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Chapter 18 Fraction Addition and Subtraction

Section 18.1 Find the Common Denominator

Many students will say that they sometimes "forget" how to do arithmetic with fractions. This often results from viewing fractions as purely a collection of symbols with particular rules for how to manipulate them. Their forgetfulness sometimes leads them to guess at how to work with fractions. Here are some examples:
\begin{equation*} \begin{array}{ccc} \dfrac{3}{5} + \dfrac{8}{5} \overset{\times}{=} \dfrac{3 + 8}{5 + 5} = \dfrac{11}{10} \amp \qquad \qquad \amp \dfrac{2}{5} + \dfrac{5}{6} \overset{\times}{=} \dfrac{2 + 5}{5 \cdot 6} = \dfrac{7}{30} \\ \\ \dfrac{4}{5} - \dfrac{1}{3} \overset{\times}{=} \dfrac{4 - 1}{5 - 3} = \dfrac{3}{2} \amp \qquad \qquad \amp \hspace{2cm} \dfrac{3}{5} - \dfrac{1}{3} \overset{\times}{=} \dfrac{3 - 1}{5 \cdot 3} = \dfrac{2}{15} \end{array} \end{equation*}
It’s not enough to simply identify that these calculations are incorrect. We have been emphasizing the importance of mathematical reasoning throughout this book. Mathematical reasoning goes beyond simply pointing at some rules and saying that they weren’t properly followed.
Let’s take a step back and just think about some basic addition and subtraction. What do we mean by \(2 + 3\text{?}\) There are several ways to think about it, but the one most people think about first is that you are combining two collections of objects:
This works well with integers because each piece is the same. But if we try to add fractions, we get a picture that doesn’t really make sense.
The basic challenge of this is that the pieces are different sizes, so it doesn’t really make sense to combine them together like this. And this is the basic explanation of a common denominator. We cannot combine the parts unless they are all the same size.
A "common denominator" is simply a choice of a denominator that works well with the fractions. Mathematically, this means that the chosen denominator is a common multiple of the denominators you’re working with. Ideally, we would use the least common denominator, but if a larger denominator is chosen, the arithmetic will still work out. It would just mean that there may be an extra step of reducing the final answer.
There are many ways to find common multiples of numbers. For fractions involving numbers, we can usually use intuition and experience to get the correct value. But there are also other methods we can use for larger numbers. One of those methods is a "brute force" method where you simply list out multiples of both numbers and look for the smallest number to appear in both lists:
This approach works and helps to reinforce the idea of common multiples, but it’s a clumsy approach that doesn’t generalize to variable expressions. A more robust and intuitive approach builds off of finding the greatest common factors.

Activity 18.1. Calculating the Least Common Multiple of Integers.

To find the least common multiple of 12 and 32, we will start off by factoring out the greatest common factor from each number:
\begin{equation*} \begin{array}{lcl} 12 \amp = \amp 4 \cdot 3 \\ 32 \amp = \amp 4 \cdot 8 \end{array} \end{equation*}
Notice that the common factor is already shared between the two numbers, so we can leave that part alone. The remaining parts are what we might call "unshared" factors. Each part is missing the other’s unshared factor, and so those are the things we need to multiply by to find the least common multiple.
\begin{equation*} \begin{array}{rclcl} 12 \cdot 8 \amp = \amp (4 \cdot 3) \cdot 8 \amp = \amp 96 \\ 32 \cdot 3 \amp = \amp (4 \cdot 8) \cdot 3 \amp = \amp 96 \end{array} \end{equation*}
Try it!
Find the least common multiple of 30 and 42.
Solution.
\begin{equation*} \left. \begin{array}{ll} 30 \amp = 6 \cdot 5 \\ 42 \amp = 6 \cdot 7 \end{array} \right\} \longrightarrow \left\{ \begin{array}{lll} 30 \cdot 7 \amp = 210 \\ 42 \cdot 5 \amp = 210 \end{array} \right. \end{equation*}

Activity 18.2. Calculating the Least Common Multiple of Monomials.

