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: $\dfrac{3}{x-2}-\dfrac{2}{x+1}$start fraction, 3, divided by, x, minus, 2, end fraction, minus, start fraction, 2, divided by, x, plus, 1, end fraction

To subtract two rational expression, each fraction must have the same denominator!
In this example, we can create a common denominator by multiplying the first fraction by left parenthesis, start fraction, x, plus, 1, divided by, x, plus, 1, end fraction, right parenthesis and the second fraction by left parenthesis, start fraction, x, minus, 2, divided by, x, minus, 2, end fraction, right parenthesis.
In order for start fraction, 3, divided by, x, minus, 2, end fraction, minus, start fraction, 2, divided by, x, plus, 1, end fraction to be defined, x, does not equal, 2, comma, minus, 1.
Therefore, start fraction, x, plus, 1, divided by, x, plus, 1, end fraction and start fraction, x, minus, 2, divided by, x, minus, 2, end fraction are both equal to 1, and multiplication by 1 does not change the value of the expression!
Then, we can subtract the numerators and write the result over the common denominator.
\begin{aligned}&\phantom{=}{\dfrac{3}{\blueD{x-2}}-\dfrac{2}{\greenD{x+1}}}\\\\\\ &=\dfrac{3}{\blueD{x-2}}{\left(\greenD{\dfrac{x+1}{x+1}}\right)}-\dfrac{2}{\greenD{x+1}}{\left(\blueD{\dfrac{x-2}{x-2}}\right)}&&\small{\gray{\text{Common denominator}}}\\\\\\ &=\dfrac{3(x+1)}{(x-2)(x+1)}-\dfrac{2(x-2)}{(x+1)(x-2)}\\ \\\\\\ &=\dfrac{3(x+1)-2(x-2)}{(x-2)(x+1)}&&\small{\gray{\text{Subtract}}}\\ \\\\\\ &=\dfrac{3x+3-2x+4}{(x-2)(x+1)}\\ \\\\\\ &=\dfrac{x+7}{(x-2)(x+1)} \end{aligned}

1) start fraction, 5, x, divided by, x, plus, 3, end fraction, plus, start fraction, 4, divided by, x, plus, 2, end fraction, equals

\begin{aligned}&\phantom{=}\dfrac{5x}{\blueD{x+3}}+\dfrac{4}{\greenD{x+2}}\\\\\\ &=\dfrac{5x}{\blueD{x+3}}\left(\greenD{\dfrac{x+2}{x+2}}\right)+\dfrac{4}{\greenD{x+2}}\left(\blueD{\dfrac{x+3}{x+3}}\right)&&\small{\gray{\text{Common denominator}}}\\ \\\\\\ &=\dfrac{5x(x+2)}{(x+3)(x+2)}+\dfrac{4(x+3)}{(x+3)(x+2)}\\ \\\\\\ &=\dfrac{5x(x+2)+4(x+3)}{(x+3)(x+2)}&&\small{\gray{\text{Add}}}\\ \\\\\\ &=\dfrac{5x^2+10x+4x+12}{(x+3)(x+2)}\\ \\\\\\ &=\dfrac{5x^2+14x+12}{(x+3)(x+2)} \end{aligned}

