Question

which expression is equivalent to the following complex fraction? \\(\frac{\frac{1}{x} - \frac{1}{a}}{\frac{1}{x} + \frac{1}{a}}\\)

Explanation:

Step1: Simplify numerator and denominator separately

The complex fraction is \(\frac{\frac{1}{x}-\frac{1}{A}}{\frac{1}{x}+\frac{1}{A}}\). First, simplify the numerator \(\frac{1}{x}-\frac{1}{A}\) and the denominator \(\frac{1}{x}+\frac{1}{A}\) by finding a common denominator. The common denominator for \(x\) and \(A\) is \(xA\).

For the numerator: \(\frac{1}{x}-\frac{1}{A}=\frac{A - x}{xA}\) (because \(\frac{1}{x}=\frac{A}{xA}\) and \(\frac{1}{A}=\frac{x}{xA}\), so \(\frac{A}{xA}-\frac{x}{xA}=\frac{A - x}{xA}\))

For the denominator: \(\frac{1}{x}+\frac{1}{A}=\frac{A + x}{xA}\) (because \(\frac{1}{x}=\frac{A}{xA}\) and \(\frac{1}{A}=\frac{x}{xA}\), so \(\frac{A}{xA}+\frac{x}{xA}=\frac{A + x}{xA}\))

Step2: Divide the simplified numerator by the simplified denominator

Now we have \(\frac{\frac{A - x}{xA}}{\frac{A + x}{xA}}\). Dividing by a fraction is the same as multiplying by its reciprocal, so this becomes \(\frac{A - x}{xA}\times\frac{xA}{A + x}\)

Step3: Cancel out common terms

The \(xA\) terms in the numerator and denominator cancel out, leaving \(\frac{A - x}{A + x}\) or we can factor out a negative sign from the numerator to get \(\frac{-(x - A)}{A + x}=\frac{x - A}{- (A + x)}=\frac{A - x}{A + x}\) (either form is correct, but typically we can write it as \(\frac{A - x}{A + x}\) or \(\frac{-(x - A)}{A + x}\))

Answer:

\(\frac{A - x}{A + x}\) (or equivalent forms like \(\frac{-(x - A)}{A + x}\) or \(\frac{x - A}{-A - x}\))