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Guides d'étude > College Algebra

Add and Subtract Rational Expressions

Learning Objectives

  • Find the LCD of two rational expressions
  • Add and subtract rational expressions
  • Simplify complex rational expressions
Adding and subtracting rational expressions works just like adding and subtracting numerical fractions. To add fractions, we need to find a common denominator. Let’s look at an example of fraction addition.
524+140=25120+3120=28120=730\begin{array}{ccc}\hfill \frac{5}{24}+\frac{1}{40}& =& \frac{25}{120}+\frac{3}{120}\hfill \\ & =& \frac{28}{120}\hfill \\ & =& \frac{7}{30}\hfill \end{array}
We have to rewrite the fractions so they share a common denominator before we are able to add. We must do the same thing when adding or subtracting rational expressions. The easiest common denominator to use will be the least common denominator, or LCD. The LCD is the smallest multiple that the denominators have in common. To find the LCD of two rational expressions, we factor the expressions and multiply all of the distinct factors. For instance, if the factored denominators were (x+3)(x+4)\left(x+3\right)\left(x+4\right) and (x+4)(x+5)\left(x+4\right)\left(x+5\right), then the LCD would be (x+3)(x+4)(x+5)\left(x+3\right)\left(x+4\right)\left(x+5\right). Once we find the LCD, we need to multiply each expression by the form of 1 that will change the denominator to the LCD. We would need to multiply the expression with a denominator of (x+3)(x+4)\left(x+3\right)\left(x+4\right) by x+5x+5\frac{x+5}{x+5} and the expression with a denominator of (x+4)(x+5)\left(x+4\right)\left(x+5\right) by x+3x+3\frac{x+3}{x+3}.

How To: Given two rational expressions, add or subtract them.

  1. Factor the numerator and denominator.
  2. Find the LCD of the expressions.
  3. Multiply the expressions by a form of 1 that changes the denominators to the LCD.
  4. Add or subtract the numerators.
  5. Simplify.

Example: Adding Rational Expressions

Add the rational expressions:
5x+6y\frac{5}{x}+\frac{6}{y}

Answer: First, we have to find the LCD. In this case, the LCD will be xyxy. We then multiply each expression by the appropriate form of 1 to obtain xyxy as the denominator for each fraction.

5x⋅yy+6y⋅xx5yxy+6xxy\begin{array}{l}\frac{5}{x}\cdot \frac{y}{y}+\frac{6}{y}\cdot \frac{x}{x}\\ \frac{5y}{xy}+\frac{6x}{xy}\end{array}
Now that the expressions have the same denominator, we simply add the numerators to find the sum.
6x+5yxy\frac{6x+5y}{xy}

Analysis of the Solution

Multiplying by yy\frac{y}{y} or xx\frac{x}{x} does not change the value of the original expression because any number divided by itself is 1, and multiplying an expression by 1 gives the original expression.  

Example: Subtracting Rational Expressions

Subtract the rational expressions:
6x2+4x+4−2x2−4\frac{6}{{x}^{2}+4x+4}-\frac{2}{{x}^{2}-4}

Answer:

6(x+2)2−2(x+2)(x−2)Factor.6(x+2)2⋅x−2x−2−2(x+2)(x−2)⋅x+2x+2Multiply each fraction to get LCD as denominator.6(x−2)(x+2)2(x−2)−2(x+2)(x+2)2(x−2)Multiply.6x−12−(2x+4)(x+2)2(x−2)Apply distributive property.4x−16(x+2)2(x−2)Subtract.4(x−4)(x+2)2(x−2)Simplify.\begin{array}{cc}\frac{6}{{\left(x+2\right)}^{2}}-\frac{2}{\left(x+2\right)\left(x - 2\right)}\hfill & \text{Factor}.\hfill \\ \frac{6}{{\left(x+2\right)}^{2}}\cdot \frac{x - 2}{x - 2}-\frac{2}{\left(x+2\right)\left(x - 2\right)}\cdot \frac{x+2}{x+2}\hfill & \text{Multiply each fraction to get LCD as denominator}.\hfill \\ \frac{6\left(x - 2\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}-\frac{2\left(x+2\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Multiply}.\hfill \\ \frac{6x - 12-\left(2x+4\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Apply distributive property}.\hfill \\ \frac{4x - 16}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Subtract}.\hfill \\ \frac{4\left(x - 4\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Simplify}.\hfill \end{array}

Q & A

Do we have to use the LCD to add or subtract rational expressions?

No. Any common denominator will work, but it is easiest to use the LCD.

Try It

Subtract the rational expressions: 3x+5−1x−3\frac{3}{x+5}-\frac{1}{x - 3}.

Answer: 2(x−7)(x+5)(x−3)\frac{2\left(x - 7\right)}{\left(x+5\right)\left(x - 3\right)}

Q & A

Can a complex rational expression always be simplified?

Yes. We can always rewrite a complex rational expression as a simplified rational expression.

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