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Study Guides > Intermediate Algebra

Add and Subtract Rational Expressions

Learning Outcomes

  • Add and subtract rational expressions
In beginning math, students usually learn how to add and subtract whole numbers before they are taught multiplication and division. However, with fractions and rational expressions, multiplication and division are sometimes taught first because these operations are easier to perform than addition and subtraction. Addition and subtraction of rational expressions are not as easy to perform as multiplication because, as with numeric fractions, the process involves finding common denominators. Add and Subtract Sign

Adding Rational Expressions

To find the least common denominator (LCD) of two rational expressions, we factor the expressions and multiply all of the distinct factors. For instance, consider the following rational expressions:

[latex]\dfrac{6}{\left(x+3\right)\left(x+4\right)},\text{ and }\frac{9x}{\left(x+4\right)\left(x+5\right)}[/latex]

The LCD would be [latex]\left(x+3\right)\left(x+4\right)\left(x+5\right)[/latex].

To find the LCD, we count the greatest number of times a factor appears in each denominator and include it in the LCD that many times. For example, in [latex]\dfrac{6}{\left(x+3\right)\left(x+4\right)}[/latex], [latex]\left(x+3\right)[/latex] is represented once and  [latex]\left(x+4\right)[/latex] is represented once, so they both appear exactly once in the LCD. In [latex]\dfrac{9x}{\left(x+4\right)\left(x+5\right)}[/latex], [latex]\left(x+4\right)[/latex] appears once and [latex]\left(x+5\right)[/latex] appears once. We have already accounted for [latex]\left(x+4\right)[/latex], so the LCD just needs one factor of [latex]\left(x+5\right)[/latex] to be complete. Once we find the LCD, we need to multiply each expression by the form of [latex]1[/latex] that will change the denominator to the LCD. What do we mean by " the form of [latex]1[/latex]"? [latex]\frac{x+5}{x+5}=1[/latex] so multiplying an expression by it will not change its value. For example, we would need to multiply the expression [latex]\dfrac{6}{\left(x+3\right)\left(x+4\right)}[/latex] by [latex]\frac{x+5}{x+5}[/latex] and the expression [latex]\frac{9x}{\left(x+4\right)\left(x+5\right)}[/latex] by [latex]\frac{x+3}{x+3}[/latex]. Hopefully this process will become clear after you practice it yourself.  As you look through the examples on this page, try to identify the LCD before you look at the answers. Also, try figuring out which "form of 1" you will need to multiply each expression by so that it has the LCD.

Example

Add the rational expressions [latex]\frac{5}{x}+\frac{6}{y}[/latex] and define the domain. State the sum in simplest form.

Answer: First, define the domain of each expression. Since we have x and y in the denominators, we can say [latex]x\ne0 ,\text{ and }y\ne0[/latex]. Now we have to find the LCD. Since x appears once and y appears once,  the LCD will be [latex]xy[/latex].  We then multiply each expression by the appropriate form of 1 to obtain [latex]xy[/latex] as the denominator for each fraction.

[latex]\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}[/latex]
Now that the expressions have the same denominator, we simply add the numerators to find the sum.
[latex]\frac{6x+5y}{xy}[/latex]
[latex-display]\frac{5}{x}+\frac{6}{y}=\frac{(6x+5y)}{xy}[/latex], [latex]x\ne0 ,\text{ and }y\ne0[/latex-display]

Here is one more example of adding rational expressions, but in this case, the expressions have denominators with multi-term polynomials. First, we will factor and then find the LCD. Note that [latex]x^2-4[/latex] is a difference of squares and can be factored using special products.

Example

Simplify[latex]\frac{2{{x}^{2}}}{{{x}^{2}}-4}+\frac{x}{x-2}[/latex] and give the domain. State the result in simplest form.

Answer: Find the least common denominator by factoring each denominator. The least common denominator includes the maximum number of times it appears in a single factorization. Remember that x cannot be [latex]2[/latex] or [latex]-2[/latex] because the denominators would be [latex]0[/latex]. [latex]\left(x+2\right)[/latex] appears a maximum of one time and so does [latex]\left(x–2\right)[/latex]. This means the LCD is [latex]\left(x+2\right)\left(x–2\right)[/latex]. Multiply each expression by the equivalent of [latex]1[/latex] that will give it the common denominator. Notice the first fraction already has the LCD and as a result remains unchanged.

[latex]\begin{array}{r}\frac{2{{x}^{2}}}{{{x}^{2}}-4}=\frac{2{{x}^{2}}}{(x+2)(x-2)}\\\frac{x}{x-2}\cdot \frac{x+2}{x+2}=\frac{x(x+2)}{(x+2)(x-2)}\end{array}[/latex]

Rewrite the original problem with the common denominator. It makes sense to keep the denominator in factored form in order to check for common factors.

[latex] \displaystyle \frac{2{{x}^{2}}}{(x+2)(x-2)}+\frac{x(x+2)}{(x+2)(x-2)}[/latex]

Combine the numerators.

