Problem 35 - Elementary and Intermediate Algebra

Question

Simplify. $$ 14 \cdot 19 \div(19 \cdot 14) $$

Step-by-Step Solution

Answer

1

Step 1 - Recognize the structure

Look at the expression: \(14 \times 19 \times (19 \times 14)\). Notice that it is a fraction with multiplication in both the numerator and the denominator.

Step 2 - Simplify the fraction

Simplify the fraction by canceling out the common terms from the numerator and the denominator. \(14 \times 19\) in the numerator and the denominator cancel each other out.

Step 3 - Write the simplified form

After canceling out \(14 \times 19\) in the numerator and the denominator, we are left with just 1.

Key Concepts

Fraction Simplification

Simplifying fractions is an essential algebraic skill. It involves reducing a fraction to its simplest form by eliminating common terms in the numerator and the denominator. In general, if you can identify a factor that is present in both the numerator and the denominator, you can divide these out to simplify the fraction. This process can make calculations easier and help you better understand the relationship between the numbers.

Numerator and Denominator

To simplify any fraction, you must understand its components: the numerator and the denominator. The numerator is the top part of a fraction. For example, in \( \frac{a}{b} \), 'a' is the numerator. The denominator is the bottom part of the fraction, so 'b' is the denominator in this case. Knowing these roles helps you to know what parts of the fraction you're looking to simplify. If the numerator and the denominator share any common factors, they can be canceled out.

Canceling Common Terms

Canceling common terms is the key step in simplifying fractions. If both the numerator and the denominator have a term in common, you can 'cancel' or divide these out to make the fraction simpler. In the given exercise, \( 14 \times 19 \div (19 \times 14) \), both terms share \( 14 \times 19 \). When we cancel \( 14 \times 19 \) in both the numerator and the denominator, we are left with 1. This final value illustrates the power of canceling common terms to bring a complex-looking expression down to something trivially simple.