8.4 Addition and Subtraction of Rational Expressions

Terrance Berg

Chapter 8: Rational Expressions

8.4 Addition and Subtraction of Rational Expressions

Adding and subtracting rational expressions is identical to adding and subtracting integers. Recall that, when adding fractions with a common denominator, you add the numerators and keep the denominator. This is the same process used with rational expressions. Remember to reduce the final answer if possible.

Example 8.4.1

Add the following rational expressions:

$\begin{array}{rl} \dfrac{x-4}{x^2-2x-8}+\dfrac{x+8}{x^2-2x-8}&\text{Same denominator, so you add the numerators and combine like terms.} \\ \\ \dfrac{2x+4}{x^2-2x-8}&\text{Factor the numerator and the denominator.} \\ \\ \dfrac{2(x+2)}{(x+2)(x-4)}&\text{Divide out }(x+2). \\ \\ \dfrac{2}{x-4}&\text{Solution.} \end{array}$

Subtraction of rational expressions with a common denominator follows the same pattern, though the subtraction can cause problems if you are not careful with it. To avoid sign errors, first distribute the subtraction throughout the numerator. Then treat it like an addition problem. This process is the same as “add the opposite,” which was seen when subtracting with negatives.

Example 8.4.2

Subtract the following rational expressions:

$\begin{array}{rl} \dfrac{6x-12}{3x-6}-\dfrac{15x-6}{3x-6}&\text{Add the opposite of the second fraction (distribute the negative).} \\ \\ \dfrac{6x-12}{3x-6}+\dfrac{-15x+6}{3x-6}&\text{Add the numerators and combine like terms.} \\ \\ \dfrac{-9x-6}{3x-6}&\text{Factor the numerator and the denominator.} \\ \\ \dfrac{-3(3x+2)}{3(x-2)}&\text{Divide out the common factor of 3.} \\ \\ \dfrac{-(3x+2)}{x-2}&\text{Solution.} \end{array}$

When there is not a common denominator, first find the least common denominator (LCD) and alter each fraction so the denominators match.

Example 8.4.3

Add the following rational expressions:

$\begin{array}{rl} \dfrac{7a}{3a^2b}+\dfrac{4b}{6ab^4}&\text{The LCD is }6a^2b^4. \\ \\ \dfrac{2b^3}{2b^3}\cdot \dfrac{7a}{3a^2b}+\dfrac{4b}{6ab^4}\cdot \dfrac{a}{a}&\text{Multiply the first fraction by }2b^3 \text { and the second by }a. \\ \\ \dfrac{14ab^3}{6a^2b^4}+\dfrac{4ab}{6a^2b^4}&\text{Add the numerators. No like terms to combine.} \\ \\ \dfrac{14ab^3+4ab}{6a^2b^4}&\text{Factor the numerator.} \\ \\ \dfrac{2ab(7b^2+2)}{6a^2b^4}&\text{Reduce, dividing out factors }2, a, \text{ and }b. \\ \\ \dfrac{7b^2+2}{3ab^3}&\text{Solution.} \end{array}$

Example 8.4.4

Subtract the following rational expressions:

$\begin{array}{rl} \dfrac{x+1}{x-4}-\dfrac{x+1}{x^2-7x+12}&\text{Add the opposite of the second fraction (distribute the negative).} \\ \\ \dfrac{x+1}{x-4}+\dfrac{-x-1}{x^2-7x+12}&\text{Factor the denominators to find the LCD}=(x-4)(x-3). \\ \\ \dfrac{(x-3)(x+1)}{(x-3)(x-4)}+\dfrac{-x-1}{(x-3)(x-4)}&\text{Only the first fraction needs to be multplied by }(x-3). \\ \\ \dfrac{x^2-2x-3}{(x-3)(x-4)}+\dfrac{-x-1}{(x-3)(x-4)}&\text{Add the numerators and combine like terms.} \\ \\ \dfrac{x^2-3x-4}{(x-3)(x-4)}&\text{Factor the numerator.} \\ \\ \dfrac{(x-4)(x+1)}{(x-3)(x-4)}&\text{Divide out the common factor of }(x-4). \\ \\ \dfrac{x+1}{x-3}&\text{Solution.} \end{array}$