Problem 23 - Elementary Algebra

Question

Simplify the complex fraction \(\frac{1-\frac{5}{x}-\frac{6}{x^{2}}}{1+\frac{6}{x}+\frac{5}{x^{2}}}\).

Step-by-Step Solution

Answer

Question: Simplify the complex fraction \(\frac{1-\frac{5}{x}-\frac{6}{x^{2}}}{1+\frac{6}{x}+\frac{5}{x^{2}}}\). Answer: \(\frac{x-6}{x+5}\)

Step 1: Identify the common denominator within the numerator and the denominator

Observe that the common denominator in both the numerator and the denominator is \(x^2\).

Step 2: Multiply the complex fraction by \(x^2\) to clear the fractions

We have \(\frac{1-\frac{5}{x}-\frac{6}{x^{2}}}{1+\frac{6}{x}+\frac{5}{x^{2}}} \cdot \frac{x^2}{x^2}\).

Step 3: Distribute the \(x^2\) to both the numerator and the denominator

Apply the distributive property and perform the multiplication: \(\frac{x^2(1-\frac{5}{x}-\frac{6}{x^{2}})}{x^2(1+\frac{6}{x}+\frac{5}{x^{2}})} = \frac{x^2-5x-6}{x^2+6x+5}\).

Step 4: Factor the numerator and the denominator

Factor the numerator and the denominator as follows: \(\frac{(x-6)(x+1)}{(x+1)(x+5)}\).

Step 5: Cancel out common factors

Cancel out the common factor \((x+1)\) from both the numerator and the denominator to get the final simplified fraction: \(\frac{x-6}{x+5}\).

Key Concepts

Distributive Property

The distributive property is a key tool in algebra that allows you to simplify expressions by spreading a factor across terms in a parenthesis. In our example, we have the complex fraction \( \frac{1-\frac{5}{x}-\frac{6}{x^{2}}}{1+\frac{6}{x}+\frac{5}{x^{2}}} \). To clear these fractions, we multiply both the numerator and the denominator by the common denominator, which is \( x^2 \).

This multiplication involves distributing \( x^2 \) to each term inside the brackets. In the numerator, the expression becomes \( x^2 \times 1 - x^2 \times \frac{5}{x} - x^2 \times \frac{6}{x^2} \). Similarly, distribute \( x^2 \) in the denominator.

Using the distributive property helps eliminate the fractions, making it easier to manage and simplify the expression further.

Factoring Polynomials

Factoring is the process of breaking down a polynomial into simpler components called factors. Once the fractions are cleared in our expression, we're left with \( x^2 - 5x - 6 \) in the numerator and \( x^2 + 6x + 5 \) in the denominator.

To factor these polynomials, we look for pairs of numbers that multiply to the constant term and add to the linear coefficient:

• The numerator \( x^2 - 5x - 6 \) factors into \((x - 6)(x + 1)\).
• The denominator \( x^2 + 6x + 5 \) factors into \((x + 1)(x + 5)\).

This factorization transforms complex expressions into products of simpler expressions, paving the way for further simplification.

Common Denominator

Finding a common denominator is crucial when dealing with complex fractions, as it allows you to combine or eliminate fractions. In our problem, both the numerator and the denominator had terms with denominators of \( x \) and \( x^2 \).

The least common denominator in this case is \( x^2 \). To eliminate the fractions entirely, we multiplied the entire fraction by \( \frac{x^2}{x^2} \), which is effectively multiplying by 1.

This didn’t change the value of the fraction but simplified the operation.

By using a common denominator, we end up with whole numbers and polynomials, making further manipulation and simplification much more straightforward.

Cancelling Common Factors

Once the fractions have been simplified and factored, the last step is to cancel out any common factors in the numerator and the denominator. In our example, after factoring, we have \( \frac{(x-6)(x+1)}{(x+1)(x+5)} \).

We notice that \((x + 1)\) is a common factor in both the numerator and the denominator. By cancelling \((x + 1)\), the expression simplifies to \( \frac{x-6}{x+5} \).

This step reduces the fraction to its simplest form, stripping away the complexity that existed initially. It's important to only cancel terms that are exactly the same, ensuring you maintain the equality of the expression.