Problem 65 - Algebra A Combined Function

Question

\(\frac{5 a+10}{18} \div \frac{a^{2}-4}{10 a}\)

Step-by-Step Solution

Answer

\( \frac{25a}{9(a-2)} \)

Step 1: Convert Division into Multiplication

When dividing by a fraction, it's equivalent to multiplying by its reciprocal. Hence, the problem \( \frac{\frac{5a+10}{18}}{\frac{a^{2}-4}{10a}} \) becomes \( \frac{5a+10}{18} \times \frac{10a}{a^2-4} \).

Step 2: Factor the Expressions

Factorize the expressions in the fractions: \( 5a + 10 \) becomes \( 5(a+2) \), and \( a^2 - 4 \) becomes \( (a-2)(a+2) \) as it is a difference of squares.

Step 3: Simplify the Fractions

Now we substitute the factored forms into the expression: \( \frac{5(a+2)}{18} \times \frac{10a}{(a-2)(a+2)} \). Cancel the common term \((a+2)\) in the numerator and the denominator.

Step 4: Multiply Simplified Fractions

After canceling, the expression simplifies to \( \frac{5}{18} \times \frac{10a}{(a-2)} \). Multiply the remaining numerators and denominators to get \( \frac{50a}{18(a-2)} \).

Step 5: Simplify the Result

Further simplify \( \frac{50a}{18(a-2)} \) by finding the greatest common divisor of 50 and 18, which is 2. Simplify to \( \frac{25a}{9(a-2)} \).

Key Concepts

Factorization

Factorization is a technique that transforms an algebraic expression into a product of simpler terms. It's used to simplify complex problems by breaking them down into more manageable parts. In our exercise, factorization is applied to the expressions \(5a+10\) and \(a^2-4\). Let's delve deeper into how this works.

• Factor \(5a + 10\): Observe that both terms share a common factor. Here, 5 is common, so we can factor it out, yielding \(5(a + 2)\). This means multiplying out \(5(a + 2)\) gives the original expression, \(5a+10\).
• Factor \(a^2 - 4\): Recognize \(a^2 - 4\) as a special type of expression called a "difference of squares." It is factored using the formula \(a^2 - b^2 = (a-b)(a+b)\). Therefore, \(a^2 - 4\) factors to \((a-2)(a+2)\).

Understanding factorization is crucial in simplifying algebraic fractions, as it allows for the reduction of terms, making manipulation of expressions easier. Always look for common factors or special patterns like difference of squares when approaching a factorization problem.

Difference of Squares

The difference of squares is a specific pattern in algebraic expressions that significantly simplifies many algebraic problems. It's easy to spot because it involves two perfect squares separated by a subtraction sign, like \(a^2 - b^2\). Recognizing this pattern is essential in efficiently breaking down equations.

How do you use it?- Identify two squares: For instance, \(a^2 - 4\) can be seen as \(a^2 - 2^2\).- Apply the formula: Once identified, apply the formula \(a^2 - b^2 = (a+b)(a-b)\) to factorize.

In the exercise, \(a^2 - 4\) is decomposed into \((a-2)(a+2)\). This is possible because 4 is a perfect square (2²), making it a typical difference of squares issue.

• This pattern allows simplification of the problem, helping cancel out terms when the expression is part of a fraction.
• Learning to recognize and factor using this pattern is vital because it recurs frequently in algebra problems, providing a quick solution path.

Simplification

Simplification is the process of reducing an algebraic expression to its simplest form. This involves canceling out terms, reducing fractions, and combining like terms, among other strategies. In our problem, once we've factored the expressions, we move on to simplify the algebraic fraction.

Here's a quick guide on how this works with the given exercise:

• Cancel common terms: After rewriting the fractions in factored form, we identify common terms that appear in both the numerator and the denominator (like \(a+2\) in our problem) and remove them. This significantly simplifies the remaining expression.
• Multiply remaining terms: With the expressions simplified, multiply the remaining terms in the numerators and denominators. For example, \(\frac{5}{18} \times \frac{10a}{(a-2)}\) results in \(\frac{50a}{18(a-2)}\).
• Simplify further: Look for any numerical simplification, such as reducing \(\frac{50a}{18(a-2)}\) to \(\frac{25a}{9(a-2)}\) by dividing both 50 and 18 by their greatest common divisor, 2.

Simplicity isn't just about making numbers smaller. It's about revealing the structure of the problem in a clearer, more manageable form. That's why simplification is such a crucial skill in algebra.