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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 18.1.1. 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 18.1.2. 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 18.1.3. 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 18.1.4. 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 18.1.5. 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*}