Basic algebra problem: $ \frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x^2}-\frac{1}{y^2}} $

Question

Basic algebra problem I can't seem to figure out: $$ \frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x^2}-\frac{1}{y^2}} $$ $x,y \in \mathbb{R}, x^2 \neq y^2, xy\neq0$.

Now I know the result is: $\frac{xy}{y-x}$, but I am not sure how to get it, I get into a mess like this: $=x+\frac{x^2}{y}-\frac{y^2}{x}-y=\frac{x(xy)+x^3-y^3-y(xy)}{xy}=?$ which doesn't seem to help me much. Halp please.

Answer 1

$$\frac { \frac { 1 }{ x } +\frac { 1 }{ y } }{ \frac { 1 }{ x^{ 2 } } -\frac { 1 }{ y^{ 2 } } } =\frac { \frac { 1 }{ x } +\frac { 1 }{ y } }{ \left( \frac { 1 }{ x } +\frac { 1 }{ y } \right) \left( \frac { 1 }{ x } -\frac { 1 }{ y } \right) } =\frac { 1 }{ \frac { 1 }{ x } -\frac { 1 }{ y } } =\frac { 1 }{ \frac { y-x }{ xy } } =\frac { xy }{ y-x } $$

Answer 2

First write $\frac1x+\frac1y = \frac{y}{xy}+\frac{x}{xy} = \frac{x+y}{xy}$.

Then write $\frac1{x^2}-\frac1{y^2} = \frac{y^2}{x^2y^2}-\frac{x^2}{x^2y^2} = \frac{y^2-x^2}{x^2y^2}=\frac{(y-x)(x+y)}{x^2y^2}$.

Therefore $$\frac{\frac1x+\frac1y}{\frac1{x^2}-\frac1{y^2}} = \frac{\frac{x+y}{xy}}{\frac{(y-x)(x+y)}{x^2y^2}} = \frac{x+y}{xy} \cdot \frac{x^2y^2}{(y-x)(x+y)} = \frac{(x+y)x^2y^2}{xy(x+y)(y-x)} = \frac{xy}{y-x}$$

Answer 3

$$a=\frac1x+\frac1y=\frac{x+y}{xy}$$

$$b=\frac1{x^2}-\frac1{y^2}=\frac{y^2-x^2}{(xy)^2}$$

$$\frac{a}b=\frac{(x+y)xy}{x^2-y^2}=\frac{(x+y)xy}{(y-x)(y+x)}$$