# Adding and Subtractracting Rational Expressions Part II

### Learning Outcomes

- Add and subtract rational expressions that share no common factors
- Add and subtract more than two rational expressions

## Add and Subtract Rational Expressions with No Common Factor

In the next example, we show how to find a common denominator when there are no common factors in the expressions.### Example

Subtract [latex] \displaystyle \frac{3y}{2y-1}-\frac{4}{y-5}[/latex], and give the domain. State the difference in simplest form.Answer:
Neither [latex]2y–1[/latex] nor [latex]y–5[/latex] can be factored. Because they* *have no common factors, the least common multiple, which will become the least common denominator, is the product of these denominators.

[latex]\text{LCM}=\left(2y-1\right)\left(y-5\right)[/latex]

Multiply each expression by the equivalent of [latex]1[/latex] that will give it the common denominator. Then rewrite the subtraction problem with the common denominator. It makes sense to keep the denominator in factored form in order to check for common factors.[latex]\begin{array}{c}\frac{3y}{2y-1}\cdot \frac{y-5}{y-5}=\frac{3y(y-5)}{(2y-1)(y-5)}\\\\\frac{4}{y-5}\cdot \frac{2y-1}{2y-1}=\frac{4(2y-1)}{(2y-1)(y-5)}\\\\\frac{3y(y-5)}{(2y-1)(y-5)}-\frac{4(2y-1)}{(2y-1)(y-5)}\end{array}[/latex]

Subtract and simplify.[latex]\begin{array}{c}\frac{3{{y}^{2}}-15y}{(2y-1)(y-5)}-\frac{8y-4}{(2y-1)(y-5)}\\\\\frac{3{{y}^{2}}-15y-(8y-4)}{(2y-1)(y-5)}\\\\\frac{3{{y}^{2}}-15y-8y+4}{(2y-1)(y-5)}\end{array}[/latex]

The domain is found by setting the original denominators equal to zero.[latex]\begin{array}{l}2y-1=0\text{ and }y-5=0\\\,\,\,y=\frac{1}{2}\,\,\,\,\,\,\,\text{ and }y=5\end{array}[/latex]

The domain is [latex]y\ne\frac{1}{2}, y\ne5[/latex]#### Answer

[latex-display] \displaystyle \frac{3y}{2y-1}-\frac{4}{y-5}=\frac{3{{y}^{2}}-23y+4}{2{{y}^{2}}-11y+5},y\ne \frac{1}{2},5[/latex-display]### Try It

[ohm_question]40252[/ohm_question]## Add and Subtract More Than Two Rational Expressions

You may need to combine more than two rational expressions. While this may seem pretty straightforward if they all have the same denominator, what happens if they do not? In the example below, notice how a common denominator is found for three rational expressions. Once that is done, the addition and subtraction of the terms looks the same as earlier, when you were only dealing with two terms.### Example

Simplify[latex]\frac{2{{x}^{2}}}{{{x}^{2}}-4}+\frac{x}{x-2}-\frac{1}{x+2}[/latex], and give the domain. State the result in simplest form.Answer:
Find the least common multiple by factoring each denominator. Multiply each factor 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, as does [latex]\left(x–2\right)[/latex]. This means the LCM is [latex]\left(x+2\right)\left(x–2\right)[/latex].

[latex]\begin{array}{l}x^{2}-4=\left(x+2\right)\left(x-2\right)\\\,\,x-2=x-2\\\,\,x+2=x+2\\\,\,\text{LCM}=\left(x+2\right)\left(x-2\right)\end{array}[/latex]

The LCM becomes the common denominator. Multiply each expression by the equivalent of [latex]1[/latex] that will give it the common denominator.[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)}\\\frac{1}{x+2}\cdot \frac{x-2}{x-2}=\frac{1(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)}-\frac{1(x-2)}{(x+2)(x-2)}[/latex]

Combine the numerators.[latex] \begin{array}{c}\frac{2{{x}^{2}}+x(x+2)-1(x-2)}{(x+2)(x-2)}\\\\\frac{2{{x}^{2}}+{{x}^{2}}+2x-x+2}{(x+2)(x-2)}\end{array}[/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}}+x+2[/latex], this expression is in simplest form.[latex]\frac{3{{x}^{2}}+x+2}{(x+2)(x-2)}[/latex]

