Problem 80 - Intermediate Algebra

Question

Perform the operations and simplify the result when possible. $$\frac{y^{2}-3 x y}{x^{3}-y^{3}}-\frac{x^{2}+4 x y}{y^{3}-x^{3}}$$

Step-by-Step Solution

Answer

\(\frac{y^2 - 7xy - x^2}{x^3 - y^3}\), no further simplification possible.

Step 1: Identify the Common Denominator

The denominators of both fractions are \(x^3 - y^3\) and \(y^3 - x^3\). Notice that \(y^3 - x^3 = -(x^3 - y^3)\). Thus, a common denominator for both fractions is \(-1(x^3 - y^3)\).

Step 2: Combine Fractions

Rewrite the subtraction as a single fraction using the common denominator: \[ \frac{y^2 - 3xy}{x^3 - y^3} - \frac{x^2 + 4xy}{y^3 - x^3} = \frac{y^2 - 3xy}{x^3 - y^3} + \frac{x^2 + 4xy}{-(x^3 - y^3)} = \frac{y^2 - 3xy - (x^2 + 4xy)}{x^3 - y^3} \]

Step 3: Simplify the Numerator

Now simplify the expression in the numerator by combining like terms:\[ y^2 - 3xy - x^2 - 4xy = y^2 - 3xy - 4xy - x^2 = y^2 - 7xy - x^2 \]

Step 4: Write the Resulting Fraction

Place the simplified numerator over the common denominator:\[ \frac{y^2 - 7xy - x^2}{x^3 - y^3} \]

Step 5: Check for Further Simplification

Since there are no common factors in the numerator \(y^2 - 7xy - x^2\) and the denominator \(x^3 - y^3\), the fraction cannot be simplified further.

Key Concepts

Common Denominator

When working with polynomial fractions, finding a common denominator is essential. It allows you to combine the fractions into a single fraction, making it easier to perform operations like addition or subtraction. The denominators in the original exercise are \(x^3 - y^3\) and \(y^3 - x^3\). You can see that \(y^3 - x^3\) is actually the negative of \(x^3 - y^3\), which gives them a special relationship.

To unify these, we recognize that a common denominator can be expressed as \(-1(x^3 - y^3)\). This technique ensures both terms have compatible denominators, paving the way for straightforward algebraic operations. By manipulating the fractions to have this common denominator, they can be effectively paired for addition or subtraction. This method is crucial in resolving polynomial fraction challenges because it simplifies the entire process of finding a single, unified expression.

Simplifying Expressions

Simplifying expressions is a key step in handling polynomial fractions. Once common denominators are identified and fractions are combined, you need to simplify the resultant expressions. This means rearranging and reducing terms where possible.

In the given solution, the combined fraction involved simplifying the numerator \(y^2 - 3xy - (x^2 + 4xy)\). Breaking this down involves distributing the negative sign and combining terms:

• First, expand: \(y^2 - 3xy - x^2 - 4xy\).
• Then combine liked terms: \(y^2 - (3xy + 4xy) - x^2\) simplifies to \(y^2 - 7xy - x^2\).

Effective simplification reduces complexity and aids in identifying further steps in solving a problem, such as factoring or recognizing irreducible terms. Therefore, mastering simplification enhances algebraic fluency.

Algebraic Operations

Algebraic operations such as addition and subtraction are fundamental tools in manipulating polynomial fractions. Once a common denominator is obtained, these operations facilitate combining fractions into a single expression. Let’s break this down step by step:

1. **Addition/Subtraction**: Once the expressions have the same denominator, you add or subtract the numerators directly. In our example:

• The subtraction turns into just another simplification of terms when we recognize the role of the negative sign with \(-1(x^3 - y^3)\).
• This results in assembling them under a nearer common roof.

2. **Combining Like Terms**: Post operations, merging like terms in the numerator ensures the expression remains as simple as possible, thereby facilitating any future operations on it.

3. **Verifying Simplicity**: The last step in our original solution, checking if further simplification is feasible, confirms you're dealing with the simplest form of the expression.

Effective use of algebraic operations streamlines solving polynomial problems and connects various concepts in algebra, reinforcing a deeper understanding of the subject.