which expression is equivalent to the following complex fraction?\n$\frac{\frac{1}{x}-\frac{1}{y}}{\frac{1}{x…

which expression is equivalent to the following complex fraction?\n$\frac{\frac{1}{x}-\frac{1}{y}}{\frac{1}{x}+\frac{1}{y}}$\n$\frac{y + x}{y - x}$\n$\frac{(y - x)(y + x)}{x^{2}y^{2}}$\n$\frac{x^{2}y^{2}}{(y + x)(y + x)}$\n$\frac{y - x}{y + x}$
Answer
Explanation:
Step1: Simplify numerator
Find a common - denominator for $\frac{1}{x}-\frac{1}{y}$. The common denominator of $x$ and $y$ is $xy$. So, $\frac{1}{x}-\frac{1}{y}=\frac{y - x}{xy}$.
Step2: Simplify denominator
Find a common - denominator for $\frac{1}{x}+\frac{1}{y}$. The common denominator of $x$ and $y$ is $xy$. So, $\frac{1}{x}+\frac{1}{y}=\frac{y + x}{xy}$.
Step3: Rewrite the complex fraction
The original complex fraction $\frac{\frac{1}{x}-\frac{1}{y}}{\frac{1}{x}+\frac{1}{y}}$ becomes $\frac{\frac{y - x}{xy}}{\frac{y + x}{xy}}$.
Step4: Divide by a fraction
Dividing by a fraction is the same as multiplying by its reciprocal. So, $\frac{\frac{y - x}{xy}}{\frac{y + x}{xy}}=\frac{y - x}{xy}\times\frac{xy}{y + x}$.
Step5: Cancel out common terms
The $xy$ terms in the numerator and denominator cancel out, leaving $\frac{y - x}{y + x}$.
Answer:
$\frac{y - x}{y + x}$