Adding and Subtracting Rational Expressions Part I
Learning Outcomes
- Add and subtract rational expressions with like denominators
- Add and subtract rational expressions with unlike denominators using a greatest common denominator
Adding and Subtracting Rational Expressions with Like Denominators
Adding rational expressions with the same denominator is the simplest place to start, so let’s begin there. To add fractions with like denominators, add the numerators and keep the same denominator. Then simplify the sum. You know how to do this with numeric fractions.[latex] \begin{array}{c}\frac{2}{9}+\frac{4}{9}=\frac{6}{9}\\\\\frac{6}{9}=\frac{3\cdot 2}{3\cdot 3}=\frac{3}{3}\cdot \frac{2}{3}=1\cdot \frac{2}{3}=\frac{2}{3}\end{array}[/latex]
Follow the same process to add rational expressions with like denominators. Let's try one.Example
Add [latex] \displaystyle \frac{2{{x}^{2}}}{x+4}+\frac{8x}{x+4}[/latex], and define the domain. State the sum in simplest form.Answer: Since the denominators are the same, add the numerators.
[latex]\frac{2{{x}^{2}}+8x}{x+4}[/latex]
Factor the numerator.[latex]\frac{2x(x+4)}{x+4}[/latex]
Simplify common factors and.[latex]\large\begin{array}{c}\frac{2x\cancel{(x+4)}}{\cancel{x+4}}\\\\=\frac{2x}{1}\end{array}[/latex]
The domain is found by setting the denominators in the original sum equal to zero.
[latex]\begin{array}{l}x+4=0\\x=-4\end{array}[/latex]
The domain is [latex]x\ne-4[/latex]
Answer
[latex-display] \displaystyle \frac{2{{x}^{2}}}{x+4}+\frac{8x}{x+4}=2x,x\ne -4[/latex-display]Example
Subtract[latex]\frac{4x+7}{x+6}-\frac{2x+8}{x+6}[/latex], and define the domain. State the difference in simplest form.Answer: Subtract the second numerator from the first and keep the denominator the same.
[latex]\frac{4x+7-(2x+8)}{x+6}[/latex]
Be careful to distribute the negative to both terms of the second numerator.[latex]\frac{4x+7-2x-8}{x+6}[/latex]
Combine like terms. This rational expression cannot be simplified any further.[latex]\frac{2x-1}{x+6}[/latex]
The domain is found from the denominators of original expression.
[latex]\begin{array}{l}x+6=0\\x=-6\end{array}[/latex]
The domain is [latex]x\ne-6[/latex]
Answer
[latex-display] \displaystyle \frac{4x+7}{x+6}-\frac{2x+8}{x+6}=\frac{2x-1}{x+6},\text{}x\ne-6[/latex-display]Try It
[ohm_question]40242[/ohm_question]Adding and Subtracting Rational Expressions with Unlike Denominators
[latex] \displaystyle \frac{5}{6}+\frac{8}{10}+\frac{3}{4}[/latex]
Since the denominators are [latex]6[/latex], [latex]10[/latex], and [latex]4[/latex], you want to find the least common denominator and express each fraction with this denominator before adding. (BTW, you can add fractions by finding any common denominator; it does not have to be the least. You focus on using the least because then there is less simplifying to do. But either way works.) Finding the least common denominator is the same as finding the least common multiple of [latex]4[/latex], [latex]6[/latex], and [latex]10[/latex]. There are a couple of ways to do this. The first is to list the multiples of each number and determine which multiples they have in common. The least of these numbers will be the least common denominator.Number |
Multiples |
|||||||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
[latex]4[/latex] | [latex]8[/latex] | [latex]12[/latex] | [latex]16[/latex] | [latex]20[/latex] | [latex]24[/latex] | [latex]28[/latex] | [latex]32[/latex] | [latex]36[/latex] | [latex]40[/latex] | [latex]44[/latex] | [latex]48[/latex] | [latex]52[/latex] | [latex]56[/latex] | [latex]\textbf{60}[/latex] | [latex]64[/latex] | |
[latex]6[/latex] | [latex]12[/latex] | [latex]18[/latex] | [latex]24[/latex] | [latex]30[/latex] | [latex]36[/latex] | [latex]42[/latex] | [latex]48[/latex] | [latex]54[/latex] | [latex]\textbf{60}[/latex] | [latex]66[/latex] | [latex]68[/latex] | |||||
[latex]10[/latex] | [latex]20[/latex] | [latex]30[/latex] | [latex]40[/latex] | [latex]50[/latex] | [latex]\textbf{60}[/latex] | [latex]70[/latex] | [latex]80[/latex] |
Example
Use prime factorization to find the least common multiple of [latex]6[/latex], [latex]10[/latex], and [latex]4[/latex].Answer: First, find the prime factorization of each denominator.
