Problem 53 - Precalculus Mathematics for Calculus

Question

Perform the addition or subtraction and simplify. $$\frac{1}{x+3}+\frac{1}{x^{2}-9}$$

Step-by-Step Solution

Answer

\( \frac{x-2}{(x-3)(x+3)} \)

Step 1: Identify the Denominators

The expression involves two fractions: \( \frac{1}{x+3} \) and \( \frac{1}{x^2 - 9} \). The denominators are \( x + 3 \) and \( x^2 - 9 \) respectively.

Step 2: Factor the Denominators

Recognize that \( x^2 - 9 \) is a difference of squares. Therefore, factor it as \( (x-3)(x+3) \). This means we have \( \frac{1}{x+3} + \frac{1}{(x-3)(x+3)} \).

Step 3: Find the Least Common Denominator (LCD)

The least common denominator for the fractions is \((x-3)(x+3)\). This is because the second fraction already has this as its denominator, and the first fraction's denominator \( x+3 \) is part of it.

Step 4: Rewrite Each Fraction with the Common Denominator

Adjust \( \frac{1}{x+3} \) to have the denominator \((x-3)(x+3)\). Multiply the numerator and denominator of the first fraction by \( x-3 \), resulting in \( \frac{x-3}{(x-3)(x+3)} \). The second fraction, \( \frac{1}{(x-3)(x+3)} \), already has the desired denominator.

Step 5: Add the Fractions

Now the expression is \( \frac{x-3}{(x-3)(x+3)} + \frac{1}{(x-3)(x+3)} \). Combine the numerators over the common denominator: \( \frac{x-3+1}{(x-3)(x+3)} = \frac{x-2}{(x-3)(x+3)} \).

Step 6: Simplify the Result

The resulting fraction \( \frac{x-2}{(x-3)(x+3)} \) cannot be simplified further, as there are no common factors between the numerator and the denominator.

Key Concepts

Difference of Squares

The difference of squares is a neat and handy mathematical trick that helps us simplify expressions, including rational expressions like the one in this exercise. An expression that looks like \( a^2 - b^2 \) fits the formula for the difference of squares. This is because it can be factored into two binomials: \( (a-b)(a+b) \).

In our exercise, the expression \( x^2 - 9 \) is a classic example of the difference of squares:

• \( x^2 \) is the square of \( x \).
• \( 9 \) is the square of \( 3 \), since \( 9 = 3^2 \).

Thus, \( x^2 - 9 \) becomes \( (x-3)(x+3) \). Recognizing and applying this can simplify many math problems, making it a valuable technique in your math toolkit.

Least Common Denominator

The least common denominator (LCD) is the smallest polynomial that can serve as the denominator for all fractions involved in solving an expression. Having a common denominator allows us to easily add or subtract rational expressions.

In this problem, we have two denominators: \( x+3 \) and \( (x-3)(x+3) \). The LCD for these fractions needs to encompass each distinct factor present in the denominators.

• The denominator \( x+3 \) is already a factor in \((x-3)(x+3)\).
• Thus, the least common denominator is \( (x-3)(x+3) \).

Using the LCD, we can rewrite the fractions to easily add or subtract them by adjusting the numerators accordingly. Simplifying becomes more straightforward with a single common denominator.

Factorization

Factorization is all about breaking down numbers or expressions into multiples that, when combined, produce the original number or expression. It's a critical step in simplifying complex algebraic expressions.

For the rational expression in the exercise, factoring helps to find the least common denominator. In this case, the expression \( x^2 - 9 \) had to be factored:

• We recognized it as a difference of squares: \( (x-3)(x+3) \).

Factoring isn't just about numbers. It's about identifying patterns and applying known factorizations like trinomials or the difference of squares. By breaking down expressions to their simplest components, we can solve equations more efficiently, and sometimes spot further ways to simplify them.