Problem 69 - Elementary Algebra

Question

Add or subtract as indicated and express your answers in simplest form. (Objective 3) $$\frac{2}{n-1}-\frac{3}{n}$$

Step-by-Step Solution

Answer

\( \frac{3-n}{n(n-1)} \)

Step 1: Identify the Problem

We need to subtract two fractions: \( \frac{2}{n-1} \) and \( \frac{3}{n} \). To do this, we need a common denominator.

Step 2: Find the Common Denominator

The denominators are \( n-1 \) and \( n \). The common denominator is the product of these, \((n-1)n\), because they have no factors in common.

Step 3: Rewrite Each Fraction with the Common Denominator

Rewrite \( \frac{2}{n-1} \) as \( \frac{2n}{(n-1)n} \) by multiplying the numerator and denominator by \( n \). Rewrite \( \frac{3}{n} \) as \( \frac{3(n-1)}{n(n-1)} \) by multiplying the numerator and denominator by \( n-1 \).

Step 4: Subtract the Fractions

Use the common denominator to subtract: \[ \frac{2n}{(n-1)n} - \frac{3(n-1)}{n(n-1)} = \frac{2n - 3(n-1)}{n(n-1)} \].

Step 5: Simplify the Expression

Simplify the numerator: \( 2n - 3(n-1) = 2n - 3n + 3 = -n + 3 \). Therefore, the expression becomes \( \frac{-n+3}{n(n-1)} \).

Step 6: Express in Simplest Form

The simplest form of the answer is \( \frac{-n+3}{n(n-1)} \), or equivalently, \( \frac{3-n}{n(n-1)} \).

Key Concepts

Subtracting Fractions

When dealing with algebraic fractions, the process of subtracting them is similar to that of numerical fractions. You need to pay close attention to the denominators. For the given exercise, we have two fractions: \( \frac{2}{n-1} \) and \( \frac{3}{n} \). These fractions need to be subtracted from one another. Just like with numerical fractions, you cannot subtract them directly unless they share the same denominator.

The procedure involves a strategic plan to make the denominators alike, as only then can the fraction subtraction proceed naturally. After finding a common denominator, you rewrite each fraction, ensuring that both have identical denominators, then proceed to subtract the numerators while maintaining the common denominator below.

Finding Common Denominators

Finding a common denominator is crucial for subtracting fractions, especially when they have distinct variables in the denominators. In our exercise, the denominators are \( n-1 \) and \( n \).

Neither of these is a factor of the other, so their least common denominator (LCD) becomes their product:

• Multiply \( (n-1) \times n \) to get \( n(n-1) \).

This common denominator allows us to express each fraction with the same denominator, enabling subtraction to proceed smoothly.

The next step is to adjust each fraction so their denominators match this common one. Multiply the denominators along with their respective numerators to achieve uniformity, ensuring the subtraction operation can then be carried out seamlessly.

Simplifying Algebraic Expressions

After rewriting the fractions with a common denominator, the next phase involves simplifying the algebraic expression. The subtraction, post-common denominator alignment, results in a new fraction. This fraction's numerator often demands further simplification to achieve its simplest form.

In our case, the original operation yields \( \frac{2n - 3(n-1)}{n(n-1)} \). The numerator, being \( 2n - 3(n-1) \), needs to be expanded and simplified:

• Distribute 3 across \( (n-1) \) to get: \( 2n - 3n + 3 \).
• Simplify this to \( -n + 3 \).

The final result is a fraction with \( -n + 3 \) as the simplified numerator, ready for expression in its simplest form.

Numerator and Denominator

Understanding the roles of the numerator and denominator helps in managing fraction operations. The numerator represents the number divided by the denominator, which acts as the divisor. In algebraic expressions, these roles remain consistent, even if variables are involved.

For our exercise, once the common denominator \( n(n-1) \) is set, attention turns to the numerators. These numerators must be subtracted and simplified:

• The first fraction's numerator becomes \( 2n \) after adjustment.
• The second fraction's numerator is transformed to \( 3(n-1) \).

The new subtracted numerator (\( -n + 3 \)) is calculated from \( 2n - 3(n-1) \). A clear grasp of numerator and denominator manipulations ensures clarity while performing operations, leading to accurate results and efficient simplification.