This same approach can be used for finding common multiples of variable expressions. We will use it to find the least common multiple of \(9x^2y\) and \(15x y^4\text{.}\)
\begin{equation*} \left. \begin{array}{lcl} 9x^2 y \amp = \amp 3xy \cdot 3x \\ 15xy^4 \amp = \amp 3xy \cdot 5y^3 \end{array} \right\} \longrightarrow \left\{ \begin{array}{lcl} 9x^2 y \cdot 5y^3 \amp = \amp 45x^2y^4 \\ 15xy^4 \cdot 3x \amp = \amp 45x^2y^4 \end{array} \right. \end{equation*}
Try it!
Find the least common multiple of \(4a^2 b^3\) and \(12 a^3 b\text{.}\)
Solution.
\begin{equation*} \left. \begin{array}{ll} 4a^2b^3 \amp = 4a^2 b \cdot b^2 \\ 12a^3 b \amp = 4a^2 b \cdot 3a \end{array} \right\} \longrightarrow \left\{ \begin{array}{lll} 4a^2b^3 \cdot 3a \amp = 12a^3 b^3 \\ 12a^3 b \cdot b^2 \amp = 12a^3 b^3 \end{array} \right. \end{equation*}
Finding the least common multiple of two expressions is simply a part of adding and subtracting fractions. Once we identify what the least common multiple of the denominators is (the least common denominator), we then need to rewrite both fractions with that denominator, and then add the results.

Activity 18.3. Adding Fractions of Numbers.

Here is the calculation of \(\frac{1}{2} + \frac{1}{3}\text{:}\)
\begin{equation*} \begin{aligned} \frac{1}{2} + \frac{1}{3} \amp = \frac{1 \cdot 3}{2 \cdot 3} + \frac{1 \cdot 2}{3 \cdot 2} \amp \eqnspacer \amp \text{Common denominator} \\ \amp = \frac{3}{6} + \frac{2}{6} \\ \amp = \frac{5}{6} \end{aligned} \end{equation*}
Try it!
Calculate \(\frac{2}{5} + \frac{3}{7}\text{.}\)
Solution.
\begin{equation*} \begin{aligned} \frac{2}{5} + \frac{3}{7} \amp = \frac{2 \cdot 7}{5 \cdot 7} + \frac{3 \cdot 5}{7 \cdot 5} \amp \eqnspacer \amp \text{Common denominator} \\ \amp = \frac{14}{35} + \frac{15}{35} \\ \amp = \frac{29}{35} \end{aligned} \end{equation*}

Activity 18.4. Subtracting Fractions of Numbers.

Subtraction is conceptually different from addition. With subtraction, you are starting with a collection of objects and then taking objects away from that collection. But the same idea for the common denominator still holds. In order for the subtraction to make sense, we need to be working with objects of the same size.
Try it!
Calculate \(\frac{5}{4} - \frac{5}{6}\text{.}\)
Solution.
\begin{equation*} \begin{aligned} \frac{5}{4} - \frac{5}{6} \amp = \frac{5 \cdot 3}{4 \cdot 3} - \frac{5 \cdot 2}{6 \cdot 2} \amp \eqnspacer \amp \text{Common denominator} \\ \amp = \frac{15}{12} - \frac{10}{12} \\ \amp = \frac{5}{12} \end{aligned} \end{equation*}

Activity 18.5. Adding Fractions of Monomials.

The same logic can be applied to adding or subtracting fractions with variables in them. Sometimes you will be able to combine like terms, and sometimes you won’t. You simply need to pay attention to the information that’s in front of you.
\begin{equation*} \begin{aligned} \frac{2x}{y} + \frac{3y}{x^2} \amp = \frac{2x \cdot x^2}{y \cdot x^2} + \frac{3y \cdot y}{x^2 \cdot y} \amp \eqnspacer \amp \text{Common denominator} \\ \amp = \frac{2x^3}{x^2y} + \frac{3y^2}{x^2y} \\ \amp = \frac{2x^3 + 3y^2}{x^2y} \end{aligned} \end{equation*}
Try it!
Calculate \(\frac{2x}{3y} + \frac{5}{7x}\text{.}\)
Solution.
\begin{equation*} \begin{aligned} \frac{2x}{3y} + \frac{5}{7x} \amp = \frac{2x \cdot 7x}{3y \cdot 7x} + \frac{5 \cdot 3y}{7x \cdot 3y} \amp \eqnspacer \amp \text{Common denominator} \\ \amp = \frac{14x^2}{21xy} + \frac{15y}{21xy} \\ \amp = \frac{14x^2 + 15y}{21xy} \end{aligned} \end{equation*}

Section 18.2 Worksheets

PDF Version of these Worksheets
 1 
external/worksheets/18-Worksheets.pdf

Worksheet Worksheet 1

1.
Find the least common multiple of 12 and 28 by writing out multiples of each number and also by applying the technique from this section.
2.
Find the least common multiple of 18 and 48 by writing out multiples of each number and also by applying the technique from this section.
3.
Calculate \(\frac{3}{5} + \frac{7}{8}\text{.}\)
4.
Check the presentation for errors. If you find one, circle it and describe the mistake in words.
\begin{equation*} \begin{aligned} \frac{4}{5} + \frac{3}{7} \amp = \frac{4}{5} \cdot 7 + \frac{3}{7} \cdot 5 \amp \amp \text{Common denominator} \\ \amp = \frac{28}{35} + \frac{15}{35} \\ \amp = \frac{43}{35} \end{aligned} \end{equation*}

