Problem 59 - Intermediate Algebra

Question

Simplify each complex fraction. $$ \frac{y}{x^{-1}-y^{-1}} $$

Step-by-Step Solution

Answer

\[ \frac{y^2x}{y-x} \]

Step 1: Rewrite Negative Exponents

Rewrite any negative exponents into fractions. Recall that a negative exponent indicates the reciprocal of the base raised to the positive of that exponent. Thus, replace \(x^{-1}\) with \(\frac{1}{x}\) and \(y^{-1}\) with \(\frac{1}{y}\). Now, our complex fraction becomes: \[ \frac{y}{\frac{1}{x} - \frac{1}{y}}. \]

Step 2: Find the Common Denominator

The denominator \(\frac{1}{x} - \frac{1}{y}\) needs to be simplified by finding a common denominator. The common denominator for fractions \(\frac{1}{x}\) and \(\frac{1}{y}\) is \(xy\). Rewrite each fraction using this common denominator: \[ \frac{1}{x} = \frac{y}{xy}, \quad \frac{1}{y} = \frac{x}{xy}. \] Thus, the expression becomes: \[ \frac{y}{\frac{y}{xy} - \frac{x}{xy}}. \]

Step 3: Simplify the Denominator

Subtract the fractions in the denominator: \[ \frac{y}{xy} - \frac{x}{xy} = \frac{y-x}{xy}. \] Now, the complex fraction is: \[ \frac{y}{\frac{y-x}{xy}}. \]

Step 4: Simplify the Complex Fraction

When simplifying a complex fraction, multiply by the reciprocal of the denominator: \[ \frac{y}{\frac{y-x}{xy}} = y \times \frac{xy}{y-x}. \] This results in: \[ = \frac{y \cdot xy}{y-x}. \] Simplify the numerator: \[ y \cdot xy = y^2x. \] Thus, we have: \[ \frac{y^2x}{y-x}. \]

Key Concepts

Negative Exponents

Negative exponents often make expressions look complicated, but they are quite manageable when understood. A negative exponent, like \(x^{-1}\), simply means the reciprocal of the base raised to the positive version of that exponent. For example, \(x^{-1}\) can be rewritten as \(\frac{1}{x}\). This transformation is crucial when dealing with algebraic expressions, especially in complex fractions.

Here's why: transforming negative exponents into fractions simplifies the expression and prepares it for further operations, such as finding a common denominator. In the original exercise, converting negative exponents \(x^{-1}\) and \(y^{-1}\) to \(\frac{1}{x}\) and \(\frac{1}{y}\) was a key step in moving towards the simplification of the complex fraction \(\frac{y}{x^{-1} - y^{-1}}\).

Remember:

• \(x^{-n} = \frac{1}{x^n}\)
• Simplify your equations by changing negative exponents to positive by using reciprocals.

These steps make the rest of the simplification process much more straightforward.

Common Denominator

Finding a common denominator is a pivotal step in simplifying expressions involving fractions. It allows you to combine and simplify fractions to a single expression. The common denominator is essentially the least common multiple (LCM) of the denominators you are working with.

In the problem, fractions \(\frac{1}{x}\) and \(\frac{1}{y}\) needed a common denominator to be subtracted from each other. For these, the LCM is simply \(xy\). Once identified, each fraction can be rewritten:

• Convert \(\frac{1}{x}\) to \(\frac{y}{xy}\)
• Convert \(\frac{1}{y}\) to \(\frac{x}{xy}\)
• This allows for the straightforward subtraction: \(\frac{y}{xy} - \frac{x}{xy} = \frac{y-x}{xy}\)

Adhering to these conversions makes combining fractions seamless, and it simplifies more complex algebraic manipulations.

Always remember, identifying and applying a common denominator simplifies the arithmetic of combining fractions, especially when moving towards solutions.

Reciprocal Multiplication

Reciprocal multiplication is a method commonly used to simplify complex fractions. When dealing with a complex fraction like \(\frac{y}{\frac{y-x}{xy}}\), you can simplify it significantly by multiplying by the reciprocal of the denominator.

The reciprocal of a fraction \(\frac{a}{b}\) is simply \(\frac{b}{a}\). Thus, for \(\frac{y}{\frac{y-x}{xy}}\), the reciprocal of the denominator \(\frac{y-x}{xy}\) is \(\frac{xy}{y-x}\). Multiplying by this reciprocal helps to eliminate the complex fraction:

• Multiply: \(y \times \frac{xy}{y-x} = \frac{y \cdot xy}{y-x}\)
• Simplify: \(y \cdot xy\) becomes \(y^2x\)
• Result: \(\frac{y^2x}{y-x}\)

Using reciprocal multiplication not only simplifies the fraction but helps to clearly see the solution, avoiding convolution. This technique is particularly useful in algebra, where expressions can appear quite daunting due to complexities foreseen in division.