Solved Exercises on Adding and Subtracting Rational Expressions in Algebra 2

Master adding and subtracting rational expressions: finding LCD, combining fractions, factoring, domain restrictions, and advanced techniques through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Same Denominator Addition
Exercise 1
Add: \(\frac{3x}{x-2} + \frac{5}{x-2}\)
Definition:

Same denominator addition: When denominators are identical, add numerators and keep the common denominator.

Addition method:
  1. Verify denominators are identical
  2. Add numerators together
  3. Keep the common denominator
  4. Simplify if possible
  5. State domain restrictions
Combine numerators
3x + 5
Keep denominator
x - 2
Step 1: Verify denominators are the same

Both fractions have the denominator \(x-2\), so we can add directly

Step 2: Add numerators and keep denominator

\(\frac{3x}{x-2} + \frac{5}{x-2} = \frac{3x + 5}{x-2}\)

Step 3: Check if further simplification is possible

The numerator \(3x + 5\) cannot be factored, and it shares no common factors with the denominator

Step 4: State domain restrictions

The expression is undefined when \(x-2 = 0\), so \(x \neq 2\)

\(\frac{3x}{x-2} + \frac{5}{x-2} = \frac{3x + 5}{x-2}\), where \(x \neq 2\)
Final answer:

\(\frac{3x}{x-2} + \frac{5}{x-2} = \frac{3x + 5}{x-2}\), where \(x \neq 2\)

Applied rules:

• Same denominator rule: \(\frac{A}{C} + \frac{B}{C} = \frac{A+B}{C}\)

• Domain preservation: Maintain restrictions from original expressions

2 Different Denominators
Exercise 2
Subtract: \(\frac{2}{x+1} - \frac{3}{x-1}\)
Definition:

Least Common Denominator (LCD): The smallest polynomial that is divisible by all denominators involved.

Find LCD
(x+1)(x-1)
Rewrite fractions
\(\frac{2(x-1)}{(x+1)(x-1)} - \frac{3(x+1)}{(x+1)(x-1)}\)
Step 1: Find the LCD

The denominators are \(x+1\) and \(x-1\), which share no common factors

Therefore, LCD = \((x+1)(x-1)\)

Step 2: Rewrite each fraction with the LCD

\(\frac{2}{x+1} = \frac{2(x-1)}{(x+1)(x-1)} = \frac{2x-2}{(x+1)(x-1)}\)

\(\frac{3}{x-1} = \frac{3(x+1)}{(x+1)(x-1)} = \frac{3x+3}{(x+1)(x-1)}\)

Step 3: Subtract the numerators

\(\frac{2x-2}{(x+1)(x-1)} - \frac{3x+3}{(x+1)(x-1)} = \frac{(2x-2) - (3x+3)}{(x+1)(x-1)}\)

Step 4: Simplify the numerator

\(\frac{2x-2-3x-3}{(x+1)(x-1)} = \frac{-x-5}{(x+1)(x-1)} = \frac{-(x+5)}{(x+1)(x-1)}\)

\(\frac{2}{x+1} - \frac{3}{x-1} = \frac{-(x+5)}{(x+1)(x-1)} = \frac{-x-5}{x^2-1}\), where \(x \neq \pm 1\)
Final answer:

\(\frac{2}{x+1} - \frac{3}{x-1} = \frac{-x-5}{x^2-1}\), where \(x \neq \pm 1\)

Applied rules:

• LCD method: Find smallest common denominator

• Distribution: Multiply numerators by necessary factors

• Subtraction: Remember to distribute the negative sign

3 Factored Denominators
Exercise 3
Add: \(\frac{x}{x^2-4} + \frac{2}{x+2}\)
Definition:

Factoring denominators: Before finding LCD, factor all denominators completely to identify common and unique factors.

Factor first denominator
\(x^2-4 = (x+2)(x-2)\)
Find LCD
\((x+2)(x-2)\)
Step 1: Factor denominators completely

\(x^2 - 4 = (x+2)(x-2)\) (difference of squares)

The second denominator is already factored: \(x+2\)

Step 2: Find the LCD

The first denominator has factors: \((x+2)(x-2)\)

The second denominator has factors: \((x+2)\)

LCD = \((x+2)(x-2)\) (includes all unique factors)

Step 3: Rewrite each fraction with LCD

\(\frac{x}{(x+2)(x-2)}\) (already has LCD)

\(\frac{2}{x+2} = \frac{2(x-2)}{(x+2)(x-2)} = \frac{2x-4}{(x+2)(x-2)}\)

Step 4: Add the numerators

\(\frac{x}{(x+2)(x-2)} + \frac{2x-4}{(x+2)(x-2)} = \frac{x + 2x - 4}{(x+2)(x-2)} = \frac{3x-4}{(x+2)(x-2)}\)