## Least common denominators

### Numerical fractions

Sometimes, the denominators of two fractions are different but have some shared factors.
For example, consider start fraction, 3, divided by, 4, end fraction, plus, start fraction, 1, divided by, 6, end fraction:
\begin{aligned} &\phantom{=}\dfrac{3}{4}+\dfrac{1}{6}\\\\\\ &=\dfrac{3}{\blueD2\cdot \greenD2}+\dfrac{1}{\blueD2\cdot \goldD3}\\ \\ &=\dfrac{3}{\blueD2\cdot \greenD2} {\left(\dfrac{\goldD3}{\goldD3}\right)}+\dfrac{1}{\blueD2\cdot\goldD3}{\left(\dfrac{\greenD2}{\greenD2}\right)}&&\small{\gray{\text{Create least common denominator}}}\\ \\ &=\dfrac{9}{12}+\dfrac{2}{12}\\ \\ &=\dfrac{11}{12} \end{aligned}
Notice that the common denominator used in this example was not the product of the two individual denominators left parenthesis, 24, right parenthesis. Instead it was the least common multiple of 4 and 6 left parenthesis, 12, right parenthesis.
The least common multiple left parenthesis, L, C, M, right parenthesis of two integers is the smallest number that is divisible by both integers.
When two integers do not share a common factor, the L, C, M is simply equal to the product of these integers. However, when two integers share a common factor, the L, C, M is not equal to the product of the integers. This is because the common factor appears only once in the L, C, M. For instance:
• 4 and 6 share a common factor of 2, so the L, C, M of 4 and 6 is 12, not 4, dot, 6, equals, 24
• 6 and 15 share a common factor of 3, so the L, C, M of 6 and 15 is 30, not 6, dot, 15, equals, 90
The least common multiple of the denominators in two or more fractions is called the least common denominator.
An easy way to find the least common denominator is to think about how to make the denominators equal without having any unnecessary factors. Writing each denominator in its prime factorization makes this easy to see.
Notice that start color blueD, 2, end color blueD, dot, start color greenD, 2, end color greenD, dot, start color goldD, 3, end color goldD, equals, 12

### Variable expressions

Now let's apply this reasoning to perform the following addition:
start fraction, 2, divided by, left parenthesis, x, minus, 2, right parenthesis, left parenthesis, x, plus, 1, right parenthesis, end fraction, plus, start fraction, 3, divided by, left parenthesis, x, plus, 1, right parenthesis, left parenthesis, x, plus, 3, right parenthesis, end fraction
First, let's find the least common denominator:
So the least common denominator is start color blueD, left parenthesis, x, minus, 2, right parenthesis, end color blueD, start color greenD, left parenthesis, x, plus, 1, right parenthesis, end color greenD, start color purpleC, left parenthesis, x, plus, 3, right parenthesis, end color purpleC.
We can add the rational expressions as follows:
\small\begin{aligned}&\phantom{=}\dfrac{2}{\blueD{(x-2)}\greenD{(x+1)}}+\dfrac{3}{\greenD{(x+1)}\purpleC{(x+3)}}\\ \\\\ &=\dfrac{2}{\blueD{(x-2)}\greenD{(x+1)}}{\left(\purpleC{\dfrac{x+3}{x+3}}\right)}+\dfrac{3}{\greenD{(x+1)}\purpleC{(x+3)}}{\left(\blueD{\dfrac{x-2}{x-2}}\right)}&&\tiny{\gray{\text{Least common denominator}}}\\\\ \\\\ &=\dfrac{2(x+3)}{(x-2)(x+1)(x+3)}+\dfrac{3(x-2)}{(x+1)(x+3)(x-2)}\\ \\\\ &=\dfrac{2(x+3)+3(x-2)}{(x-2)(x+1)(x+3)}&&\small{\gray{\text{Add}}}\\ \\\\ &=\dfrac{2x+6+3x-6}{(x-2)(x+1)(x+3)}\\ \\\\ &=\dfrac{5x}{(x-2)(x+1)(x+3)} \end{aligned}

2) start fraction, 1, divided by, x, left parenthesis, x, minus, 6, right parenthesis, end fraction, plus, start fraction, 3, divided by, left parenthesis, x, plus, 1, right parenthesis, left parenthesis, x, minus, 6, right parenthesis, end fraction, equals

