Problem 46 - College Algebra

Question

For the following exercises, simplify the rational expression. $$ \frac{\frac{3}{x+1}+\frac{2}{x-1}}{\frac{x-1}{x+1}} $$

Step-by-Step Solution

Answer

The simplified expression is \( \frac{5x - 1}{x-1} \).

Step 1: Simplify the Numerator

The numerator is \( \frac{3}{x+1} + \frac{2}{x-1} \). We need a common denominator to combine the fractions. The common denominator is \((x+1)(x-1)\). Rewriting each fraction: \( \frac{3(x-1)}{(x+1)(x-1)} + \frac{2(x+1)}{(x+1)(x-1)} \). Simplifying further, we get \( \frac{3(x-1) + 2(x+1)}{(x+1)(x-1)} \).

Step 2: Expand the Numerator Terms

Expand the terms in the numerator. The expression becomes \( \frac{3x - 3 + 2x + 2}{(x+1)(x-1)} \). Simplify by combining like terms: \( \frac{5x - 1}{(x+1)(x-1)} \).

Step 3: Analyze the Denominator

The denominator is \( \frac{x-1}{x+1} \). Since this is already a simplified fraction, we do not need to change anything here.

Step 4: Simplify the Entire Expression

Now, treat the entire expression as division: \( \frac{5x - 1}{(x+1)(x-1)} \div \frac{x-1}{x+1} \). In order to divide by a fraction, multiply by its reciprocal: \( \frac{5x - 1}{(x+1)(x-1)} \times \frac{x+1}{x-1} \).

Step 5: Multiply and Simplify

Multiply across: \( \frac{(5x - 1)(x+1)}{((x+1)(x-1))(x-1)} \). The \((x+1)\) terms in numerator and denominator cancel out, leaving \( \frac{5x - 1}{x-1} \).

Key Concepts

Algebraic Simplification

Simplifying algebraic expressions involves reducing them to their most basic form. It's like cleaning up a messy room—removing what's unnecessary and organizing what's left. In this exercise, our goal is to simplify a complex rational expression. This requires several steps:

• Combining fractions
• Expanding terms
• Canceling out like terms

The objective is to make the expression as uncomplicated as possible so that it's easier to work with in future equations or problem-solving scenarios. Remember to always look for common factors that might cancel out, just as how you reduce regular fractions to their simplest form.

Common Denominators

Finding a common denominator is a crucial step when adding or subtracting fractions because it allows us to combine them into a single expression. For instance, if you have the fractions \( \frac{3}{x+1} \) and \( \frac{2}{x-1} \), you can't combine them directly. Instead, you first find a common denominator. The common denominator is found by multiplying the two different denominators:

• For \( \frac{3}{x+1} + \frac{2}{x-1} \), it's \((x+1)(x-1)\).

Once you have the common denominator,

• Rewrite each fraction so they share this denominator \( \frac{3(x-1)}{(x+1)(x-1)} + \frac{2(x+1)}{(x+1)(x-1)} \).

This turns them into "like terms" that are easily added together.

Fraction Division

Dividing fractions is often simplified by multiplying by the reciprocal. This means flipping the second fraction and then proceeding with multiplication. When faced with \( \frac{a}{b} \div \frac{c}{d} \), you simplify it to \( \frac{a}{b} \times \frac{d}{c} \). In our exercise, the expression \( \frac{5x - 1}{(x+1)(x-1)} \div \frac{x-1}{x+1} \) becomes \( \frac{5x - 1}{(x+1)(x-1)} \times \frac{x+1}{x-1} \). Now, you simply multiply

• Both the numerators
• Both the denominators

These steps effectively turn division problems into multiplication, which generally are more straightforward to solve. Additionally, it allows you to cancel out like terms, simplifying even further.

Like Terms

Combining like terms is an essential technique in algebra that helps to simplify expressions. Like terms are terms that contain the same variables raised to the same power. Only the coefficients (the numerical parts) are different. In the numerator of our rational expression, we expanded \(3(x-1) + 2(x+1) = 3x - 3 + 2x + 2\). By identifying and combining like terms

• \(3x\) and \(2x\)
• \(-3\) and \(+2\)

We get \(5x - 1\).Combining like terms not only reduces the number of terms but also simplifies the expression significantly, making it manageable and easier to handle in calculations or further algebraic manipulations.