=13\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,q=1\] |
| Substituting \[q=1\] in eq. (2), we get |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p=7-2(1)=5\] |
| Now \[\frac{1}{x}=5\Rightarrow x=\frac{1}{5}\] and \[\frac{1}{y}=1\Rightarrow y=1\] |
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