Step 5: Check for further simplification

The numerator \(3x-4\) cannot be factored and shares no common factors with the denominator

\(\frac{x}{x^2-4} + \frac{2}{x+2} = \frac{3x-4}{x^2-4}\), where \(x \neq \pm 2\)
Final answer:

\(\frac{x}{x^2-4} + \frac{2}{x+2} = \frac{3x-4}{x^2-4}\), where \(x \neq \pm 2\)

Applied rules:

• Factor first: Always factor denominators before finding LCD

• LCD construction: Include each unique factor the maximum number of times it appears

• Common factors: Don't duplicate factors that appear in multiple denominators

Key Concepts: Definitions, Rules, and Methods
\(\frac{A}{C} + \frac{B}{C} = \frac{A+B}{C}, \quad \frac{A}{B} + \frac{C}{D} = \frac{AD + BC}{BD}\)
Fundamental Addition Rules
Key definitions:

Rational Expression: A ratio of two polynomials where the denominator is not zero.

Least Common Denominator (LCD): The smallest polynomial divisible by all denominators.

Domain: The set of all real numbers for which the expression is defined.

Equivalent Expressions: Expressions that yield the same value for all values in their common domain.

Addition methodology:
  1. Factor denominators: Factor all denominators completely
  2. Find LCD: Identify the least common denominator
  3. Rewrite fractions: Express each fraction with the LCD
  4. Add/subtract numerators: Combine numerators over common denominator
  5. Simplify: Factor and cancel if possible
  6. State domain restrictions: Identify values that make original denominators zero
Tip 1: Always factor denominators completely before finding LCD.
Tip 2: When subtracting, distribute the negative sign to every term in the second numerator.
Tip 3: LCD includes each unique factor the maximum number of times it appears.
Tip 4: Always state domain restrictions from the original expressions.
Common errors: Forgetting to factor, not distributing negative signs, missing domain restrictions.
Exam preparation: Practice factoring patterns, memorize LCD rules, verify results.
Essential rules to remember:

• Same denominator: \(\frac{A}{C} + \frac{B}{C} = \frac{A+B}{C}\)

• Different denominators: Find LCD and rewrite fractions

• Subtraction: Distribute negative sign to all terms in numerator

• LCD: Include each unique factor the maximum number of times it appears

• Domain preservation: State restrictions from original expressions

\(a^2 - b^2 = (a+b)(a-b)\)
Difference of Squares
\(a^2 \pm 2ab + b^2 = (a \pm b)^2\)
Perfect Square Trinomial
\(ax^2 + bx + c = (x+p)(x+q) \text{ where } pq=c, p+q=b\)
Quadratic Factoring
Solution: Exercises 4 to 5
4 Complex LCD
Exercise 4
Subtract: \(\frac{3x}{x^2+x-6} - \frac{2}{x^2-4}\)
Definition:

Complex LCD: When denominators have multiple factors, LCD includes each unique factor the maximum number of times it appears.

Factor denominators
\(x^2+x-6 = (x+3)(x-2)\), \(x^2-4 = (x+2)(x-2)\)
Find LCD
\((x+3)(x+2)(x-2)\)
Step 1: Factor both denominators completely

For \(x^2+x-6\): Find two numbers that multiply to -6 and add to 1 → numbers are 3 and -2

So \(x^2+x-6 = (x+3)(x-2)\)

For \(x^2-4\): This is difference of squares → \(x^2-4 = (x+2)(x-2)\)

Step 2: Find the LCD

First denominator: \((x+3)(x-2)\)

Second denominator: \((x+2)(x-2)\)

LCD = \((x+3)(x+2)(x-2)\) (includes all unique factors)

Step 3: Rewrite each fraction with LCD

\(\frac{3x}{(x+3)(x-2)} = \frac{3x(x+2)}{(x+3)(x-2)(x+2)} = \frac{3x(x+2)}{(x+3)(x+2)(x-2)}\)

\(\frac{2}{(x+2)(x-2)} = \frac{2(x+3)}{(x+2)(x-2)(x+3)} = \frac{2(x+3)}{(x+3)(x+2)(x-2)}\)

Step 4: Expand numerators

\(\frac{3x(x+2)}{(x+3)(x+2)(x-2)} = \frac{3x^2+6x}{(x+3)(x+2)(x-2)}\)

\(\frac{2(x+3)}{(x+3)(x+2)(x-2)} = \frac{2x+6}{(x+3)(x+2)(x-2)}\)

