Problem 69 - College Algebra

Question

Simplify each complex rational expression. $$\frac{\frac{1}{x+1}}{\frac{1}{x^{2}-2 x-3}+\frac{1}{x-3}}$$

Step-by-Step Solution

Answer

The simplified form of the complex rational expression \(\frac{\frac{1}{x+1}}{\frac{1}{x^{2}-2 x-3}+\frac{1}{x-3}}\) is \(\frac{x-1}{(x+1)(x-2)}\).

Step 1: Simplify the denominator

To simplify the denominator, one has to find the common denominator of the fractions \(\frac{1}{x^{2}-2 x-3}\) and \(\frac{1}{x-3}\). This is found by factoring the quadratic expression as \((x-1)(x+3)\) and identifying that \(x-3\) is already a factor. So, the common denominator is \((x-1)(x+3)\). Multiply each fraction by the missing factors in order to combine them into a single fraction:

Step 2: Combine the fractions in the denominator

Obtain a single fraction in the denominator by adding the fractions, giving \(\frac{1}{x-1}+\frac{x-3}{x-1}= \frac{x-3+1}{x-1}= \frac{x-2}{x-1}\).

Step 3: Simplify complex fraction

Simplify the complex fraction by multiplying the numerator with the reciprocal of the denominator. This yields: \(\frac{\frac{1}{x+1}}{\frac{x-2}{x-1}} = \frac{1}{x+1} * \frac{x-1}{x-2} = \frac{x-1}{(x+1)(x-2)}\). This is the simplified form of the complex fraction.

Step 4: Check result

Check the simplified fraction is equivalent to the original complex fraction by substituting some values for \(x\). Do not use \(x = -1, 2, 1\) since they make denominator equal to zero (undefined). Choose \(x = 0\). Original complex fraction: \(\frac{\frac{1}{1}}{\frac{1}{-3}+\frac{1}{-3}} = 1\). Simplified fraction: \(\frac{-1}{(1)(-2)} = 1\). The results are equal, confirming the simplification is correct.