Worksheet Worksheet 2

1.
Find the least common multiple of 40 and 72.
2.
Find the least common multiple of \(6x^2 y\) and \(9x y^3\text{.}\)
3.
Calculate \(\frac{11}{6} - \frac{7}{20}\text{.}\)
4.
Calculate \(\frac{3}{x} + \frac{7}{y}\text{.}\)

Worksheet Worksheet 3

1.
Find the least common multiple of 8 and 32.
2.
Find the least common multiple of 8 and \(3p^2 q\text{.}\)
3.
Calculate \(\frac{3a}{4b} + \frac{4b}{3a}\text{.}\)
4.
Calculate \(\frac{11}{6} + \frac{7}{20}\text{.}\)

Worksheet Worksheet 4

1.
Calculate \(\frac{13}{8} + \frac{17}{20}\text{.}\)
2.
Calculate \(\frac{3x}{8y^2} - \frac{5y}{6x}\text{.}\)
3.
Calculate \(\frac{5x}{6y^2} + \frac{8}{10x^2y^3}\text{.}\)
4.
Calculate \(\frac{x}{y^2} - \frac{y}{x^3}\text{.}\)

Worksheet Worksheet 5

1.
Calculate \(\frac{7}{8} + \frac{5}{24}\text{.}\)
2.
Calculate \(\frac{22}{15} - \frac{3}{35}\text{.}\)
3.
Calculate \(\frac{a}{b} + \frac{c}{d}\text{.}\)
4.
In the previous problem, you derived a general formula for adding two fractions together. Apply the formula to calculating \(\frac{11}{32} + \frac{23}{48}\) and then explain why it’s not a good idea to use the formula in every situation.

Section 18.3 Deliberate Practice: Adding and Subtracting Fractions

Algebra is a skill, which means it requires practice to become proficient. But it will take more than rote repetition to get there. Deliberate practice is the thoughtful repetition of a task. For each of these sections, you will be given a list of specific skills or ideas to focus on as you practice thinking through the problems.
Focus on these skills:
  • Write the original expression.
  • Show the multiplication for the common denominator step.
  • Try to use the least common denominator rather than applying the general formula.
  • Present your work legibly.

Worksheet Worksheet

Instructions: Perform the indicated calculation..
1.
Calculate \(\displaystyle \frac{7}{12} + \frac{11}{18}\text{.}\)
2.
Calculate \(\displaystyle \frac{13}{10} - \frac{7}{25}\text{.}\)
3.
Calculate \(\displaystyle \frac{5x}{4y} + \frac{8y}{3x}\text{.}\)
4.
Calculate \(\displaystyle \frac{4a}{3b^2} - \frac{2}{7ab}\text{.}\)
5.
Calculate \(\displaystyle \frac{6}{5p^2q} + \frac{2q}{3p}\text{.}\)
6.
Calculate \(\displaystyle \frac{4}{7x^2 y} - \frac{3x}{2y^2}\text{.}\)
7.
Calculate \(\displaystyle \frac{3n^2}{m^3} + \frac{4m}{3n^2}\text{.}\)
8.
Calculate \(\displaystyle \frac{4}{xy} + \frac{3}{x^2z}\text{.}\)
9.
Calculate \(\displaystyle \frac{5a}{3b^2 c^3} - \frac{4c}{7b^2}\text{.}\)
10.
Calculate \(\displaystyle \frac{3mn}{5p^2} + \frac{5m^2}{8n^2 p}\text{.}\)

Section 18.4 Closing Ideas

On the last worksheet, you derived the general formula for adding two fractions. There’s a similar one for subtraction.
On the very last problem of the worksheet, you also saw the downside to having such a formula. If you had no concept of common denominators, you would have ended up making the problem unnecessarily difficult for yourself because the numbers in your calculation would end up being quite large.
Some students simply learn the "rule" for adding and subtracting fractions with different denominators. These are also the students that tend to forget over time, and they will often start to guess wildly at solutions. Here are some non-examples of adding fractions together:
\begin{equation*} \frac{a}{b} + \frac{c}{d} \overset{\times}{=} \frac{a + c}{b + d} \qquad \frac{a}{b} + \frac{c}{d} \overset{\times}{=} \frac{a + c}{bd} \end{equation*}
The major trapping is for students to fall into the habit of simply manipulating the symbols without understanding what’s happening. It’s very easy to think you understand something because you can execute the proper procedure. But mathematical reasoning goes beyond execution.
Take another look at the two errors. Suppose that a friend showed you that work and asked you for help. Would you just tell them that they did it wrong and show them the right way to do it? Would be able to explain the right ideas to them? In the mathematical world, the goal is to both be able to execute the calculation and explain the reasoning behind it. This is a theme that we will keep returning to throughout the book.

