Problem 39 - Elementary and Intermediate Algebra

Question

Simplify each complex fraction. See Example 4. $$ \frac{\frac{a}{7}-\frac{7}{a}}{\frac{1}{a}+\frac{1}{7}} $$

Step-by-Step Solution

Answer

The simplified form is \( a - 7 \).

Step 1: Identify the Inner Fractions

The complex fraction has a fraction in the numerator and in the denominator. Identify them: - Numerator: \( \frac{a}{7} - \frac{7}{a} \)- Denominator: \( \frac{1}{a} + \frac{1}{7} \)

Step 2: Find a Common Denominator for the Numerator

For the numerator \( \frac{a}{7} - \frac{7}{a} \), find a common denominator, which is \( 7a \). Rewrite each fraction:- \( \frac{a}{7} = \frac{a \cdot a}{7a} = \frac{a^2}{7a} \)- \( \frac{7}{a} = \frac{7 \cdot 7}{7a} = \frac{49}{7a} \)Combine them: \( \frac{a^2 - 49}{7a} \).

Step 3: Find a Common Denominator for the Denominator

For the denominator \( \frac{1}{a} + \frac{1}{7} \), find a common denominator, which is \( 7a \). Rewrite each fraction:- \( \frac{1}{a} = \frac{7}{7a} \)- \( \frac{1}{7} = \frac{a}{7a} \)Combine them: \( \frac{7 + a}{7a} \).

Step 4: Divide the Numerator by the Denominator

The complex fraction becomes:\[ \frac{\frac{a^2 - 49}{7a}}{\frac{7 + a}{7a}} \]To divide, multiply by the reciprocal of the denominator:\[ \frac{a^2 - 49}{7a} \times \frac{7a}{7 + a} \]The \( 7a \) terms cancel out, resulting in:\[ \frac{a^2 - 49}{7 + a} \].

Step 5: Factor the Numerator

Notice that \( a^2 - 49 \) is a difference of squares, which factors to:\[ (a - 7)(a + 7) \].Substitute back into the fraction to get:\[ \frac{(a - 7)(a + 7)}{7 + a} \].

Step 6: Simplify the Fraction

Since \( 7 + a = a + 7 \), the \( a + 7 \) terms cancel out, leaving:\( a - 7 \).

Key Concepts

Common Denominator

When dealing with complex fractions, one of the most important steps is finding a common denominator. This makes it easier to combine the fractions in both the numerator and the denominator of the complex fraction. In our exercise, we first look at the fractions inside the numerator and denominator separately.

• For the numerator, which involves fractions like \( \frac{a}{7} \) and \( \frac{7}{a} \), the common denominator is the product of 7 and \( a \), giving us \( 7a \).
• Similarly, for the denominator, with terms \( \frac{1}{a} \) and \( \frac{1}{7} \), the common denominator is again \( 7a \).

Using a common denominator allows us to rewrite each fraction so they can be combined into one. For instance, \( \frac{a}{7} = \frac{a^2}{7a} \) and \( \frac{7}{a} = \frac{49}{7a} \). This simplification helps in the process of further operations like addition or subtraction. In the denominator, we rewrite \( \frac{1}{a} \) as \( \frac{7}{7a} \) and \( \frac{1}{7} \) as \( \frac{a}{7a} \), making the expressions compatible for further simplification.

Difference of Squares

The term "difference of squares" refers to a mathematical expression that can be factored into two binomial expressions. It is a useful tool in simplifying expressions, particularly in this exercise. Recognizing the difference of squares makes it easier to simplify the resulting expression.A difference of squares takes the form \( a^2 - b^2 \) and can be factored into \( (a - b)(a + b) \). In our example, after simplifying the numerator of the complex fraction, we end up with \( a^2 - 49 \). The number 49 is a perfect square (since \( 49 = 7^2 \)), which makes this a classic case of the difference of squares.

• We factor \( a^2 - 49 \) as \( (a - 7)(a + 7) \).

This factorization enables us to handle further simplification steps more easily. By expressing the numerator in its factored form, it becomes clearer how parts of the expression can be cancelled out later with the denominator if possible, as seen in the steps of simplifying \( \frac{(a - 7)(a + 7)}{7 + a} \).

Simplifying Fractions

Simplifying fractions is about reducing the expression to its simplest form, making it cleaner and easier to interpret or compute. In this exercise, once we have the complex fraction in a clearer format, simplifying involves canceling out common terms.Once we have reached the expression \( \frac{(a - 7)(a + 7)}{7 + a} \), we notice that \( 7 + a \) is equivalent to \( a + 7 \). This observation allows us to cancel the \( a + 7 \) terms from the numerator and the denominator.

• Canceling these terms simplifies the fraction to just \( a - 7 \).

This final step of simplification demonstrates how recognizing equivalent expressions in the numerator and denominator can greatly reduce complexity. By understanding and applying these principles, we simplify the original cumbersome expression into a concise and manageable result.