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# Adding & subtracting rational expressions

Have you learned the basics of rational expression addition/subtraction? Great! Now dig deeper with some advanced examples.

#### What we need to know before this lesson

A rational expression is a ratio of two polynomials.
To add or subtract two rational expressions with the same denominator, we simply add or subtract the numerators and write the result over the common denominator.
When the denominators are not the same, we must manipulate them so that they become the same. In other words, we must find a common denominator.
If this is new to you, you may want to check out the following articles first:

#### What you will learn in this lesson

In this lesson, you will practice adding and subtracting rational expressions with different denominators. You will use the least common denominator as your common denominator in these examples and explore why it is beneficial to do so.

## Warm-up: $\frac{3}{x-2}-\frac{2}{x+1}$‍

To subtract two rational expressions, each fraction must have the same denominator.
In this example, we can create a common denominator by multiplying the first fraction by $\left(\frac{x+1}{x+1}\right)$ and the second fraction by $\left(\frac{x-2}{x-2}\right)$.
Then, we can subtract the numerators and write the result over the common denominator.
$\begin{array}{rl}& \phantom{=}\frac{3}{x-2}-\frac{2}{x+1}\\ \\ & =\frac{3}{x-2}\left(\frac{x+1}{x+1}\right)-\frac{2}{x+1}\left(\frac{x-2}{x-2}\right)\\ \\ & =\frac{3\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}-\frac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}\\ \\ & =\frac{3\left(x+1\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+1\right)}\\ \\ & =\frac{3x+3-2x+4}{\left(x-2\right)\left(x+1\right)}\\ \\ & =\frac{x+7}{\left(x-2\right)\left(x+1\right)}\end{array}$

Problem 1
The numerator should be expanded and simplified. The denominator should be either expanded or factored.
$\frac{5x}{x+3}+\frac{4}{x+2}=$

## Least common denominators

### Numerical fractions

Sometimes, the denominators of two fractions are different but have some shared factors.
For example, consider $\frac{3}{4}+\frac{1}{6}$:
$\begin{array}{rl}& \phantom{=}\frac{3}{4}+\frac{1}{6}\\ \\ & =\frac{3}{2\cdot 2}+\frac{1}{2\cdot 3}\\ \\ & =\frac{3}{2\cdot 2}\left(\frac{3}{3}\right)+\frac{1}{2\cdot 3}\left(\frac{2}{2}\right)\\ \\ & =\frac{9}{12}+\frac{2}{12}\\ \\ & =\frac{11}{12}\end{array}$
Notice that the common denominator used in this example was not the product of the two individual denominators ($24$). Instead it was the least common multiple of $4$ and $6$ ($12$).
The least common multiple of the denominators in two or more fractions is called the least common denominator.

### Variable expressions

Now let's apply this reasoning to perform the following addition:
$\frac{2}{\left(x-2\right)\left(x+1\right)}+\frac{3}{\left(x+1\right)\left(x+3\right)}$
First, let's find the least common denominator:
So the least common denominator is $\left(x-2\right)\left(x+1\right)\left(x+3\right)$.
We can add the rational expressions as follows:
$\begin{array}{rl}& \phantom{=}\frac{2}{\left(x-2\right)\left(x+1\right)}+\frac{3}{\left(x+1\right)\left(x+3\right)}\\ \\ & =\frac{2}{\left(x-2\right)\left(x+1\right)}\left(\frac{x+3}{x+3}\right)+\frac{3}{\left(x+1\right)\left(x+3\right)}\left(\frac{x-2}{x-2}\right)\\ \\ & =\frac{2\left(x+3\right)}{\left(x-2\right)\left(x+1\right)\left(x+3\right)}+\frac{3\left(x-2\right)}{\left(x+1\right)\left(x+3\right)\left(x-2\right)}\\ \\ & =\frac{2\left(x+3\right)+3\left(x-2\right)}{\left(x-2\right)\left(x+1\right)\left(x+3\right)}\\ \\ & =\frac{2x+6+3x-6}{\left(x-2\right)\left(x+1\right)\left(x+3\right)}\\ \\ & =\frac{5x}{\left(x-2\right)\left(x+1\right)\left(x+3\right)}\end{array}$

Problem 2
The numerator should be expanded and simplified. The denominator should be either expanded or factored.
$\frac{1}{x\left(x-6\right)}+\frac{3}{\left(x+1\right)\left(x-6\right)}=$

Problem 3
Subtract.
The numerator should be expanded and simplified. The denominator should be either expanded or factored.
$\frac{3x}{2\left(x-1\right)}-\frac{4}{\left(x-1\right)\left(x+2\right)}=$