#### Answer

[latex-display] \displaystyle \frac{2{{x}^{2}}}{{{x}^{2}}-4}+\frac{x}{x-2}-\frac{1}{x+2}=\frac{3{{x}^{2}}+x+2}{(x+2)(x-2)}[/latex][latex] \displaystyle x\ne 2,-2[/latex-display]### Example

Simplify[latex]\frac{{{y}^{2}}}{3y}-\frac{2}{x}-\frac{15}{9}[/latex], and give the domain. State the result in simplest form.Answer: Find the least common multiple by factoring each denominator. Multiply each factor the maximum number of times it appears in a single factorization.

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,3y=3\cdot{y}\\\,\,\,\,\,\,\,\,\,\,x=x\\\,\,\,\,\,\,\,\,\,\,9=3\cdot3\\\text{LCM}=3\cdot3\cdot{x}\cdot{y}\\\text{LCM}=9xy\end{array}[/latex]

The LCM becomes the common denominator. Multiply each expression by the equivalent of [latex]1[/latex] that will give it the common denominator.[latex]\begin{array}{r}\frac{{{y}^{2}}}{3y}\cdot \frac{3x}{3x}=\frac{3x{{y}^{2}}}{9xy}\\\\\frac{2}{x}\cdot \frac{9y}{9y}=\frac{18y}{9xy}\,\,\\\\\frac{15}{9}\cdot \frac{xy}{xy}=\frac{15xy}{9xy}\end{array}[/latex]

Rewrite the original problem with the common denominator.[latex]\frac{3x{{y}^{2}}}{9xy}-\frac{18y}{9xy}-\frac{15xy}{9xy}[/latex]

Combine the numerators.[latex]\frac{3x{{y}^{2}}-18y-15xy}{9xy}[/latex]

Check for simplest form.[latex]\large\begin{array}{c}\frac{3y(xy-6-5x)}{9xy}\\\\=\frac{3y(xy-6-5x)}{3y(3x)}\\\\=\frac{\cancel{3y}(xy-6-5x)}{\cancel{3y}(3x)}\\\\=\frac{xy-6-5x}{3x}\end{array}[/latex]

The domain is found by setting the denominators equal to zero. [latex]9=0[/latex] is nonsense, so we don't need to worry about that denominator.[latex]\begin{array}{l}3y=0\text{ and }x=0\\y=0\text{ and }x=0\end{array}[/latex]

The domain is [latex]y\ne0, x\ne0[/latex]#### Answer

[latex-display]\frac{{{y}^{2}}}{3y}-\frac{2}{x}-\frac{15}{9}=\frac{xy-5x-6}{3x},y\ne 0,x\ne 0[/latex-display]### Try It

[ohm_question]189261[/ohm_question]## Summary

The methods shown here will help you when you are solving rational equations later on. To add and subtract rational expressions that share common factors, you first identify which factors are missing from each expression, and build the LCD with them. To add and subtract rational expressions with no common factors, the LCD will be the product of all the factors of the denominators.## Contribute!

## Licenses & Attributions

### CC licensed content, Original

- Image: No common factors..
**Provided by:**Lumen Learning**License:**CC BY: Attribution. - Screenshot: Add and subtract.
**Provided by:**Lumen Learning**License:**CC BY: Attribution. - Revision and Adaptation.
**Provided by:**Lumen Learning**License:**CC BY: Attribution. - Subtract Rational Expressions with UnLike Denominators - 3 Expressions.
**Authored by:**James Sousa (Mathispower4u.com) for Lumen Learning.**License:**CC BY: Attribution. - Add and Subtract Rational Expressions with UnLike Denominators - 3 Expressions.
**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:**Monterey Institute of Technology and Education**Located at:**https://www.nroc.org/.**License:**CC BY: Attribution. - Ex: Add Rational Expressions with Unlike Denominators.
**Authored by:**James Sousa (Mathispower4u.com) .**License:**CC BY: Attribution.