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Study Guides > College Algebra CoRequisite Course

Adding and Subtracting Rational Expressions

Learning Outcomes

  • Find the least common denominator 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 \dfrac{5}{24}+\dfrac{1}{40}& =& \dfrac{25}{120}+\dfrac{3}{120}\hfill \\ & =& \dfrac{28}{120}\hfill \\ & =& \dfrac{7}{30}\hfill \end{array}

how did we know what number to use for the denominator?

In the example above, we rewrote the fractions as equivalent fractions with a common denominator of 120. Recall that we use the least common multiple of the original denominators. To find the LCM of 24 and 40, rewrite 24 and 40 as products of primes, then select the largest set of each prime appearing. 24=23⋅324 = 2^3\cdot3 40=23⋅540=2^3\cdot5 We choose 23⋅3⋅5=1202^3\cdot3\cdot5=120 as the LCM, since that's the largest number of factors of 2, 3, and 5 we see. The LCM is 120. We multiply each numerator with just enough of the LCM to make each denominator 120 to get the equivalent fractions. When referring to fractions, we call the LCM the least common denominator, or the LCD.
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\dfrac{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\dfrac{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\dfrac{5}{x}+\dfrac{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}\dfrac{5}{x}\cdot \dfrac{y}{y}+\dfrac{6}{y}\cdot \dfrac{x}{x}\\ \dfrac{5y}{xy}+\dfrac{6x}{xy}\end{array}
Now that the expressions have the same denominator, we simply add the numerators to find the sum.
6x+5yxy\dfrac{6x+5y}{xy}

Analysis of the Solution

Multiplying by yy\dfrac{y}{y} or xx\dfrac{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.

Try It

[ohm_question]110918-110919[/ohm_question]

Example: Subtracting Rational Expressions

Subtract the rational expressions:
6x2+4x+4−2x2−4\dfrac{6}{{x}^{2}+4x+4}-\dfrac{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 the LCD as the 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}\dfrac{6}{{\left(x+2\right)}^{2}}-\dfrac{2}{\left(x+2\right)\left(x - 2\right)}\hfill & \text{Factor}.\hfill \\ \dfrac{6}{{\left(x+2\right)}^{2}}\cdot \dfrac{x - 2}{x - 2}-\dfrac{2}{\left(x+2\right)\left(x - 2\right)}\cdot \dfrac{x+2}{x+2}\hfill & \text{Multiply each fraction to get the LCD as the denominator}.\hfill \\ \dfrac{6\left(x - 2\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}-\dfrac{2\left(x+2\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Multiply}.\hfill \\ \dfrac{6x - 12-\left(2x+4\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Apply distributive property}.\hfill \\ \dfrac{4x - 16}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Subtract}.\hfill \\ \dfrac{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\dfrac{3}{x+5}-\dfrac{1}{x - 3}.

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

[ohm_question]39519[/ohm_question]

Simplifying Complex Rational Expressions

A complex rational expression is a rational expression that contains additional rational expressions in the numerator, the denominator, or both. We can simplify complex rational expressions by rewriting the numerator and denominator as single rational expressions and dividing. The complex rational expression a1b+c\dfrac{a}{\dfrac{1}{b}+c} can be simplified by rewriting the numerator as the fraction a1\dfrac{a}{1} and combining the expressions in the denominator as 1+bcb\dfrac{1+bc}{b}. We can then rewrite the expression as a multiplication problem using the reciprocal of the denominator. We get a1⋅b1+bc\dfrac{a}{1}\cdot \dfrac{b}{1+bc} which is equal to ab1+bc\dfrac{ab}{1+bc}.

How To: Given a complex rational expression, simplify it

  1. Combine the expressions in the numerator into a single rational expression by adding or subtracting.
  2. Combine the expressions in the denominator into a single rational expression by adding or subtracting.
  3. Rewrite as the numerator divided by the denominator.
  4. Rewrite as multiplication.
  5. Multiply.
  6. Simplify.

Example: Simplifying Complex Rational Expressions

Simplify: y+1xxy\dfrac{y+\dfrac{1}{x}}{\dfrac{x}{y}} .

Answer: Begin by combining the expressions in the numerator into one expression.

y⋅xx+1xMultiply by xxto get LCD as denominator.xyx+1xxy+1xAdd numerators.\begin{array}{cc}y\cdot \dfrac{x}{x}+\dfrac{1}{x}\hfill & \text{Multiply by }\dfrac{x}{x}\text{to get LCD as denominator}.\hfill \\ \dfrac{xy}{x}+\dfrac{1}{x}\hfill & \\ \dfrac{xy+1}{x}\hfill & \text{Add numerators}.\hfill \end{array}
Now the numerator is a single rational expression and the denominator is a single rational expression.
xy+1xxy\dfrac{\dfrac{xy+1}{x}}{\dfrac{x}{y}}
We can rewrite this as division and then multiplication.
xy+1x÷xyxy+1x⋅yxRewrite as multiplication.y(xy+1)x2Multiply.\begin{array}{cc}\dfrac{xy+1}{x}\div \dfrac{x}{y}\hfill & \\ \dfrac{xy+1}{x}\cdot \dfrac{y}{x}\hfill & \text{Rewrite as multiplication}\text{.}\hfill \\ \dfrac{y\left(xy+1\right)}{{x}^{2}}\hfill & \text{Multiply}\text{.}\hfill \end{array}

Try It

Simplify: xy−yxy\dfrac{\dfrac{x}{y}-\dfrac{y}{x}}{y}

Answer: x2−y2xy2\dfrac{{x}^{2}-{y}^{2}}{x{y}^{2}}

[ohm_question]3078-3080-59554[/ohm_question]

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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