Question

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. $x=1,y=2$
• B. $x=2,y=1$
• C. $x=2,y=\frac{1}{2}$
• D. $x=\frac{1}{2},y=2$

Answer 1

The correct option is D $x=\frac{1}{2},y=2$

$\frac{2}{x+2y}+\frac{1}{2x-y}+\frac{5}{9}=\mathrm{0......}\left(1\right)$

$\frac{9}{x+2y}+\frac{6}{2x-y}+4=\mathrm{0......}\left(2\right)$

Let $\frac{1}{x+2y}=a$ and $\frac{1}{2x-y}=b$

Substituting the assumed values in equations $1$ and $2$, we get:

$2a+b=\frac{-5}{9}$ or

$18a+9b=-\mathrm{5......}\left(3\right)$

and

$9a+6b=-\mathrm{4......}\left(4\right)$

Solving equations $\left(3\right)$ and $\left(4\right)$ using elimination method:

Multiplying equation $\left(4\right)$ by $2$ and subtracting $\left(3\right)$ from it, we get

$3b=-3$ or

$b=-1$

$⟹2x-y=-1......\left(5\right)$

Substituting $b=-1$ in $\left(4\right)$

$9a-6=-4$

$a=\frac{2}{9}$

$\Rightarrow x+2y=\frac{9}{2}......\left(6\right)$

Solving equations $\left(5\right)$ and $\left(6\right)$ using elimination method:

Multiplying equation $\left(6\right)$ by $2$ and subtracting $\left(5\right)$ from it, we get

$2x+4y-2x+y=9+1$

$\therefore y=2$

Substituting in $\left(5\right)$, we get

$x=\frac{1}{2}$

$y=2x+1=2.\frac{1}{2}+1=2$