Challenge problem
The numerator should be expanded and simplified. The denominator should be either expanded or factored.
$\frac{2}{{x}^{2}-1}+\frac{1}{{x}^{2}-3x-4}=$

## Why use the least common denominator?

You may be wondering why it is so important to use the least common denominator to add or subtract rational expressions.
After all, this is not a requirement, and it is easy enough to use other denominators with numerical fractions.
For example, the table below calculates $\frac{3}{4}+\frac{1}{6}$ using two different common denominators: one using the least common denominator ($12$) and the other using the product of the two denominators ($24$).
Least common denominator ($12$)Common denominator ($24$)
$\begin{array}{rl}\frac{3}{4}+\frac{1}{6}& =\frac{3}{4}\left(\frac{6}{6}\right)+\frac{1}{6}\left(\frac{4}{4}\right)\\ \\ & =\frac{18}{24}+\frac{4}{24}\\ \\ & =\frac{22}{24}\\ \\ & =\frac{11}{12}\end{array}$
Notice that when using $24$ as the common denominator, more work was required. The numbers were larger and the resulting fraction needed to be simplified.
This will also happen if you do not use the least common denominator when adding or subtracting rational expressions.
However, with rational expressions, this process is much more difficult because the numerators and denominators will be polynomials instead of integers! You will have to perform arithmetic with higher degree polynomials and factor polynomials to simplify the fraction.
All of this extra work can be avoided by using the least common denominator when adding or subtracting rational expressions.

## Want to join the conversation?

• I'm trying to figure out how to simplify a Rational Expression with subtraction, 3 equations, and they all have different denominators.
• You would have to multiply each of the fractions by the other two denominators in order to make one same denominator
For example,
x/a + y/b + z/c would become (x*b*c + y*a*c + z*a*b) / a*b*c in order to be added or subtracted
Does that make sense?
• How do you do it when it's like this:
m^3 - 3m
(5m+3)(2m-1) - (3m-1)(2m-1)

Because this is confusing me big time.
• m^3/(5m+3)(2m-1) - 3m/(3m-1)(2m-1)

1. *find LCD*; LCD is (5m+3)(3m-1)(2m-1)
2. multiply both num and denom of both fractions by the missing factors to get LCD (first fraction by (3m-1) second fraction by (5m+3))
3. expand numerator if needed
4. *distribute negative sign* to the numerator of the fraction you're subtracting from the other fraction -(15m^2+9m) = +(15m^2-9m)
5. simplify / add like terms

we get: 3m^4-m^3-15m^2-9m/(5m+3)(2m-1)(3m-1)

remember to expand the *numerator*
• These 4 exercises have problems because they do not want to put the exponents
• Use a caret: ^ (Shift + 6) For Example, if you wanted 5x squared, type 5x^2 into the answer box, it should automatically change it.
• Not a question but a comment: the square function is not showing up when I try and enter an answer.
• unless you put it for denominator, you can use the up arrow key instead
• I dont get how you do the exponents! I am so confussed
• (If you are still alive on Khan Academy) For the exponents, break it down into x times something else, like 9x^2 - 45 could be simplified into 9x(x-5). Then you just add/subtract the rational expressions. Hope this helps!
• I still don't understand the technique here.
• In the Check Your Understandings,there needs to be a bottom row of buttons when answering, which includes one to show an exponent.
• Actually, there usually is, only on some problems you don't need it to answer the question, so they don't put it.
• I have a problem that looks like this: 5/(x-1) + 8/(x-1)^2 - 3/(x-1)^3. How should I start this? Should I multiple the first term by (x-1)^2 and (x-1)^3 since those are its two missing denominators, or is there another step I should take?
• You need a common denominator. Since all your denominators have the same factor of (x-1), you common denominator = the one with the highest exponent: LCD = (x-1)^3
Thus, you multiply the first fraction's numerator & denominator by (x-1)^2 to get it to a denominator of (x-1)^3. And, you multiply the 2nd fraction's numerator & denominator by (x-1) to get it to a denominator of (x-1)^3.

Hope this helps. If you have more questions, comment back.
• I believe you did something wrong. In the answer above you said the final answer was 5x^2+14x+12/(x+2)(x+3), however, this could further be simplified. We could do 5x^2 + 10x + 4x + 12, and that would factor too (5x+4)(x+2)/(x+2)(x+3). We then could cancel out the (x+2) and we would get (5x+4)/(x+3), where x can't be -2 or -3.
(1 vote)
• Sorry, but your factors don't work.
(5x+4)(x+2) = 5x^2 +10x +4x + 8 = 5x^2+14x+8
You need factors that create 5x^2+14x+12.

To factor 5x^2+14x+12, you would need 2 factors of 5(12) = 60, that add to 14. There are none. So, it is not factorable.

Hope this helps.