# Least Common Denominators (LCDs)

It is difficult to add or subtract fractions when the denominators are not the same. So, we use a common denominator. It is usually easiest to use the
**
least
**
common denominator. The least common denominator is simply the
least common multiple
(
LCM
) of the two denominators.

**
Example 1:
**

Find the common denominator of the fractions.

$\frac{1}{6}$ and $\frac{3}{8}$

We need to find the least common multiple of $6$ and $8$ . One way to do this is to list the multiples:

$\begin{array}{l}6,12,18,\underset{\_}{24},30,36,42,48,\mathrm{...}\\ 8,16,\underset{\_}{24},32,40,48,\mathrm{...}\end{array}$

The first number that occurs in both lists is $24$ , so $24$ is the LCM. So we use this as our common denominator.

Listing multiples is impractical for large numbers. Another way to find the LCM of two numbers is to divide their product by their greatest common factor ( GCF ).

**
Example 2:
**

Find the common denominator of the fractions.

$\frac{5}{12}$ and $\frac{2}{15}$

The greatest common factor of $12$ and $15$ is $3$ .

So, to find the least common multiple, divide the product by $3$ .

$\frac{12\cdot 15}{3}=\frac{\overline{)3}\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}4\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}15}{\overline{)3}}=60$

If you can find a least common denominator, then you can rewrite the problem using equivalent fractions that have like denominators, so they are easy to add or subtract.

**
Example 3:
**

Add.

$\frac{5}{12}+\frac{2}{15}$

In the previous example, we found that the least common denominator was $60$ .

Write each fraction as an equivalent fraction with the denominator $60$ . To do this, we multiply both the numerator and denominator of the first fraction by $5$ , and the numerator and denominator of the second fraction by $4$ . (This is the same as multiplying by $1=\frac{5}{5}=\frac{4}{4}$ , so it doesn't change the value.)

$\begin{array}{l}\frac{5}{12}=\frac{5}{12}\cdot \frac{5}{5}=\frac{25}{60}\\ \frac{2}{15}=\frac{2}{15}\cdot \frac{4}{4}=\frac{8}{60}\end{array}$

$\begin{array}{l}\frac{5}{12}+\frac{2}{15}=\frac{25}{60}+\frac{8}{60}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{33}{60}\end{array}$

Note that this method may not always give the result in lowest terms. In this case, we have to simplify.

$=\frac{11}{30}$

**
The same idea
**
can be used when there are variables in the fractions—that is, to add or subtract
rational expressions
.

**
Example 4:
**

Subtract.

$\frac{1}{2a}-\frac{1}{3b}$

The two expressions $2a$ and $3b$ have no common factors, so their least common multiple is simply their product: $2a\cdot 3b=6ab$ .

Rewrite the two fractions with $6ab$ in the denominator.

$\begin{array}{l}\frac{1}{2a}\cdot \frac{3b}{3b}=\frac{3b}{6ab}\\ \frac{1}{3b}\cdot \frac{2a}{2a}=\frac{2a}{6ab}\end{array}$

Subtract.

$\begin{array}{l}\frac{1}{2a}-\frac{1}{3b}=\frac{3b}{6ab}-\frac{2a}{6ab}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3b\text{\hspace{0.17em}}-\text{\hspace{0.17em}}2a}{6ab}\end{array}$

**
Example 5:
**

Subtract.

$\frac{x}{16}-\frac{3}{8x}$

$16$ and $8x$ have a common factor of $8$ . So, to find the least common multiple, divide the product by $8$ .

$\frac{16\cdot 8x}{8}=16x$

The LCM is $16x$ . So, multiply the first expression by $1$ in the form $\frac{x}{x}$ , and multiply the second expression by $1$ in the form $\frac{2}{2}$ .

$\begin{array}{l}\frac{x}{16}\cdot \frac{x}{x}=\frac{{x}^{2}}{16x}\\ \frac{3}{8x}\cdot \frac{2}{2}=\frac{6}{16x}\end{array}$

Subtract.

$\begin{array}{l}\frac{x}{16}-\frac{3}{8x}=\frac{{x}^{2}}{16x}-\frac{6}{16x}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{{x}^{2}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}6}{16x}\end{array}$