Problem 61 - Elementary and Intermediate Algebra

Question

Simplify each complex fraction. $$ \frac{4-\frac{1}{8 h}}{12+\frac{3}{4 h}} $$

Step-by-Step Solution

Answer

Simplified form is \(\frac{32h - 1}{2(48h + 3)}\).

Step 1: Identify the Complex Fraction

A complex fraction has a fraction in the numerator and/or the denominator. In this case, the complex fraction is \( \frac{4-\frac{1}{8h}}{12+\frac{3}{4h}} \). We need to simplify this by treating both the numerator and the denominator.

Step 2: Simplify the Numerator

The numerator is \(4 - \frac{1}{8h}\). To combine terms, multiply the first term by \(\frac{8h}{8h}\) to have the same denominator: \(\frac{32h}{8h} - \frac{1}{8h} = \frac{32h - 1}{8h}\).

Step 3: Simplify the Denominator

The denominator is \(12 + \frac{3}{4h}\). Similarly, multiply the first term by \(\frac{4h}{4h}\) to have the same denominator: \(\frac{48h}{4h} + \frac{3}{4h} = \frac{48h + 3}{4h}\).

Step 4: Divide the Simplified Numerator by the Simplified Denominator

The expression with both simplified parts is: \(\frac{\frac{32h - 1}{8h}}{\frac{48h + 3}{4h}}\). To divide, multiply the numerator by the reciprocal of the denominator: \(\frac{32h - 1}{8h} \times \frac{4h}{48h + 3}\).

Step 5: Simplify the Resulting Expression

Multiply the numerators together and the denominators: \((32h - 1) \times 4h = (32h - 1)4h\) ; \(8h(48h + 3) = 8h(48h + 3)\). The fraction simplifies to \(\frac{(32h - 1)4h}{8h(48h + 3)}\).

Step 6: Simplify by Cancelling Common Terms

Factor out the constants: divide both the numerator and denominator by \(4h\): \(\frac{32h - 1}{2(48h + 3)}\). Feel free to distribute and simplify further if applicable or needed.

Key Concepts

Simplifying Fractions

Simplifying fractions is an important skill in mathematics as it helps make expressions easier to understand and compute. When we simplify a complex fraction, we are reducing it to its simplest form. This involves finding a common denominator, combining like terms, or cancelling out common factors. Here’s how we approach it in the given complex fraction:

• Look at both the numerator and the denominator separately and aim to express each component as a single fraction.
• Once both are simplified, the next step is dividing the numerator by the denominator, making this fraction as simple as possible.
• Remember, simplifying often involves visualizing the fraction in smaller, more manageable parts.

By using these steps, we transform complex fractions into simpler ones that are much easier to manipulate or evaluate in further calculations.

Numerator and Denominator

In a fraction, the numerator and the denominator are its two foundational components. Understanding their roles is critical for dealing with any kind of fraction, including complex fractions.

• The Numerator: It’s the top part of a fraction that represents how many parts are being considered.
• The Denominator: This is the bottom part which signifies the total number of parts the whole is divided into.

In our exercise, to simplify the complex fraction:

• We first focus on the numerator, which in our case is expressed as a subtraction involving a fraction: \(4 - \frac{1}{8h}\).
• We combine terms by converting them into a single fraction with a common denominator, \(\frac{32h - 1}{8h}\).
• Then we work on the denominator, \(12 + \frac{3}{4h}\), and similarly convert it to \(\frac{48h + 3}{4h}\).

By addressing each part independently, we're able to simplify the entire complex fraction more effectively.

Reciprocal in Division

The concept of reciprocals is a key tool used in simplifying complex fractions. The reciprocal of a fraction is achieved by swapping its numerator and denominator.
In division involving fractions, multiplying by the reciprocal often makes the problem much simpler:

• Instead of dividing by a fraction, you multiply by its reciprocal.
• For our exercise, we took the reciprocal of the denominator \(\frac{48h + 3}{4h}\) which becomes \(\frac{4h}{48h + 3}\) and multiplied it by the simplified numerator \(\frac{32h - 1}{8h}\).

This flips the fraction and converts what might seem like a daunting division problem into a straightforward multiplication one.
Using reciprocals in such manner helps streamline the process of dealing with complex fractions and leads to much cleaner results.