Lesson Objectives
  • Demonstrate an understanding of how to find the LCD for a group of rational expressions
  • Learn how to add rational expressions
  • Learn how to subtract rational expressions

How to Add & Subtract Rational Expressions


When we add or subtract rational expressions, we follow the same rules we learned with fractions. When there is a common denominator present, we can perform the given operation with the numerators and place the result over the common denominator. When there is not a common denominator, we first find the LCD for the rational expressions. This is the LCM of the denominators. Once this is done, we will transform each rational expression into an equivalent rational expression where the LCD is its denominator. We can then perform operations with the numerators and place the result over the common denominator. In the end, we want to leave a simplified answer. This means we will factor our numerator and denominator and cancel any common factors. It is common to leave a rational expression in factored form to show that no other factors can be canceled. Let's look at a few examples.
Example 1: Perform each indicated operation $$\frac{3}{x + 6} - \frac{3x}{3x + 24}$$ $$\frac{3}{x + 6} - \frac{3x}{3x + 24}$$ Step 1) Find the LCD
LCD » 3(x + 6)(x + 8)
Step 2) Transform each rational expression into an equivalent rational expression with the LCD as its denominator
$$\frac{3}{x + 6} \cdot \frac{3(x+8)}{3(x+8)} = \frac{9(x+8)}{3(x + 8)(x+6)}$$ $$\frac{3}{x + 6} \cdot \frac{3(x+8)}{3(x+8)} = $$$$\frac{9(x+8)}{3(x + 8)(x+6)}$$ $$\frac{3x}{3(x + 8)} \cdot \frac{(x + 6)}{(x + 6)} = \frac{3x(x + 6)}{3(x + 8)(x + 6)}$$ $$\frac{3x}{3(x + 8)} \cdot \frac{(x + 6)}{(x + 6)} = $$$$\frac{3x(x + 6)}{3(x + 8)(x + 6)}$$ We will leave our denominators in factored form, but simplify the numerators. This will be necessary to combine like terms: $$\frac{9(x+8)}{3(x + 8)(x+6)} = \frac{9x + 72}{3(x + 8)(x + 6)}$$ $$\frac{9(x+8)}{3(x + 8)(x+6)} = $$$$\frac{9x + 72}{3(x + 8)(x + 6)}$$ $$\frac{3x(x + 6)}{3(x + 8)(x + 6)} = \frac{3x^2 + 18x}{3(x + 8)(x + 6)}$$ $$\frac{3x(x + 6)}{3(x + 8)(x + 6)} = $$$$\frac{3x^2 + 18x}{3(x + 8)(x + 6)}$$ Step 3) Perform the given operation with the numerators. We have subtraction here, so we must be very careful of our signs: $$\frac{9x + 72}{3(x + 8)(x + 6)} - \frac{3x^2 + 18x}{3(x + 8)(x + 6)}$$ $$\frac{9x + 72}{3(x + 8)(x + 6)} - $$$$\frac{3x^2 + 18x}{3(x + 8)(x + 6)}$$ We will change our subtraction to addition of the opposite: $$\frac{9x + 72}{3(x + 8)(x + 6)} + \frac{-3x^2 - 18x}{3(x + 8)(x + 6)}$$ $$\frac{9x + 72}{3(x + 8)(x + 6)} + $$$$\frac{-3x^2 - 18x}{3(x + 8)(x + 6)}$$ Now we can combine like terms between numerators: $$\frac{9x + 72}{3(x + 8)(x + 6)} + \frac{-3x^2 - 18x}{3(x + 8)(x + 6)} = \frac{-3x^2 -9x + 72}{3(x + 8)(x + 6)}$$ $$\frac{9x + 72}{3(x + 8)(x + 6)} + $$$$\frac{-3x^2 - 18x}{3(x + 8)(x + 6)} = $$$$\frac{-3x^2 -9x + 72}{3(x + 8)(x + 6)}$$ Step 4) Look to see if we can simplify. Factor the numerator and cancel any common factors: $$\frac{-3x^2 -9x + 72}{3(x + 8)(x + 6)} = \frac{-3(x^2 + 3x - 24)}{3(x + 8)(x + 6)}$$ $$\frac{-3x^2 -9x + 72}{3(x + 8)(x + 6)} = $$$$\frac{-3(x^2 + 3x - 24)}{3(x + 8)(x + 6)}$$ Cancel the common factor of 3: $$\require{cancel}\frac{-1\cancel{3}(x^2 + 3x - 24)}{\cancel{3}(x + 8)(x + 6)} = -\frac{x^2 + 3x - 24}{(x + 8)(x + 6)}$$ $$\require{cancel}\frac{-1\cancel{3}(x^2 + 3x - 24)}{\cancel{3}(x + 8)(x + 6)} = $$$$ -\frac{x^2 + 3x - 24}{(x + 8)(x + 6)}$$ Example 2: Perform each indicated operation $$\frac{x - 8}{x - 2} + \frac{x - 3}{x + 6}$$ $$\frac{x - 8}{x - 2} + \frac{x - 3}{x + 6}$$ Step 1) Find the LCD
LCD » (x - 2)(x + 6)
Step 2) Transform each rational expression into an equivalent rational expression with the LCD as its denominator
$$\frac{(x - 8)}{(x - 2)} \cdot \frac{(x + 6)}{(x + 6)} = \frac{(x - 8)(x + 6)}{(x - 2)(x + 6)}$$ $$\frac{(x - 8)}{(x - 2)} \cdot \frac{(x + 6)}{(x + 6)} = $$$$\frac{(x - 8)(x + 6)}{(x - 2)(x + 6)}$$ $$\frac{(x-3)}{(x + 6)} \cdot \frac{(x-2)}{(x-2)} = \frac{(x-3)(x-2)}{(x+6)(x-2)}$$ $$\frac{(x-3)}{(x + 6)} \cdot \frac{(x-2)}{(x-2)} = $$$$\frac{(x-3)(x-2)}{(x+6)(x-2)}$$ We will simplify the numerators, but leave our denominators in factored form: $$\frac{(x - 8)(x + 6)}{(x - 2)(x + 6)} = \frac{x^2 - 2x - 48}{(x - 2)(x + 6)}$$ $$\frac{(x - 8)(x + 6)}{(x - 2)(x + 6)} = $$$$\frac{x^2 - 2x - 48}{(x - 2)(x + 6)}$$ $$\frac{(x - 3)(x - 2)}{(x - 2)(x + 6)} = \frac{x^2 - 5x + 6}{(x - 2)(x + 6)}$$ $$\frac{(x - 3)(x - 2)}{(x - 2)(x + 6)} = $$$$\frac{x^2 - 5x + 6}{(x - 2)(x + 6)}$$ Step 3) Perform the given operation with the numerators: $$\frac{x^2 - 2x - 48}{(x - 2)(x + 6)} + \frac{x^2 - 5x + 6}{(x - 2)(x + 6)} = \frac{2x^2 - 7x - 42}{(x - 2)(x + 6)}$$ $$\frac{x^2 - 2x - 48}{(x - 2)(x + 6)} + $$$$\frac{x^2 - 5x + 6}{(x - 2)(x + 6)} = $$$$\frac{2x^2 - 7x - 42}{(x - 2)(x + 6)}$$ Step 4) Look to see if we can simplify. Factor the numerator and cancel any common factors: $$\frac{2x^2 - 7x - 42}{(x - 2)(x + 6)}$$ $$\frac{2x^2 - 7x - 42}{(x - 2)(x + 6)}$$ The numerator is a prime polynomial. We can't simplify our rational expression any further.