Problem 68 - College Algebra

Question

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

Step-by-Step Solution

Answer

The simplified complex rational expression is \(\frac{2(x+1)}{(x-1)}\).

Step 1: Simplify the Numerator

First start with the numerator, \(\frac{x}{x-2}+1\). To combine these fractions, they need a common denominator, which is \(x-2\). Rewrite 1 as \(\frac{(x-2)}{(x-2)}\) to get a common denominator. Now the numerator becomes \(\frac{x}{x-2} + \frac{(x-2)}{(x-2)}\), which simplifies to \(\frac{2x - 2}{x - 2}\).

Step 2: Simplify the Denominator

Now simplify the denominator, \(\frac{3}{x^{2}-4}+1\). The denominator \(x^{2}-4\) can be factored as \((x-2)(x+2)\). Rewrite 1 as \(\frac{(x-2)(x+2)}{(x-2)(x+2)}\) to get a common denominator. This gives \(\frac{3 + (x-2)(x+2)}{(x-2)(x+2)}\), which simplifies to \(\frac{3 + x^{2} - 4}{(x-2)(x+2)} = \frac{x^{2}-1}{(x-2)(x+2)}\). Here, \(x^{2}-1\) can also be factored as \((x-1)(x+1)\), so we get \(\frac{(x-1)(x+1)}{(x-2)(x+2)}\).

Step 3: Simplify the Entire Fraction

Now divide the fraction of step 1 by the fraction of step 2, \(\frac{\frac{2x - 2}{x - 2}}{\frac{(x-1)(x+1)}{(x-2)(x+2)}}\). This is the same as \(\frac{2x - 2}{x - 2} \times \frac{(x-2)(x+2)}{(x-1)(x+1)}\). Here, \(x-2\) simplifies out, and the final result is \(\frac{2(x+1)}{(x-1)}\).