Step 5: Subtract numerators

\(\frac{3x^2+6x}{(x+3)(x+2)(x-2)} - \frac{2x+6}{(x+3)(x+2)(x-2)} = \frac{(3x^2+6x) - (2x+6)}{(x+3)(x+2)(x-2)}\)

= \(\frac{3x^2+6x-2x-6}{(x+3)(x+2)(x-2)} = \frac{3x^2+4x-6}{(x+3)(x+2)(x-2)}\)

Step 6: Check for further simplification

Try to factor \(3x^2+4x-6\): Looking for factors of -18 that add to 4 → None found easily

So the expression is in simplest form

\(\frac{3x}{x^2+x-6} - \frac{2}{x^2-4} = \frac{3x^2+4x-6}{(x+3)(x+2)(x-2)}\), where \(x \neq -3, -2, 2\)
Final answer:

\(\frac{3x}{x^2+x-6} - \frac{2}{x^2-4} = \frac{3x^2+4x-6}{(x+3)(x+2)(x-2)}\), where \(x \neq -3, -2, 2\)

Applied rules:

• Complex LCD: Include each unique factor the maximum number of times it appears

• Distribution: Multiply numerators by necessary factors

• Subtraction: Remember to distribute the negative sign to all terms

5 Three-Term Expression
Exercise 5
Simplify: \(\frac{1}{x-1} + \frac{2}{x+1} - \frac{3}{x^2-1}\)
Definition:

Multiple denominators: When working with three or more fractions, find LCD that accommodates all denominators.

Factor denominators
\(x^2-1 = (x+1)(x-1)\)
Find LCD
\((x+1)(x-1)\)
Step 1: Factor all denominators completely

First denominator: \(x-1\) (already factored)

Second denominator: \(x+1\) (already factored)

Third denominator: \(x^2-1 = (x+1)(x-1)\) (difference of squares)

Step 2: Find the LCD

The LCD must include both \((x-1)\) and \((x+1)\)

LCD = \((x+1)(x-1) = x^2-1\)

Step 3: Rewrite each fraction with LCD

\(\frac{1}{x-1} = \frac{1(x+1)}{(x-1)(x+1)} = \frac{x+1}{(x+1)(x-1)}\)

\(\frac{2}{x+1} = \frac{2(x-1)}{(x+1)(x-1)} = \frac{2x-2}{(x+1)(x-1)}\)

\(\frac{3}{(x+1)(x-1)} = \frac{3}{(x+1)(x-1)}\) (already has LCD)

Step 4: Combine all numerators

\(\frac{x+1}{(x+1)(x-1)} + \frac{2x-2}{(x+1)(x-1)} - \frac{3}{(x+1)(x-1)}\)

= \(\frac{(x+1) + (2x-2) - 3}{(x+1)(x-1)}\)

= \(\frac{x+1+2x-2-3}{(x+1)(x-1)} = \frac{3x-4}{(x+1)(x-1)}\)

Step 5: Check for further simplification

The numerator \(3x-4\) cannot be factored and shares no common factors with the denominator

\(\frac{1}{x-1} + \frac{2}{x+1} - \frac{3}{x^2-1} = \frac{3x-4}{x^2-1}\), where \(x \neq \pm 1\)
Final answer:

\(\frac{1}{x-1} + \frac{2}{x+1} - \frac{3}{x^2-1} = \frac{3x-4}{x^2-1}\), where \(x \neq \pm 1\)

Applied rules:

• Multiple fractions: Find LCD that works for all denominators

• Order of operations: Process all additions and subtractions from left to right

• Sign handling: Carefully manage positive and negative terms

Adding and Subtracting Rational Expressions: Complete Guide
\(\frac{A}{C} + \frac{B}{C} = \frac{A+B}{C}, \quad \frac{A}{B} + \frac{C}{D} = \frac{AD + BC}{BD}\)
Fundamental Operations
Key definitions:

Rational Expression: A ratio of two polynomials where the denominator is not zero.

Least Common Denominator (LCD): The smallest polynomial divisible by all denominators in the expression.

Simplified Form: When the numerator and denominator share no common factors other than constants.

Domain: The set of all real numbers for which the expression is defined.

