Problem 31 - Elementary Algebra

Question

Find the sum: \(\frac{2}{a-3}+\frac{3}{a+3}+\frac{18}{a^{2}-9}\).

Step-by-Step Solution

Answer

Answer: The sum of the three fractions is \(\frac{5a+15}{(a-3)(a+3)}\).

Step 1: Find the LCM of the denominators

We already have the denominators: \((a-3)\), \((a+3)\), and \((a^{2}-9)\). Notice that \((a^{2}-9)\) is of difference of squares form, so we can factor it as \((a-3)(a+3)\). Therefore, the LCM of the denominators is \((a-3)(a+3)\).

Step 2: Convert each fraction to have the LCM as the denominator

We now need to convert each fraction so they all have the LCM, \((a-3)(a+3)\), as their denominators: For the first fraction: \(\frac{2}{a-3} \cdot \frac{a+3}{a+3} = \frac{2(a+3)}{(a-3)(a+3)}\) For the second fraction: \(\frac{3}{a+3} \cdot \frac{a-3}{a-3} = \frac{3(a-3)}{(a-3)(a+3)}\) For the third fraction: \(\frac{18}{a^{2}-9} = \frac{18}{(a-3)(a+3)}\) Now we have the three fractions as: \(\frac{2(a+3)}{(a-3)(a+3)}\), \(\frac{3(a-3)}{(a-3)(a+3)}\), and \(\frac{18}{(a-3)(a+3)}\).

Step 3: Add the numerators

Adding the numerators of the fractions, we get: \(\frac{2(a+3)+3(a-3)+18}{(a-3)(a+3)}\) Simplify the numerator by distributing the constants and combining like terms: \(\frac{2a+6+3a-9+18}{(a-3)(a+3)} = \frac{5a+15}{(a-3)(a+3)}\)

Step 4: Simplify the final fraction

We now have the final fraction, \(\frac{5a+15}{(a-3)(a+3)}\). It can be further simplified if there is any common factor between the numerator and denominator. However, there are no common factors, so the simplified sum is: \(\frac{5a+15}{(a-3)(a+3)}\)

Key Concepts

Least Common Multiple

When working with algebraic fractions, having a common denominator is crucial for operations like addition or subtraction. To achieve this, we need the least common multiple (LCM) of the denominators. The LCM is the smallest expression that all denominators can divide into without leaving a remainder.

Let's consider the exercise: the denominators are \((a-3)\), \((a+3)\), and \((a^2-9)\). Recognize that \((a^2-9)\) is a difference of squares, which we can factor as \((a-3)(a+3)\). Thus, the LCM is \((a-3)(a+3)\), since:

• Both \((a-3)\) and \((a+3)\) are factors of \((a^2-9)\).
• The product \((a-3)(a+3)\) includes all necessary components from each denominator.

Finding the LCM allows us to rewrite each fraction with the same denominator, enabling us to easily add them later.

Difference of Squares

A difference of squares is an expression of the form \(a^2 - b^2\), which can be factored into \((a-b)(a+b)\). Identifying and factoring these expressions is a key skill in simplifying algebraic fractions.

In the given solution, \(a^2 - 9\) is a classic difference of squares. We can rewrite \(a^2 - 9\) as \((a^2 - 3^2)\), which factors into \((a-3)(a+3)\).

Understanding this concept is essential because:

• It simplifies the process of finding the LCM of the denominators.
• It helps in simplifying expressions later, as we'll see the factors might cancel out with terms in the numerator.

Factoring differences of squares turns complex expressions into simpler, more manageable factors.

Simplifying Expressions

Simplifying expressions is about reducing them to their most basic form without changing their value. For expressions involving fractions, this often involves finding a common denominator and combining like terms in the numerator.

In the exercise:

We first rewrote each term to have the LCM, \((a-3)(a+3)\), as a common denominator.

Next, we focused on the numerators, distributing and combining like terms.

Our target expression became \(\frac{2a+6+3a-9+18}{(a-3)(a+3)}\), which after simplification, resulted in \(\frac{5a+15}{(a-3)(a+3)}\).

Simplifying ensures our final expression is easier to understand and use. Always check if further simplification is possible, such as factoring the numerator to see if any terms cancel out with the denominator. Here, the numerator and denominator share no common factors, so \(\frac{5a+15}{(a-3)(a+3)}\) is as simple as it gets.