• # question_answer Solve for x and y in the following question. $\frac{2}{x+2y}+\frac{1}{2x-y}+\frac{5}{9}=0,$ $\frac{9}{x+2y}+\frac{6}{2x-y}+4=0$ A)  $x=1,\text{ }y=2$               B)  $x=2,\text{ }y=1$   C)  $x=2,y=\frac{1}{2}$   D)         $x=\frac{1}{2},y=2$

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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