Addition and Subtraction of Algebraic Fractions#
Algebraic Fractions Revision#
Theory#
Before we dive into adding and subtracting algebraic fractions, let’s take a moment to revisit what makes algebraic fractions special. An algebraic fraction is simply a fraction where the numerator, denominator, or both contain algebraic expressions. Just like regular fractions, they follow the same fundamental principles, but with an important twist - we need to be mindful of domain restrictions.
Fundamental Principle: An algebraic fraction has the form \(\frac{P(x)}{Q(x)}\) where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x) \neq 0\).
Key Domain Considerations: For the fraction \(\frac{x+3}{x-2}\), we must exclude \(x = 2\) since this would make the denominator zero.
Equivalent Fractions: Just as \(\frac{2}{4} = \frac{1}{2}\) in arithmetic, we can create equivalent algebraic fractions by multiplying both numerator and denominator by the same non-zero expression:
Application#
Examples#
Example 1: Simple Domain Identification#
Let’s identify the domain restrictions for \(\frac{2x+1}{x^2-9}\):
\(x^2 - 9 = 0 \quad \text{(Set denominator equal to zero)}\)
\((x-3)(x+3) = 0 \quad \text{(Factor the denominator)}\)
\(x = 3 \text{ or } x = -3 \quad \text{(Find the restricted values)}\)
Domain: All real numbers except \(x = 3\) and \(x = -3\)
Example 2: Creating Equivalent Fractions#
Express \(\frac{x}{x+2}\) with denominator \((x+2)(x-1)\):
\(\frac{x}{x+2} = \frac{x \cdot (x-1)}{(x+2) \cdot (x-1)} = \frac{x(x-1)}{(x+2)(x-1)} \quad \text{(Multiply by } \frac{x-1}{x-1}\text{)}\)
Interactive Visualization: Algebraic Fractions Explorer#
Multiple Choice Questions#
Addition and Subtraction of Algebraic Fractions#
Theory#
Now, let’s explore how to add and subtract algebraic fractions. Here’s the beautiful thing - the process follows exactly the same logic as adding and subtracting numerical fractions, but with algebraic expressions. The key insight is that we need a common denominator before we can combine fractions.
The Golden Rule: To add or subtract fractions, they must have the same denominator:
The Challenge: When denominators are different, we need to find the Least Common Denominator (LCD).
Finding the LCD - A Systematic Approach:
Factor each denominator completely - This reveals the building blocks we’re working with
Identify all unique factors - Look for common factors and unique factors
Use the highest power of each factor - The LCD contains each factor raised to its highest power across all denominators
Why This Works: The LCD is the smallest expression that contains all factors from each denominator, ensuring that each original fraction can be rewritten with this common denominator.
Key Strategies for Different Scenarios:
Simple Common Factors: When denominators share factors like \(x\) and \(x^2\), the LCD is the highest power: \(x^2\).
Different Linear Factors: When denominators like \((x-1)\) and \((x+2)\) share no common factors, the LCD is their product: \((x-1)(x+2)\).
Factorable Denominators: Always factor first! What looks like \(x^2-4\) and \(x-2\) is really \((x+2)(x-2)\) and \(x-2\), so the LCD is \((x+2)(x-2)\).
Opposite Factors: Remember that \((a-b) = -(b-a)\). This relationship often simplifies problems significantly.
Interactive Visualization: Common Denominator Explorer#
Application#
Examples#
Example 1: Simple Common Factors#
Let’s work through this step by step: \(\frac{3}{x} + \frac{5}{x^2}\)
Step 1: Analyze the denominators
\(\frac{3}{x} + \frac{5}{x^2} \quad \text{(Denominators are } x \text{ and } x^2\text{)}\)
Step 2: Find the LCD
\(\text{LCD} = x^2 \quad \text{(The highest power of } x \text{ is } x^2\text{)}\)
Step 3: Rewrite with common denominator
\(\frac{3 \cdot x}{x \cdot x} + \frac{5}{x^2} = \frac{3x}{x^2} + \frac{5}{x^2} \quad \text{(Multiply first fraction by } \frac{x}{x}\text{)}\)
Step 4: Add the numerators
\(\frac{3x + 5}{x^2} \quad \text{(Combine numerators over common denominator)}\)
Domain: \(x \neq 0\)
Example 2: Different Linear Factors#
Here’s a typical challenge: \(\frac{2}{x-1} - \frac{3}{x+2}\)
Step 1: Identify that denominators share no common factors
\(\frac{2}{x-1} - \frac{3}{x+2} \quad \text{(} (x-1) \text{ and } (x+2) \text{ are different linear factors)}\)
Step 2: The LCD is their product
\(\text{LCD} = (x-1)(x+2) \quad \text{(Product of distinct linear factors)}\)
Step 3: Rewrite each fraction with the LCD
\(\frac{2(x+2)}{(x-1)(x+2)} - \frac{3(x-1)}{(x+2)(x-1)} \quad \text{(Multiply to get common denominator)}\)
Step 4: Expand the numerators
\(\frac{2x + 4}{(x-1)(x+2)} - \frac{3x - 3}{(x-1)(x+2)} \quad \text{(Distribute in each numerator)}\)
Step 5: Combine carefully (watch the subtraction!)
\(\frac{2x + 4 - (3x - 3)}{(x-1)(x+2)} = \frac{2x + 4 - 3x + 3}{(x-1)(x+2)} = \frac{-x + 7}{(x-1)(x+2)} \quad \text{(Distribute the negative sign)}\)
Domain: \(x \neq 1, x \neq -2\)
Example 3: Factoring Required#
This might look tricky at first, but notice what happens when we factor: \(\frac{x}{x^2-4} + \frac{2}{x-2}\)
Step 1: Factor all denominators completely
\(x^2 - 4 = (x+2)(x-2) \quad \text{(Difference of squares pattern)}\)
\(\frac{x}{(x+2)(x-2)} + \frac{2}{x-2} \quad \text{(Rewrite with factored form)}\)
Step 2: Find the LCD
\(\text{LCD} = (x+2)(x-2) \quad \text{(Already present in first fraction)}\)
Step 3: Rewrite second fraction
\(\frac{x}{(x+2)(x-2)} + \frac{2(x+2)}{(x-2)(x+2)} \quad \text{(Multiply second fraction by } \frac{x+2}{x+2}\text{)}\)
Step 4: Add the numerators
\(\frac{x + 2(x+2)}{(x+2)(x-2)} = \frac{x + 2x + 4}{(x+2)(x-2)} = \frac{3x + 4}{(x+2)(x-2)} \quad \text{(Combine and simplify)}\)
Domain: \(x \neq 2, x \neq -2\)
Multiple Choice Questions#
Sector Specific Questions: Addition and Subtraction Applications#
Key Takeaways#
Important
Adding and subtracting algebraic fractions requires a common denominator
The LCD contains all factors from each denominator, each raised to its highest power
Always factor denominators completely before finding the LCD
Domain restrictions must be preserved throughout the process
Watch for opposite factors: \((a-b) = -(b-a)\)
When subtracting, distribute the negative sign through the entire numerator
Verify your answer by checking domain restrictions and substituting test values
These skills are essential for electrical circuits, optics, economics, and many other applications