Problem 48 - College Algebra Essentials

Question

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

Step-by-Step Solution

Answer

The answer to the exercise is \[ \frac{2x^2 + 18}{(x-3)(x+3)} \]

Step 1: Find the common denominator

Here, the common denominator will be the product of the two individual denominators since they are not identical. Therefore, the common denominator is \((x - 3)(x + 3)\)

Step 2: Rewrite fractions with common denominator

Re-write the two fractions with the common denominator, by multiplying each fraction by the excluded denominator of the other, so we get: \[ \frac{(x+3)(x+3)}{(x-3)(x+3)} + \frac{(x-3)(x-3)}{(x+3)(x-3)} \]

Step 3: Simplify the numerators

We should apply the distributive property (FOIL Method) to simplify the numerators in the fractions we have: \[ \frac{x^2 + 6x + 9}{(x-3)(x+3)} + \frac{x^2 - 6x + 9}{(x+3)(x-3)} \]

Step 4: Add the fractions

Now that the fractions have the same denominators we can add the numerators to get: \[ \frac{x^2 + 6x + 9 + x^2 - 6x + 9}{(x-3)(x+3)} \]

Step 5: Simplify the fraction

Simplify by combining like terms in the numerator to get: \[ \frac{2x^2 + 18}{(x-3)(x+3} \]