Adding and Subtracting Rational Expressions

In beginning math, students usually learn how to add and subtract whole numbers before they're taught multiplication and division. However, with fractions and rational expressionsA fraction with a polynomial in the numerator and/or denominator., 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 trickier than multiplication because, like with numeric fractions, the process involves finding common denominators. By working carefully and writing down the steps along the way, we can keep track of all of the numbers and variables and perform the operations accurately.

We follow the same process for adding rational expressions that we do to combine numeric fractions. To add fractions with like denominators, we add the numerators and keep the same denominator. After adding, we express the fraction in simplest terms:

We follow the same process to add rational expressions with like denominators, but we also have to describe the domainThe set of all possible inputs of a function which allow the function to work., the set of all possible values for the variables. The excluded valuesA value for a variable that is not allowed in an expression, such as a variable in a rational expression that would make the denominator equal zero. of the domain are any values of the variable(s) that result in any denominator being equal to `0`.

Example
Problem	Add, simplify, and state the domain of `(2x^2)/(x+4)+(8x)/(x+4)`.
	`x + 4 = 0` `x = -4`	Determine the excluded values by setting the denominator equal to `0` and solving for `x`.
	`(2x^2+8x)/(x+4)`	Since the denominators are the same, add the numerators.
	`(2x(x+4))/(x+4)`	Factor the numerator.
	`(x+4)/(x+4)*2x`	Rewrite the common factor as multiplication by `1`.
Answer	`2x, x !=-4`

To subtract rational expressions with like denominators, we follow the same process we use to subtract fractions with like denominators. The process is just like addition of rational expressions, except that we subtract.

Example
Problem	Subtract, simplify, and state the domain of `(4x-7)/(x+6)-(2x-8)/(x+6)`.
	`x + 6 = 0` `x = -6` `-6` is an excluded value.	Determine the excluded values by setting the denominator equal to `0` and solving for `x`.
	`(4x-7-(2x-8))/(x+6)`	Subtract the second numerator from the first and keep the denominator the same.
	`(4x-7-2x-(-8))/(x+6)` `(4x-2x-7-(-8))/(x+6)`	Be careful to distribute the negative to both terms of the second numerator.
	`(2x+1)/(x+6)`	Note that `-7 - (-8) = -7 + 8 = 1`. This is the final answer because this rational expression cannot be simplified.
Answer	`(2x+1)/(x+6), x!=-6`

Before adding and subtracting rational expressions with unlike denominators, we need to find a common denominator. This process is once again similar to the one used for adding and subtracting numeric fractions with unlike denominators. Let’s look at a numeric example to start.

Since the denominators are `6`, `10`, and `4`, we want to find the least common denominatorThe smallest number or expression that is a multiple of all the denominators in a group of fractions or rational expressions. and express each fraction with this denominator before adding. (By the way, you can add fractions by finding any common denominator; it does not have to be the least. We 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 multipleThe smallest number or expression that is a multiple of a group of numbers or expressions. of `4`, `6`, and `10`. 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 smallest of these numbers will be the least common denominator.

The other method is to use prime factorizationThe process of breaking a number down into its prime factors., the process of finding the prime number factors of a number. This is how the method works with numbers:

Example
Problem	Use prime factorization to find the least common multiple of `6`, `10`, and `4`.
	`6 = bb32` `10 = bb52` `4 = bb2*bb2`	First, find the prime factors of each denominator.
	`352*2`	Multiply all of the prime factors. Use each number the maximum number of times it appears in a single factorization. In this case, `2` is used twice because it appears twice in the prime factorization of `4`.
Answer	`60`

We found the same least common multiple with both methods. Prime factorization was faster though, because we didn't have to make a chart full of multiples.

Let's move on. Now that we have found the least common multiple, we'll use that number as the least common denominator of our fractions. We'll multiply each fraction by the fractional form of `1` that will produce a denominator of `60`:

We can find least common denominators for rational expressions, too, and use them to allow us to add rational expressions with unlike denominators:

Example
Problem	Add: `(2m)/(15m^2n^3)+(3n)/(21mn^2)`
	`15m^2n^3 = 0` `m = 0` or `n = 0` `21mn^2 = 0` `m = 0` or `n = 0`	Find excluded values.
	`15m^2n^3 = 35mmnnn` `21mn^2 = 37mnn`	Find the prime factors of each denominator.
	` 357mmnn*n` `105m^2n^3`	Find the least common multiple. `3` appears exactly once in both of the expressions, so it will appear once in the least common denominator. Both `5` and `7` appear at most once. For the variables, the most `m` appears is twice, and the most `n` appears is three times.
	`15m^2n^3 = 35mmnnn` `105m^2n^3 = 35bb7mmnnn` `(2m)/(15m^2n^3)7/7=(14m)/(105m^2n^3)` `21mn^2 = 37mnn` `105m^2n^3 = 3bb57bbmmbbnnn` `(3n)/(21mn^2)(5mn)/(5mn)=(15mn^2)/(105m^2n^3)`	Rewrite the rational expressions to each have a denominator of `105m^2n^3`. Compare the prime factors of each denominator and the common denominator; to get the common denominator, multiply the original denominator by whatever factors are missing.
	`(14m)/(105m^2n^3)+(15mn^2)/(105m^2n^3)` `(14m+15mn^2)/(105m^2n^3)`	Add the numerators and keep the denominator the same.
	`(m\ (14+15n^2))/(m\ (105mn^3))` `(14+15n^2)/(105mn^3)` `(15n^2+14)/(105mn^3)`	Simplify by finding common factors in the numerator and denominator.
Answer	`(15n^2+14)/(105mn^3), m!=0" and "n!=0`

That took a while, but we got through it. Adding rational expressions can be a lengthy process, but if we take it one step at a time, we'll get there.

Ready to try subtracting rational expressions? We'll use the same basic technique of finding the least common denominator and rewriting each rational expression to have that denominator:

Example
Problem	Subtract `2/(t+1)-(t-2)/(t^2-t-2)` and state any excluded values.
	`t + 1 = 0` `t = -1` `t^2 - t - 2 = (t - 2)(t + 1) = 0` `t = -1, 2`	Determine the excluded values by setting each denominator equal to `0` and solving.
	`t + 1 = t + 1` `t^2 - t - 2 = (t-2)(t + 1)`	Find a least common multiple by factoring each denominator. Since `t + 1` is already a factor of `t^2 - t - 2`, the least common denominator is `(t + 1)(t-2)`.
	`2/(t+1)*(t-2)/(t-2)-(t-2)/(t^2-t-2)` `(2(t-2))/((t+1)(t-2))-(t-2)/((t+1)(t-2))`	Multiply the first expression by the equivalent of `1` to give it the common denominator. Then rewrite the rational expressions with the common denominator.
	`(2(t-2)-(t-2))/((t+1)(t-2))` `(2t-4-(t-2))/((t+1)(t-2))` `(2t-4-t+2)/((t+1)(t-2))` `(t-2)/((t+1)(t-2))`	Subtract the numerators and simplify.
	`(t-2)/((t-2)(t+1))` `(t-2)/(t-2)*1/(t+1)` `1/(t+1)`	The numerator and denominator have a common factor of `t - 2`, and so the rational expression can be simplified.
Answer	`1/(t+1), t!=-1" or "2`

So far, all the rational expressions we've added and subtracted have shared some factors. What happens when they don't have factors in common?

Example
Problem	Subtract `(3y)/(2y-1)-4/(y-5)` and state the excluded values.
	`2y-1 = 0` `y=1/2` `y - 5 = 0` `y = 5` `1/2` and `5` are excluded values.	Find the excluded values.
	`2y-1` `y-5` common denominator `=` `(2y - 1)(y - 5)`	Find a least common multiple by factoring each denominator. Neither `2y - 1` or `y - 5` can be factored. Because they have no common factors, the least common denominator is the product of these denominators.
	`(3y)/(2y-1)(y-5)/(y-5)-4/(y-5)(2y-1)/(2y-1)` `(3y(y-5))/(2y^2-11y+5)-(4(2y-1))/(2y^2-11y+5)`	Multiply each expression by the equivalent of `1` that will give it the common denominator. Then rewrite the rational expressions with the common denominator.
	`(3y^2-15y-(8y-4))/(2y^2-11y+5)`	Subtract the numerators.
	`(3y^2-23y+4)/(2y^2-11y+5)`	Simplify.
Answer	`(3y^2-23y+4)/(2y^2-11y+5), y!=1/2, 5`

To add and subtract rational expressions, we apply the same idea used for adding and subtracting fractions: first find a common denominator. The least common denominator is the same as the least common multiple and can be found by listing multiples of each denominator or through prime factorization.

When working with rational expressions, it is always important to include the excluded values of the domain with the answer.

Learning Objective

Introduction

Adding and Subtracting Rational Expressions with Like Denominators

Adding and Subtracting Rational Expressions with Unlike Denominators

Summary