[latex]\begin{array}{r}6=3\cdot2\\10=5\cdot2\\4=2\cdot2\end{array}[/latex]
The LCM will contain factors of [latex]2[/latex],[latex]3[/latex], and [latex]5[/latex]. Multiply each number the maximum number of times it appears in a single factorization. In this case, [latex]3[/latex] appears once, [latex]5[/latex] appears once, and [latex]2[/latex] is used twice because it appears twice in the prime factorization of [latex]4[/latex]. Therefore, the LCM of [latex]6[/latex], [latex]10[/latex], and [latex]4[/latex] is [latex]3\cdot5\cdot2\cdot2[/latex], or [latex]60[/latex].[latex]\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,6=3\cdot2\\\,\,\,\,\,\,\,\,10=5\cdot2\\\,\,\,\,\,\,\,\,\,\,\,4=2\cdot2\\\text{LCM}=3\cdot5\cdot2\cdot2\end{array}[/latex]
Answer
The least common multiple of [latex]6, 10[/latex], and [latex]4[/latex] is [latex]60[/latex].[latex]\begin{array}{r}\frac{5}{6}\cdot \frac{10}{10}=\frac{50}{60}\\\\\frac{8}{10}\cdot\frac{6}{6}=\frac{48}{60}\\\\\frac{3}{4}\cdot\frac{15}{15}=\frac{45}{60}\end{array}[/latex]
Now that you have like denominators, add the fractions:[latex]\frac{50}{60}+\frac{48}{60}+\frac{45}{60}=\frac{143}{60}[/latex]
In the next example, we show how to find the least common multiple of a rational expression with a monomial in the denominator.Example
Add[latex]\frac{2n}{15m^{2}}+\frac{3n}{21m}[/latex], and give the domain. State the sum in simplest form.Answer: Find the prime factorization of each denominator.
[latex]\begin{array}{l}15m^{2}\,=\,3\cdot5\cdot{m}\cdot{m}\\\,\,\,21m\,=\,3\cdot7\cdot{m}\end{array}[/latex]
Find the least common multiple. [latex]3[/latex] appears exactly once in both of the expressions, so it will appear once in the least common multiple. Both [latex]5[/latex] and [latex]7[/latex] appear at most once. For the variables, the most m appears is twice. Use the least common multiple for your new common denominator, it will be the LCD.[latex]\begin{array}{l}15m^{2}\,=\,3\cdot5\cdot{m}\cdot{m}\\21m\,\,\,=\,3\cdot7\cdot{m}\\\text{LCM}:3\cdot5\cdot7\cdot{m}\cdot{m}\\\text{LCM}:105m^{2}\end{array}[/latex]
Compare each original denominator and the new common denominator. Now rewrite the rational expressions to each have the common denominator of [latex]105m^{2}[/latex]. Remember that m cannot be 0 because the denominators would be [latex]0[/latex]. The first denominator is [latex]15m^{2}[/latex] and the LCD is [latex]105m^{2}[/latex]. You need to multiply [latex]15m^{2}[/latex] by [latex]7[/latex] to get the LCD, so multiply the entire rational expression by [latex]\frac{7}{7}[/latex]. The second denominator is [latex]21m[/latex] and the LCD is [latex]105m^{2}[/latex]. You need to multiply [latex]21m[/latex] by [latex]5m[/latex] to get the LCD, so multiply the entire rational expression by [latex]\frac{5m}{5m}[/latex].[latex]\begin{array}{c}\frac{2n}{15m^{2}}\cdot\frac{7}{7}=\frac{14n}{105m^{2}}\\\\\frac{3n}{21m}\cdot\frac{5m}{5m}=\frac{15mn}{105m^{2}}\end{array}[/latex]
Add the numerators and keep the denominator the same.[latex]\frac{14n}{105{{m}^{2}}}+\frac{15mn}{105{{m}^{2}}}=\frac{14n+15mn}{105{{m}^{2}}}[/latex]
If possible, simplify by finding common factors in the numerator and denominator. This rational expression is already in simplest form because the numerator and denominator have no factors in common.[latex] \displaystyle \frac{n(14+15m)}{105{{m}^{2}}}[/latex]
Answer
[latex-display] \displaystyle \frac{2n}{15{{m}^{2}}}+\frac{3n}{21m}=\frac{n(14+15m)}{105{{m}^{2}}},m\ne 0[/latex-display][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.
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]Subtracting Rational Expressions
Now let’s try subtracting rational expressions. You'll use the same basic technique of finding the least common denominator and rewriting each rational expression to have that denominator.Example
Subtract[latex]\frac{2}{t+1}-\frac{t-2}{{{t}^{2}}-t-2}[/latex], define the domain. State the difference in simplest form.Answer: Find the prime factorization of each denominator. [latex]t+1[/latex] cannot be factored any further, but [latex]{{t}^{2}}-t-2[/latex] can be. Remember that t cannot be [latex]-1[/latex] or [latex]2[/latex] because the denominators would be [latex]0[/latex].
[latex]\begin{array}{c}t+1=t+1\\t^{2}-t-2=\left(t-2\right)\left(t+1\right)\end{array}[/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 least common multiple. In this case, it is easier to leave the common multiple in terms of the factors, so you will not multiply it out. Use the least common multiple for your new common denominator, it will be the LCD.[latex]\begin{array}{c}t+1=t+1\\t^{2}-t-2=\left(t-2\right)\left(t+1\right)\\\text{LCM}:\left(t+1\right)\left(t-1\right)\end{array}[/latex]
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]. 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]
Then rewrite the subtraction problem 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 t.[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]
Answer
[latex-display] \displaystyle \frac{2}{t+1}-\frac{t-2}{{{t}^{2}}-t-2}=\frac{1}{t+1},t\ne -1,2[/latex-display]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]Try it
[ohm_question]3434[/ohm_question]Contribute!
Licenses & Attributions
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.
- Add and Subtract Rational Expressions with Like Denominators and Give the Domain. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Add Rational Expressions with Unlike Denominators and Give the Domain (Mono Denom). 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.