First, let's find the least common denominator:
So the least common denominator is start color blueD, x, end color blueD, start color greenD, left parenthesis, x, minus, 6, right parenthesis, end color greenD, start color purpleC, left parenthesis, x, plus, 1, right parenthesis, end color purpleC.
We can add the rational expressions as follows:
\small\begin{aligned}&\phantom{=}\dfrac{1}{\blueD x\greenD{(x-6)}}+\dfrac{3}{\purpleC{(x+1)}\greenD{(x-6)}}\\\\\\ &=\dfrac{1}{\blueD x\greenD{(x-6)}}\left(\purpleC{\dfrac{x+1}{x+1}}\right)+\dfrac{3}{\purpleC{(x+1)}\greenD{(x-6)}}\left(\blueD{\dfrac{x}{x}}\right)&&\tiny{\gray{\text{Least common denominator}}}\\\\ \\ &=\dfrac{1(x+1)}{x(x-6)(x+1)}+\dfrac{3x}{x(x-6)(x+1)}\\ \\\\ &=\dfrac{1(x+1)+3x}{x(x-6)(x+1)}&&\small{\gray{\text{Add}}}\\ \\\\ &=\dfrac{x+1+3x}{x(x-6)(x+1)}\\ \\\\ &=\dfrac{4x+1}{x(x-6)(x+1)} \end{aligned}
3) start fraction, 3, x, divided by, 2, left parenthesis, x, minus, 1, right parenthesis, end fraction, minus, start fraction, 4, divided by, left parenthesis, x, minus, 1, right parenthesis, left parenthesis, x, plus, 2, right parenthesis, end fraction, equals

First, let's find the least common denominator:
So the least common denominator is start color blueD, 2, end color blueD, start color greenD, left parenthesis, x, minus, 1, right parenthesis, end color greenD, start color purpleC, left parenthesis, x, plus, 2, right parenthesis, end color purpleC.
We can subtract the rational expressions as follows:
\small\begin{aligned}&\phantom{=}\dfrac{3x}{\blueD 2\greenD{(x-1)}}-\dfrac{4}{\greenD{(x-1)}\purpleC{(x+2)}}\\\\\\ &=\dfrac{3x}{\blueD 2\greenD{(x-1)}}\left(\purpleC{\dfrac{x+2}{x+2}}\right)-\dfrac{4}{\greenD{(x-1)}\purpleC{(x+2)}}\left(\blueD{\dfrac{2}{2}}\right)&&\tiny{\gray{\text{Least common denominator}}}\\ \\\\ &=\dfrac{3x(x+2)}{2(x-1)(x+2)}-\dfrac{8}{2(x-1)(x+2)}\\ \\\\ &=\dfrac{3x(x+2)-8}{2(x-1)(x+2)}&&\small{\gray{\text{Subtract}}}\\ \\ &=\dfrac{3x^2+6x-8}{2(x-1)(x+2)} \end{aligned}

### Challenge Problem

4*) start fraction, 2, divided by, x, start superscript, 2, end superscript, minus, 1, end fraction, plus, start fraction, 1, divided by, x, start superscript, 2, end superscript, minus, 3, x, minus, 4, end fraction, equals

Notice that in this case, the denominators are not in factored form. So let's factor first.
start fraction, 2, divided by, x, start superscript, 2, end superscript, minus, 1, end fraction, plus, start fraction, 1, divided by, x, start superscript, 2, end superscript, minus, 3, x, minus, 4, end fraction, equals, start fraction, 2, divided by, left parenthesis, x, minus, 1, right parenthesis, left parenthesis, x, plus, 1, right parenthesis, end fraction, plus, start fraction, 1, divided by, left parenthesis, x, minus, 4, right parenthesis, left parenthesis, x, plus, 1, right parenthesis, end fraction
Now it is easier to find the least common denominator.
The least common denominator is start color blueD, left parenthesis, x, minus, 1, right parenthesis, end color blueD, start color greenD, left parenthesis, x, plus, 1, right parenthesis, end color greenD, start color purpleC, left parenthesis, x, minus, 4, right parenthesis, end color purpleC.
We can add the rational expressions as follows:
\small\begin{aligned}&\phantom{=}\dfrac{2}{\blueD {(x-1)}\greenD{(x+1)}}+\dfrac{1}{\purpleC{(x-4)}\greenD{(x+1)}}\\\\\\ &=\dfrac{2}{\blueD {(x-1)}\greenD{(x+1)}}\left(\purpleC{\dfrac{x-4}{x-4}}\right)+\dfrac{1}{\purpleC{(x-4)}\greenD{(x+1)}}\left(\blueD{\dfrac{x-1}{x-1}}\right)&&\tiny{\gray{\text{Least common denominator}}}\\ \\\\ &=\dfrac{2(x-4)}{(x-1)(x+1)(x-4)}+\dfrac{1(x-1)}{(x-4)(x+1)(x-1)}\\ \\\\ &=\dfrac{2(x-4)+1(x-1)}{(x-1)(x+1)(x-4)}&&\small{\gray{\text{Add}}}\\ \\\\ &=\dfrac{2x-8+x-1}{(x-1)(x+1)(x-4)}\\\\\\ &=\dfrac{3x-9}{(x-1)(x+1)(x-4)} \end{aligned}

