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Study Guides > ALGEBRA / TRIG I

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
So far all the rational expressions you've added and subtracted have shared some factors. What happens when they don't have factors in common?
fish, screwdriver, socks and soccer ball to show no common factors No Common Factors

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 3y2y14y5 \displaystyle \frac{3y}{2y-1}-\frac{4}{y-5}, and give the domain. State the difference in simplest form.

Answer: Neither 2y12y–1 nor y5y–5 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.

LCM=(2y1)(y5)\text{LCM}=\left(2y-1\right)\left(y-5\right)

Multiply each expression by the equivalent of 11 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.

3y2y1y5y5=3y(y5)(2y1)(y5)4y52y12y1=4(2y1)(2y1)(y5)3y(y5)(2y1)(y5)4(2y1)(2y1)(y5)\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}

Subtract and simplify.

3y215y(2y1)(y5)8y4(2y1)(y5)3y215y(8y4)(2y1)(y5)3y215y8y+4(2y1)(y5)\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}

The domain is found by setting the original denominators equal to zero.

2y1=0 and y5=0   y=12        and y=5\begin{array}{l}2y-1=0\text{ and }y-5=0\\\,\,\,y=\frac{1}{2}\,\,\,\,\,\,\,\text{ and }y=5\end{array}

The domain is y12,y5y\ne\frac{1}{2}, y\ne5

Answer

3y2y14y5=3y223y+42y211y+5,y12,5 \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

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

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

Simplify2x2x24+xx21x+2\frac{2{{x}^{2}}}{{{x}^{2}}-4}+\frac{x}{x-2}-\frac{1}{x+2}, 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 22 or 2-2 because the denominators would be 00. (x+2)\left(x+2\right) appears a maximum of one time, as does (x2)\left(x–2\right). This means the LCM is (x+2)(x2)\left(x+2\right)\left(x–2\right).

x24=(x+2)(x2)  x2=x2  x+2=x+2  LCM=(x+2)(x2)\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}

The LCM becomes the common denominator. Multiply each expression by the equivalent of 11 that will give it the common denominator.

2x2x24=2x2(x+2)(x2)xx2x+2x+2=x(x+2)(x+2)(x2)1x+2x2x2=1(x2)(x+2)(x2)\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}

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.

2x2(x+2)(x2)+x(x+2)(x+2)(x2)1(x2)(x+2)(x2) \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)}

Combine the numerators.

2x2+x(x+2)1(x2)(x+2)(x2)2x2+x2+2xx+2(x+2)(x2) \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}

Check for simplest form. Since neither (x+2)\left(x+2\right) nor (x2)\left(x-2\right) is a factor of 3x2+x+23{{x}^{2}}+x+2, this expression is in simplest form.

3x2+x+2(x+2)(x2)\frac{3{{x}^{2}}+x+2}{(x+2)(x-2)}

Answer

2x2x24+xx21x+2=3x2+x+2(x+2)(x2)[/latex][latex]x2,2 \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

In the video that follows we present an example of subtracting 33 rational expressions with unlike denominators. One of the terms being subtracted is a number, so the denominator is 11. https://www.youtube.com/watch?v=c-8xQyU0ch0&feature=youtu.be

Example

Simplifyy23y2x159\frac{{{y}^{2}}}{3y}-\frac{2}{x}-\frac{15}{9}, 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.

        3y=3y          x=x          9=33LCM=33xyLCM=9xy\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}

The LCM becomes the common denominator. Multiply each expression by the equivalent of 11 that will give it the common denominator.

y23y3x3x=3xy29xy2x9y9y=18y9xy  159xyxy=15xy9xy\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}

Rewrite the original problem with the common denominator.

3xy29xy18y9xy15xy9xy\frac{3x{{y}^{2}}}{9xy}-\frac{18y}{9xy}-\frac{15xy}{9xy}

Combine the numerators.

3xy218y15xy9xy\frac{3x{{y}^{2}}-18y-15xy}{9xy}

Check for simplest form.

3y(xy65x)9xy=3y(xy65x)3y(3x)=3y(xy65x)3y(3x)=xy65x3x\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}

The domain is found by setting the denominators equal to zero. 9=09=0 is nonsense, so we don't need to worry about that denominator.

3y=0 and x=0y=0 and x=0\begin{array}{l}3y=0\text{ and }x=0\\y=0\text{ and }x=0\end{array}

The domain is y0,x0y\ne0, x\ne0

Answer

y23y2x159=xy5x63x,y0,x0\frac{{{y}^{2}}}{3y}-\frac{2}{x}-\frac{15}{9}=\frac{xy-5x-6}{3x},y\ne 0,x\ne 0

In this last video, we present another example of adding and subtracting three rational expressions with unlike denominators. https://www.youtube.com/watch?v=43xPStLm39A&feature=youtu.be
Add and Subtract Add and Subtract

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.

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