Problem 5 - Algebra for College Students

Question

Simplify each complex fraction. Use either method. $$\frac{\frac{1}{2}-\frac{1}{3}}{\frac{1}{4}-\frac{1}{5}}$$

Step-by-Step Solution

Answer

\(\frac{10}{3}\)

Step 1: Identify the Numerator and Denominator

The given complex fraction is \(\frac{\frac{1}{2}-\frac{1}{3}}{\frac{1}{4}-\frac{1}{5}}\). The numerator is \(\frac{1}{2}-\frac{1}{3}\) and the denominator is \(\frac{1}{4}-\frac{1}{5}\).

Step 2: Simplify the Numerator

To simplify \(\frac{1}{2}-\frac{1}{3}\), find a common denominator. The common denominator for 2 and 3 is 6. \[\frac{1}{2} = \frac{3}{6} \quad \text{and} \quad \frac{1}{3} = \frac{2}{6}\] Thus, \[\frac{1}{2}-\frac{1}{3}=\frac{3}{6}-\frac{2}{6} = \frac{1}{6}\]

Step 3: Simplify the Denominator

To simplify \(\frac{1}{4}-\frac{1}{5}\), find a common denominator. The common denominator for 4 and 5 is 20. \[\frac{1}{4}=\frac{5}{20} \quad \text{and} \quad \frac{1}{5}=\frac{4}{20}\] Thus, \[\frac{1}{4}-\frac{1}{5}=\frac{5}{20}-\frac{4}{20}=\frac{1}{20}\]

Step 4: Divide the Simplified Numerator by the Simplified Denominator

Now, divide \(\frac{1}{6}\) by \(\frac{1}{20}\). Dividing fractions involves multiplying by the reciprocal: \[\frac{\frac{1}{6}}{\frac{1}{20}} = \frac{1}{6} \cdot \frac{20}{1}=\frac{20}{6}\]

Step 5: Simplify the Final Fraction

Simplify \(\frac{20}{6}\) by dividing the numerator and denominator by their greatest common divisor, which is 2: \[\frac{20}{6}=\frac{20 \div 2}{6 \div 2} = \frac{10}{3}\]

Key Concepts

Fraction Subtraction

Fraction subtraction involves finding the difference between two fractions. You need common denominators to subtract fractions effectively. If the denominators are different, you first have to find a common denominator. For instance, \(\frac{1}{2} - \frac{1}{3}\) requires converting both fractions to have the same denominator. The least common multiple (LCM) of 2 and 3 is 6.

Thus, \(\frac{1}{2} = \frac{3}{6}\) and \(\frac{1}{3} = \frac{2}{6}\).

Now, subtracting these gives:
\(\frac{3}{6} - \frac{2}{6} = \frac{1}{6}\).

Finding a common denominator makes it simpler to handle fraction subtraction.

Common Denominator

A common denominator is a shared multiple of the denominators of two or more fractions. It's crucial for performing arithmetic operations like addition and subtraction.

For example, with the fractions \(\frac{1}{2}\) and \(\frac{1}{3}\), the common denominator is 6. This is because 6 is the smallest number that both 2 and 3 can divide without a remainder.

To convert each fraction:

• \(\frac{1}{2} = \frac{3}{6}\)
• \(\frac{1}{3} = \frac{2}{6}\)

You now have fractions with a common denominator, \(\frac{3}{6} - \frac{2}{6}\). This makes subtraction straightforward.

Similarly, the common denominator for \(\frac{1}{4}\) and \(\frac{1}{5}\) is 20 since it is the smallest number that both 4 and 5 can divide evenly:

• \(\frac{1}{4} = \frac{5}{20}\)
• \(\frac{1}{5} = \frac{4}{20}\)

Now, you can subtract, \(\frac{5}{20} - \frac{4}{20} = \frac{1}{20}\).

Fraction Simplification

Simplifying fractions means reducing them to their lowest possible terms. This happens by dividing the numerator and the denominator by their greatest common divisor (GCD).

In our main example, we reached \(\frac{20}{6}\) after dividing the simplified numerator by the simplified denominator.

The GCD of 20 and 6 is 2. Dividing both by their GCD:

• \(\frac{20}{6} = \frac{20 \(\div\) 2}{6 \(\div\) 2} = \frac{10}{3}\)

This results in the lowest terms of the fraction, simplifying it to \(\frac{10}{3}\).

Reciprocals

Reciprocals are simply the flipped versions of fractions. They are crucial when dividing fractions. In other words, the reciprocal of a fraction \(\frac{a}{b}\) is \(\frac{b}{a}\).

For instance, to divide by \(\frac{1}{20}\), we multiply by its reciprocal, which is 20:

• \(\frac{\frac{1}{6}}{\frac{1}{20}} = \frac{1}{6} \(\times\) 20 = \frac{20}{6}\)

In this case, dividing by \(\frac{1}{20}\) turns into multiplying by 20.

This makes it easier to handle operations involving division of fractions. Understanding reciprocals is key to solving complex fraction problems efficiently.