Problem 11 - Elementary Algebra

Question

Simplify each algebraic fraction. $$\frac{32 x y^{2} z^{3}}{72 y z^{4}}$$

Step-by-Step Solution

Answer

The simplified fraction is \(\frac{4x}{9z}\).

Step 1: Factor the Numerator and Denominator

First, we need to factor the numerator and the denominator of the fraction. The numerator is \(32 x y^2 z^3\), which can be factored as \(2^5 \cdot x \cdot y^2 \cdot z^3\). The denominator is \(72 y z^4\), which can be factored as \(2^3 \cdot 3^2 \cdot y \cdot z^4\).

Step 2: Cancel Common Factors

Identify and cancel the common factors between the numerator and the denominator. The common factors are \(2^3\), \(y\), and \(z^3\). Removing these from both the numerator and the denominator, we are left with \(2^2 x y z^0\) in the numerator and \(3^2 z\) in the denominator.

Step 3: Simplify Remaining Expression

After canceling the common factors, simplify the remaining expression. The fraction reduces to \(\frac{4x}{9z}\), since \(2^2 = 4\) and \(3^2 = 9\). The \(z^0\) becomes \(1\), making it irrelevant in the numerator, and no \(y\) remains.

Key Concepts

Factorization

Let's dive into the process of factorization, which is crucial when simplifying algebraic fractions. In mathematics, factorization involves splitting an expression into a product of simpler factors that, when multiplied together, give back the original expression.
For instance, consider the algebraic expression within the fraction provided: the numerator is \(32 x y^2 z^3\) and the denominator is \(72 y z^4\).

• The number 32 can be factored into its prime components as \(2^5\), which means \(32 = 2 \times 2 \times 2 \times 2 \times 2\).
• Similarly, 72 can be factorized into \(2^3 \times 3^2\) because \(72 = 2 \times 2 \times 2 \times 3 \times 3\).
• Variables \(y^2\) and \(z^3\) remain as they are because they are already primary factors for this specific problem.

The goal here is to turn a complex expression into smaller, more manageable components. This makes it easier to identify and eliminate what is common on both sides of the fraction. Factorization is the first fundamental step in simplifying algebraic fractions.

Common Factors

Once you've factored both the numerator and the denominator of a fraction, the next step is to identify common factors. These are factors that appear in both parts of the fraction. Understanding common factors is essential in reducing expressions to their simplest form.
Take the fraction we are working with. Having factored it, the common factors between the numerator \(32 x y^2 z^3\) and the denominator \(72 y z^4\) are:\(2^3\), \(y\), and \(z^3\).

• \(2^3\) appears as a factor in both the numerator and the denominator.
• The variable \(y\) is in both expressions, appearing as \(y^2\) in the numerator and \(y^1\) in the denominator.
• \(z^3\) can be found in both the numerator and the denominator, which means it can be cancelled out entirely on both sides.

By identifying these common factors, you pave the way for simplification by cancelling these out, thereby reducing the fraction to a simpler form. Identifying common factors makes the process of simplification straightforward and efficient.

Simplifying Fractions

The ultimate goal in this exercise is to simplify the given algebraic fraction. Simplifying a fraction means making it as simple as possible by eliminating unnecessary factors.
After removing the common factors, our original expression \(\frac{32 x y^2 z^3}{72 y z^4}\) transforms. We are left with \(\frac{2^2 x y z^0}{3^2 z}\).

• \(2^2 = 4\), so the numerator becomes \(4x\).
• \(3^2 = 9\), which turns the denominator into \(9z\).
• The term \(z^0 = 1\), meaning it is effectively removed from the numerator, as multiplying by 1 does not change the value.

Thus, the simplified version of our original fraction is \(\frac{4x}{9z}\). Simplifying makes it more manageable and easier to understand. Moreover, it becomes much more convenient to work with in further analyses or real-world applications. This process enhances clarity and reduces complexity in mathematical expressions, making it a key skill in algebra.