Complete methodology:
  1. Factor denominators: Factor all denominators completely into irreducible polynomials
  2. Find LCD: Identify the least common denominator by including each unique factor the maximum number of times it appears
  3. Rewrite fractions: Express each fraction with the LCD as denominator
  4. Combine numerators: Add or subtract numerators according to the operation
  5. Simplify: Factor the resulting numerator and cancel any common factors
  6. State domain restrictions: Identify all values that made original denominators zero
  7. Verify: Check that no further simplification is possible
Tip 1: Always factor denominators completely before finding LCD to identify common factors.
Tip 2: When subtracting, distribute the negative sign to every term in the second numerator.
Tip 3: LCD includes each unique factor the maximum number of times it appears across all denominators.
Tip 4: Always state domain restrictions from the original expressions, even after simplification.
Common errors: Forgetting to factor, not distributing negative signs properly, missing domain restrictions, incorrect LCD.
Exam preparation: Practice factoring patterns, memorize LCD rules, verify results by substitution.
Essential formulas and rules:

• Same denominator: \(\frac{A}{C} + \frac{B}{C} = \frac{A+B}{C}\)

• Different denominators: \(\frac{A}{B} + \frac{C}{D} = \frac{AD + BC}{BD}\)

• Subtraction: \(\frac{A}{B} - \frac{C}{D} = \frac{AD - BC}{BD}\)

• Difference of squares: \(a^2 - b^2 = (a+b)(a-b)\)

• Perfect square trinomial: \(a^2 \pm 2ab + b^2 = (a \pm b)^2\)

\(a^2 - b^2 = (a+b)(a-b)\)
Difference of Squares
\(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\)
Difference of Cubes
\(a^2 \pm 2ab + b^2 = (a \pm b)^2\)
Perfect Square Trinomial
Common factoring patterns:

GCF: Always factor out the greatest common factor first

Quadratic trinomials: Find two numbers that multiply to c and add to b

Grouping: For 4-term polynomials, group pairs and factor out common terms

Higher degree: May require multiple rounds of factoring

LCD construction techniques:
  1. Factor all denominators: Identify all unique factors
  2. Count occurrences: Determine maximum number of times each factor appears
  3. Build LCD: Include each unique factor the maximum number of times it appears
  4. Verify: Ensure LCD is divisible by each original denominator
Key note: The simplified form and original form are equivalent everywhere except at excluded values.
Key note: When subtracting, be especially careful with the distributive property for the negative sign.

Questions & Answers

Question: How do I find the LCD when denominators have multiple factors? I get confused about which factors to include.

Answer: To find the LCD when denominators have multiple factors, follow this systematic approach:

  1. Factor all denominators: Completely factor each denominator into irreducible factors
  2. Identify unique factors: List all different factors that appear across all denominators
  3. Count occurrences: For each unique factor, find the maximum number of times it appears in any single denominator
  4. Build LCD: Include each unique factor the maximum number of times it appears

Example: Find LCD of \((x+2)^2(x-1)\) and \((x+2)(x-1)^3(x+3)\)

  • Unique factors: \((x+2)\), \((x-1)\), \((x+3)\)
  • Max occurrences: \((x+2)^2\), \((x-1)^3\), \((x+3)^1\)
  • LCD = \((x+2)^2(x-1)^3(x+3)\)

This ensures the LCD is divisible by each original denominator.

Question: I always make mistakes when subtracting rational expressions. What's the trick to getting the signs right?

Question: I always make mistakes when subtracting rational expressions. What's the trick to getting the signs right?

Answer: The key to correctly handling signs in subtraction is to remember that you're subtracting the entire second numerator:

  1. Find LCD and rewrite fractions
  2. Write the subtraction as: first numerator - (second numerator)
  3. Distribute the negative sign to EVERY term in the second numerator
  4. Combine like terms carefully

Example: \(\frac{3x}{x-2} - \frac{x+1}{x-2} = \frac{3x - (x+1)}{x-2} = \frac{3x - x - 1}{x-2} = \frac{2x-1}{x-2}\)

Notice how the negative sign was distributed to both \(x\) AND \(1\), changing both signs.

A helpful tip: Write the subtraction as \(\frac{A}{C} - \frac{B}{C} = \frac{A + (-B)}{C}\) to remember that you're adding the opposite of the second fraction.

Question: Why do we need to keep track of domain restrictions? What happens if we ignore them after simplifying?

Answer: Domain restrictions are crucial because they preserve the original function's behavior and ensure mathematical correctness.

Consider: \(\frac{x}{x-1} + \frac{1}{x-1} = \frac{x+1}{x-1}\)

  • The original expression is undefined when \(x = 1\) (division by zero)
  • The simplified expression \(\frac{x+1}{x-1}\) is also undefined when \(x = 1\)
  • But if we had something like \(\frac{(x-1)(x+2)}{x-1}\), after simplifying to \(x+2\), the simplified form would appear to be defined at \(x = 1\), but the original wasn't

Even though the simplified form might appear to allow values that were originally restricted, those restrictions must be preserved because the original expression was undefined there. The complete answer is: \(\frac{x+1}{x-1}\), where \(x \neq 1\).

Ignoring domain restrictions can lead to incorrect evaluations and loss of important mathematical properties.