Section 18.5 Going Deeper: Adding and Subtracting Rational Expressions

In some ways, the exercises in this section give a false impression of working with rational expressions. It is not often the case that we’re adding and subtracting the types of monomial expressions that were presented. The reason the exercises look that way is because the emphasis is on learning to recognize when the denominators have common factors and when they don’t, which leads to knowing what to multiply by to create the common denominators. By using different variables, it’s much easier for students to train themselves to recognize the common factors.
It’s likely that you will only ever encounter single variable rational expressions. But the skill of identifying different terms will still come into play because you will need to recognize when polynomial factors are the same or different. For example, instead of working with fractions of the form
\begin{equation*} \frac{1}{a} + \frac{1}{b} \end{equation*}
you will be working with fractions of the form
\begin{equation*} \frac{1}{x + 2} + \frac{1}{x - 3}. \end{equation*}
In the previous "Going Deeper" section, we emphasized the importance of multiplicative factors when reducing rational expressions. The same idea holds for adding and subtracting rational expressions. We need to think in terms of multiplicative factors. The way we accomplish this is to think of the whole binomial as a single object. Basically, we want to envision a set of parentheses around each binomial.
\begin{equation*} \frac{1}{x + 2} + \frac{1}{x - 3} = \frac{1}{(x + 2)} + \frac{1}{(x - 3)} \end{equation*}
And once we have done this, all of the experience that we were developing in this section can come into play. All of the previous logic applies to this calculation.
\begin{equation*} \begin{aligned} \frac{1}{x + 2} + \frac{1}{x - 3} \amp = \frac{1 \cdot (x - 3)}{(x + 2) \cdot (x - 3)} + \frac{1 \cdot (x + 2)}{(x - 3) \cdot (x + 2)} \amp \eqnspacer \amp \text{Common denominator} \\ \amp = \frac{x - 3}{(x + 2)(x - 3)} + \frac{x + 2}{(x + 2)(x - 3)} \\ \amp = \frac{(x - 3) + (x + 2)}{(x + 2)(x - 3)} \\ \amp = \frac{2x - 1}{(x + 2)(x - 3)} \end{aligned} \end{equation*}
When subtracting rational expressions, it’s extremely important to keep the parentheses around terms when subtracting. This did not come up in the section because we were only working with monomial terms. But when working with binomials, those parentheses are critical to avoid errors.
\begin{equation*} \begin{aligned} \frac{x}{x - 1} - \frac{3}{x + 3} \amp = \frac{x \cdot (x + 3)}{(x - 1) \cdot (x + 3)} - \frac{3 \cdot (x - 1)}{(x + 3) \cdot (x - 1)} \amp \eqnspacer \amp \text{Common denominator} \\ \amp = \frac{x^2 + 3x}{(x - 1)(x + 3)} - \frac{3x - 3}{(x - 1)(x + 3)} \\ \amp = \frac{(x^2 + 3x) - (3x - 3)}{(x - 1)(x + 3)} \\ \amp = \frac{x^2 + 3x - 3x + 3}{(x - 1)(x + 3)} \\ \amp = \frac{x^2 + 3}{(x - 1)(x + 3)} \end{aligned} \end{equation*}
In some expressions, you may find that after adding or subtracting, it may be possible to factor the numerator. When you can see a way to factor, it’s best to do it. The reason is that sometimes these factorization steps will reveal that the fraction can be reduced or simplified, which can be very helpful.
\begin{equation*} \begin{aligned} \frac{1}{x - 2} + \frac{x - 5}{(x - 2)(x + 1)} \amp = \frac{1 \cdot (x + 1)}{(x - 2) \cdot (x + 1)} + \frac{x - 5}{(x - 2)(x + 1)} \amp \eqnspacer \amp \text{Common denominator} \\ \amp = \frac{x + 1}{(x - 2)(x + 1)} + \frac{x - 5}{(x - 2)(x + 1)} \\ \amp = \frac{(x + 1) + (x - 5)}{(x - 2)(x + 1)} \\ \amp = \frac{2x - 4}{(x - 2)(x + 1)} \\ \amp = \frac{2(x - 2)}{(x - 2)(x + 1)} \\ \amp = \frac{2 \cdot \cancel{(x - 2)}}{(x + 1) \cdot \cancel{(x - 2)}} \\ \amp = \frac{2}{x + 1} \end{aligned} \end{equation*}
It is hard to know in advance which calculations will lead to fractions that reduce and which ones don’t, so you will just want to get into the habit of factoring the numerator when you can.