Problem 69 - College Algebra Essentials

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 given complex rational expression is \(\frac{x-3}{2(x-1)}\).

Step 1: Find Common Denominators in the Denominator

In the denominator, the expressions \(\frac{1}{x^{2}-2x-3}\) and \(\frac{1}{x-3}\) have the different denominators \(x^{2}-2x-3\) and \(x-3\). First, factorize the quadratic expression, which results in \(x^{2}-2x-3 = (x-3)(x+1)\). The common denominator for the two expressions, thus, is \(x^{2}-2x-3 = (x-3)(x+1)\).

Step 2: Perform Addition in the Denominator

Rewrite the expression as \(\frac{\frac{1}{x+1}}{\frac{(x+1)+(x-3)}{(x-3)(x+1)}}\). By adding the numerators, the expression becomes \(\frac{\frac{1}{x+1}}{\frac{2x-2}{(x-3)(x+1)}}\). Thus, simplifying the denominator further equals to \(\frac{\frac{1}{x+1}}{\frac{2(x-1)}{(x-3)(x+1)}}\).

Step 3: Simplify the Complex Fraction

Apply the 'division is multiplication by the reciprocal' rule to the fraction. This leads to \(\frac{1}{x+1} \times \frac{(x-3)(x+1)}{2(x-1)}\). Now, cancel out the common factors to simplify it further to \(\frac{x-3}{2(x-1)}\).