[latex] \begin{array}{c}\frac{2{{x}^{2}}+x(x+2)}{(x+2)(x-2)}\\\\\frac{2{{x}^{2}}+{{x}^{2}}+2x}{(x+2)(x-2)}\end{array}[/latex]

[latex]\frac{3{{x}^{2}}+2x}{(x+2)(x-2)}[/latex]

Check for simplest form. Since neither [latex]\left(x+2\right)[/latex] nor [latex]\left(x-2\right)[/latex] is a factor of [latex]3{{x}^{2}}+2x[/latex], this expression is in simplest form. [latex-display] \displaystyle \frac{2{{x}^{2}}}{{{x}^{2}}-4}+\frac{x}{x-2}=\frac{3{{x}^{2}}+2x}{(x+2)(x-2)}[/latex]    [latex] \displaystyle x\ne 2,-2[/latex-display]

In the video that follows, we present an example of adding two rational expression whose denominators are binomials with no common factors. https://www.youtube.com/watch?v=CKGpiTE5vIg&feature=youtu.be

Subtracting Rational Expressions

To subtract rational expressions, follow the same process you use to add rational expressions. You will need to be careful with signs though.

Example

Subtract[latex]\frac{2}{t+1}-\frac{t-2}{{{t}^{2}}-t-2}[/latex] and define the domain. State the difference in simplest form.

Answer: Find the LCD of each expression. [latex]t+1[/latex] cannot be factored any further but [latex]{{t}^{2}}-t-2[/latex] can be. Note that t cannot be [latex]-1[/latex] or [latex]2[/latex] because the denominators would be [latex]0[/latex]. Find the least common multiple. [latex]t+1[/latex] appears exactly once in both of the expressions, so it will appear once in the least common denominator. [latex]t–2[/latex] also appears once. This means that [latex]\left(t-2\right)\left(t+1\right)[/latex] is the LCD. In this case, it is easier to leave the common denominator in terms of the factors. You will not need to multiply it out. Use the LCD as your new common denominator. Compare each original denominator and the new common denominator. Now rewrite the rational expressions to each have the common denominator of [latex]\left(t+1\right)\left(t–2\right)[/latex]. In the first expression, you need to multiply [latex]t+1[/latex] by [latex]t–2[/latex] to get the LCD, so multiply the entire rational expression by [latex] \displaystyle \frac{t-2}{t-2}[/latex]. The second expression already has a denominator of [latex]\left(t+1\right)\left(t–2\right)[/latex], so you do not need to multiply it by anything.

[latex] \begin{array}{c}\frac{2}{t+1}\cdot \frac{t-2}{t-2}=\frac{2(t-2)}{(t+1)(t-2)}\\\\\,\,\,\frac{t-2}{{{t}^{2}}-t-2}=\frac{t-2}{(t+1)(t-2)}\end{array}[/latex]

Now rewrite the subtraction problem using the expressions with the common denominator.

[latex] \frac{2\left(t-2\right)}{\left(t+1\right)\left(t-2\right)}-\frac{t-2}{\left(t+1\right)\left(t-2\right)}[/latex]

Subtract the numerators and simplify. Remember that parentheses need to be included around the second [latex]\left(t–2\right)[/latex] in the numerator because the whole quantity is subtracted. Otherwise, you would be subtracting just the [latex]t[/latex].

[latex] \begin{array}{c}\frac{2(t-2)-(t-2)}{(t+1)(t-2)}\\\\\frac{2t-4-t+2}{(t+1)(t-2)}\\\\\frac{t-2}{(t+1)(t-2)}\end{array}[/latex]

The numerator and denominator have a common factor of [latex]t–2[/latex], so the rational expression can be simplified.

[latex]\large\begin{array}{c}\frac{\cancel{t-2}}{(t+1)\cancel{(t-2)}}=\frac{1}{t+1}\end{array}[/latex]

[latex-display] \displaystyle \frac{2}{t+1}-\frac{t-2}{{{t}^{2}}-t-2}=\frac{1}{t+1}[/latex],   [latex]t\ne -1,2[/latex-display]

In the next example, we will give less instruction. See if you can find the LCD yourself before you look at the answer.

Example

Subtract the rational expressions: [latex]\frac{6}{{x}^{2}+4x+4}-\frac{2}{{x}^{2}-4}[/latex], and define the domain. State the difference in simplest form.

Answer: Note that the denominator of the first expression is a perfect square trinomial, and the denominator of the second expression is a difference of squares so they can be factored using special products. [latex-display]\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}[/latex-display]

The domain is [latex]x\ne-2,2[/latex]

[latex-display]\frac{6}{{x}^{2}+4x+4}-\frac{2}{{x}^{2}-4}=\frac{4(x-4)}{(x+2)^2(x-2)}[/latex],   [latex]x\ne-2,2[/latex-display]

In the previous example, the LCD was  [latex]\left(x+2\right)^2\left(x-2\right)[/latex].  The reason we need to include [latex]\left(x+2\right)[/latex] two times is because it appears two times in the expression [latex]\frac{6}{{x}^{2}+4x+4}[/latex]. The video that follows contains an example of subtracting rational expressions. https://www.youtube.com/watch?v=MMlNtCrkakI&feature=youtu.be

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CC licensed content, Original

  • Screenshot: Caution. Provided by: Lumen Learning License: CC BY: Attribution.
  • Image: What do they have in common?. Provided by: Lumen Learning License: CC BY: Attribution.
  • Revision and Adaptation. Provided by: Lumen Learning License: CC BY: Attribution.
  • Ex: Add Rational Expressions with Unlike Denominators. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Subtract Rational Expressions with Unlike Denominators and Give the Domain. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Subtract Rational Expressions with Unlike Denominators and Give the Domain. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.

CC licensed content, Shared previously

  • Unit 15: Rational Expressions, from Developmental Math: An Open Program. Provided by: Lumen Learning License: CC BY: Attribution.