Problem 48 - College Algebra

Question

Add or subtract as indicated. $$\frac{x+3}{x-3}+\frac{x-3}{x+3}$$

Step-by-Step Solution

Answer

The answer is \(\frac{2x^2+18}{x^2-9}\)

Step 1: Multiply the Fractions

Multiply each fraction by the other fraction's denominator over itself. This gives: \(\frac{(x+3)(x+3)}{(x-3)(x+3)}+\frac{(x-3)(x-3)}{(x+3)(x-3)}\). This makes the denominators the same, allowing us to add the fractions in the next step.

Step 2: Simplifying the Fractions and Addition

Now we simplify each fraction and add them, like so: \(\frac{x^2+6x+9}{x^2-9}+\frac{x^2-6x+9}{x^2-9}=\frac{x^2+6x+9+x^2-6x+9}{x^2-9}\). Notice that \(+6x\) and \(-6x\) cancel each other.

Step 3: Final Simplification

Combine like-terms in the numerator to find the final result: \(\frac{2x^2+18}{x^2-9}\).