A) \[x=1,\text{ }y=2\]
B) \[x=2,\text{ }y=1\]
C) \[x=2,y=\frac{1}{2}\]
D) \[x=\frac{1}{2},y=2\]
Correct Answer: D
Solution :
We have, \[\frac{2}{x+2y}+\frac{1}{2x-y}+\frac{5}{9}=0\] and \[\frac{9}{x+2y}+\frac{6}{2x-y}+4=0\] Let \[\frac{1}{x+2y}=a\] and \[\frac{1}{2x-y}=b\] Thus equations would reduce to \[2a+b=-\frac{5}{9}\] .....(1) and \[9a+6b=-\text{ }4\] .....(2) Solving (1) and (2), we get \[a=\frac{2}{9}\] and \[b=-1\] \[\frac{2}{9}=\frac{1}{x+2y}\] and \[-1=\frac{1}{2x-y}\] \[\Rightarrow \] \[2x+4y=9\] ?..(3) and \[2x-y=-1\] ?..(4) Solving (3) and (4), we get \[y=2\]and \[x=\frac{1}{2}\].
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