 # Basic algebra problem: 1 x </mfrac> + 1 Arraryeldergox2 2022-06-26 Answered
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: $=x+\frac{{x}^{2}}{y}-\frac{{y}^{2}}{x}-y=\frac{x\left(xy\right)+{x}^{3}-{y}^{3}-y\left(xy\right)}{xy}=?$ which doesn't seem to help me much. Halp please.
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$\frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{{x}^{2}}-\frac{1}{{y}^{2}}}=\frac{\frac{1}{x}+\frac{1}{y}}{\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x}-\frac{1}{y}\right)}=\frac{1}{\frac{1}{x}-\frac{1}{y}}=\frac{1}{\frac{y-x}{xy}}=\frac{xy}{y-x}$

We have step-by-step solutions for your answer! aligass2004yi
First write $\frac{1}{x}+\frac{1}{y}=\frac{y}{xy}+\frac{x}{xy}=\frac{x+y}{xy}$
Then write $\frac{1}{{x}^{2}}-\frac{1}{{y}^{2}}=\frac{{y}^{2}}{{x}^{2}{y}^{2}}-\frac{{x}^{2}}{{x}^{2}{y}^{2}}=\frac{{y}^{2}-{x}^{2}}{{x}^{2}{y}^{2}}=\frac{\left(y-x\right)\left(x+y\right)}{{x}^{2}{y}^{2}}$
Therefore
$\frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{{x}^{2}}-\frac{1}{{y}^{2}}}=\frac{\frac{x+y}{xy}}{\frac{\left(y-x\right)\left(x+y\right)}{{x}^{2}{y}^{2}}}=\frac{x+y}{xy}\cdot \frac{{x}^{2}{y}^{2}}{\left(y-x\right)\left(x+y\right)}=\frac{\left(x+y\right){x}^{2}{y}^{2}}{xy\left(x+y\right)\left(y-x\right)}=\frac{xy}{y-x}$

We have step-by-step solutions for your answer!