## 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 start fraction, 3, divided by, 4, end fraction, plus, start fraction, 1, divided by, 6, end fraction using two different common denominators — one using the least common denominator left parenthesis, 12, right parenthesis and the other using the product of the two denominators left parenthesis, 24, right parenthesis.
Least common denominator left parenthesis, 12, right parenthesisspace, space, space, space, spaceCommon denominator left parenthesis, 24, right parenthesis
\begin{aligned}~\dfrac{3}{4}+\dfrac{1}{6}&=\dfrac{3}{\blueD4} {\left(\dfrac{\goldD3}{\goldD3}\right)}+\dfrac{1}{\greenD6}{\left(\dfrac{\purpleD2}{\purpleD2}\right)}\\\\&=\dfrac{9}{12}+\dfrac{2}{12}\\\\&=\dfrac{11}{12}\\\\&\phantom{\dfrac{1}{2}}\end{aligned}~~~~~~~\begin{aligned}\dfrac{3}{\blueD4}+\dfrac{1}{\greenD6}&=\dfrac{3}{\blueD4}\left(\greenD{\dfrac{6}{6}}\right)+\dfrac{1}{\greenD6}\left(\blueD{\dfrac{4}{4}}\right)\\\\&=\dfrac{18}{24}+\dfrac{4}{24}\\\\&=\dfrac{22}{24}\\\\&=\dfrac{11}{12}\end{aligned}

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.
Recall the following problem.
start fraction, 2, divided by, left parenthesis, x, minus, 2, right parenthesis, left parenthesis, x, plus, 1, right parenthesis, end fraction, plus, start fraction, 3, divided by, left parenthesis, x, plus, 1, right parenthesis, left parenthesis, x, plus, 3, right parenthesis, end fraction
Suppose we obtained a common denominator by multiplying the first fraction by the denominator of the second and the second fraction by the denominator of the first. Note that this is a common denominator but not the least common denominator.
\scriptsize\begin{aligned}&\phantom{=}\dfrac{2}{\blueD{(x-2){(x+1)}}}+\dfrac{3}{\greenD{(x+1)(x+3)}}\\ \\\\ &=\dfrac{2}{\blueD{(x-2){(x+1)}}}{\left(\greenD{\dfrac{(x+1)(x+3)}{(x+1)(x+3)}}\right)}+\dfrac{3}{\greenD{(x+1)(x+3)}}{\left(\blueD{\dfrac{(x-2)(x+1)}{(x-2)(x+1)}}\right)} \end{aligned}
\scriptsize\begin{aligned}&=\dfrac{2(x+1)(x+3)}{(x-2)(x+1)^2(x+3)}+\dfrac{3(x-2)(x+1)}{(x+1)^2(x+3)(x-2)}\\ \\\\\\ &=\dfrac{2(x+1)(x+3)+3(x-2)(x+1)}{(x-2)(x+1)^2(x+3)}&&\small{\gray{\text{Add}}}\\ \\\\ &=\dfrac{2(x^2+4x+3)+3(x^2-x-2)}{(x-2)(x+1)^2(x+3)}\\ \\\\ &=\dfrac{2x^2+8x+6+3x^2-3x-6}{(x-2)(x+1)^2(x+3)}\\ \\\\ &=\dfrac{5x^2+5x}{(x-2)(x+1)^2(x+3)}\\ \\\\ &=\dfrac{5x(x+1)}{(x-2)(x+1)^2(x+3)}&&\small{\gray{\text{Factor out }5x}}\\\\\\ &=\dfrac{5x}{(x-2)(x+1)(x+3)}&&\small{\gray{\text{Cancel a factor of }x+1}} \end{aligned}
Notice that while this gave you the correct answer in the end, much more work was required!
All of this extra work can be avoided by using the least common denominator when adding or subtracting rational expressions.