Simplify
[tex]E=\dfrac{\frac{x}{y}-\frac{y}{x}}{\frac{1}{y}+\frac{1}{x}}[/tex]
Conditions for the existance of [tex]E:[/tex]
[tex]x\neq 0~~\text{ and}~~y\neq 0~~\text{ and }~~x\neq -y.[/tex]
(otherwise, the one of the denominators would be zero.)
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Simplifying:
[tex]E=\dfrac{\frac{x}{y}-\frac{y}{x}}{\frac{1}{y}+\frac{1}{x}}[/tex]
Multiply both the numerator and the denominator by [tex]xy:[/tex]
[tex]E=\dfrac{xy\cdot \left(\frac{x}{y}-\frac{y}{x} \right )}{xy\cdot \left(\frac{1}{y}+\frac{1}{x} \right )}\\\\\\ E=\dfrac{\frac{xy\cdot x}{y}-\frac{xy\cdot y}{x}}{\frac{xy}{y}+\frac{xy}{x}}\\\\\\ E=\dfrac{x^{2}-y^{2}}{x+y}\\\\\\ E=\dfrac{(x+y)(x-y)}{x+y}[/tex]
Simplify common factors:
[tex]E=x-y\\\\\\ \therefore~~\boxed{\begin{array}{c}\dfrac{\frac{x}{y}-\frac{y}{x}}{\frac{1}{y}+\frac{1}{x}}=x-y \end{array}}[/tex]
