Arraryeldergox2

2022-06-26

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

$x,y\in \mathbb{R},{x}^{2}\ne {y}^{2},xy\ne 0$

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.

$x,y\in \mathbb{R},{x}^{2}\ne {y}^{2},xy\ne 0$

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.

Jake Mcpherson

Beginner2022-06-27Added 23 answers

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

aligass2004yi

Beginner2022-06-28Added 7 answers

First write $\frac{1}{x}+\frac{1}{y}=\frac{y}{xy}+\frac{x}{xy}=\frac{x+y}{xy}$

Then write $\frac{1}{{x}^{2}}-\frac{1}{{y}^{2}}=\frac{{y}^{2}}{{x}^{2}{y}^{2}}-\frac{{x}^{2}}{{x}^{2}{y}^{2}}=\frac{{y}^{2}-{x}^{2}}{{x}^{2}{y}^{2}}=\frac{(y-x)(x+y)}{{x}^{2}{y}^{2}}$

Therefore

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

Then write $\frac{1}{{x}^{2}}-\frac{1}{{y}^{2}}=\frac{{y}^{2}}{{x}^{2}{y}^{2}}-\frac{{x}^{2}}{{x}^{2}{y}^{2}}=\frac{{y}^{2}-{x}^{2}}{{x}^{2}{y}^{2}}=\frac{(y-x)(x+y)}{{x}^{2}{y}^{2}}$

Therefore

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

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