Problem 68 - College Algebra Essentials

Question

Simplify each complex rational expression. $$\frac{\frac{x}{x-2}+1}{\frac{3}{x^{2}-4}+1}$$

Step-by-Step Solution

Answer

After simplifying the complex rational expression, the result is \(\frac{2}{x + 1}\).

Step 1: Recognize common factor

First, recognize that \(x^2 - 4\) in the denominator of the bigger fraction can be factored as \((x-2)(x+2)\). This will help when locating a common denominator.

Step 2: Combine fractions in numerator and denominator

In the top fraction, or the numerator of the big fraction, combine \(\frac{x}{x-2} + 1\) as \(\frac{x + x - 2}{x-2} = \frac{2x - 2}{x - 2}\). In the bottom fraction, or the denominator of the big fraction, combine \(\frac{3}{x^2 - 4} + 1\) as \(\frac{3 + x^2 - 4}{x^2 - 4} = \frac{x^2 - 1}{x^2 - 4}\). The big fraction is now \(\frac{2x - 2}{x - 2}\) over \(\frac{x^2 - 1}{x^2 - 4}\)

Step 3: Flip and Multiply fractions

In order to simplify, compute the division as multiplication of the top fraction, \(\frac{2x - 2}{x-2}\), by the reciprocal of the bottom fraction, changing \(\frac{x^2 - 1}{x^2 - 4}\) to \(\frac{x^2 - 4}{x^2 - 1}\). So the division becomes multiplication: \(\frac{2x - 2}{x - 2} * \frac{x^2 - 4}{x^2 - 1}\).

Step 4: Cancel common terms and simplify

Cancellation can now be done because \((x - 2)\) is a factor in both the numerator and the denominator. This simplifies the expression to \(\frac{2x - 2}{x^2 - 1}\). We can then factor the expressions in both the numerator and the denominator to get: \(\frac{2(x - 1)}{(x - 1)(x + 1)}\). We can then cancel out the \((x - 1)\) term from the numerator and the denominator, simplifying the expression to \(\frac{2}{x + 1}\).