# Solving $\frac{x+y}{xy}=2$, $\frac{x-y}{xy}=6$

$$\frac{x+y}{xy}=2,\ \ \frac{x-y}{xy}=6$$

I am not understanding how to solve the equation. I tried dividing the whole equation by $xy$, but, that didn't work too. Any hint or help would be much appreciated.

• Take the 1st eqn divide 2nd eqn then apply Componendo and dividendo. Can you take it from here? May 3, 2015 at 12:07

Note that

$$\frac 1y+\frac 1x=2\tag 1$$ $$\frac 1y-\frac 1x=6\tag 2$$ Now $(1)+(2)$ gives you $$\frac 2y=8\Rightarrow y=\frac 14.$$

We have a system of equations:

$$\frac{1}{x}+\frac{1}{y}=2$$ $$\frac{1}{y}-\frac{1}{x}=6$$

From equation 1 we have,

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

putting it in equation 2 we have,

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

Putting $x=-\frac{1}{2}$ in $y=\frac{x}{2x-1}$, we get $y$ in terms of a compound fraction,

$$y=\frac{-\frac{1}{2}}{-\frac{2}{2}-1}$$

$$\implies y=\frac{1}{4}$$

• You can verify them by putting them in original equations and see that whether they satisfies your system or not. I have checked already so I would place a solution set, called System's solution set. $$S=(-\frac{1}{2}, \frac{1}{4})$$ May 3, 2015 at 12:15
• $x+y=2xy$
• $x-y=6xy$

so:

$x-y=6xy=3*2xy= 3(x+y)$

so

$2x=-4y$

and

$x=-2y$

By replacing above equation in the first equation, we have:

$-2y+y=2*(-2y)*y$

so

$y=4y^2$

and finally:

$y=1/4$ and $x=-1/2$

$$\{^{\frac{x+y}{xy}=2}_ {\frac{x-y}{xy}=6}<=>$$ $$x=-\frac{1}{2}, y=\frac{1}{4}$$