Back
Finding Common Denominators
> ... Math > Fractions > Finding Common Denominators

In order to add fractions, the denominators must be the same. In the example of pizza, this means each slice has to be the same size. If we are working with pizza slices that have been cut into different sizes, we need to find a way to cut all of them into slices that are the same size. This is done using Least Common Multiples. LCM’s are how we change the denominator of a fraction. We multiply by 1, but the 1 doesn’t look like a 1. It is in the form of something like \(\dfrac{3}{3}\) or \(\dfrac{7}{7}\). (Remember, anything divided by itself equals 1.) The following video will give more details and work through some examples.

Finding Common Denominators

Video Source (09:55 mins) | Transcript

Example of Adding Fractions with Different Denominators (inch)

Video Source (04:15 mins) | Transcript

To find the common denominator, find the LCM of the existing denominators. To find the new numerator, multiply the existing numerator by the same number multiplied to its denominator to get the LCM.

Additional Resources

Practice Problems

  1. What is the common denominator you would use if you wanted to add the fractions \(\dfrac{1}{4}\) and \(\dfrac{1}{3}\)? (
    Video Solution
    x
    Solution: 12

    Details:

    (Video Source | Transcript)
    )
  2. What is the common denominator you would use if you wanted to add the fractions \(\dfrac{1}{6}\) and \(\dfrac{1}{9}\)? (
    Solution
    x
    Solution:
    18
    )
  3. What do you get when you add the fractions \(\dfrac{1}{4}\) and \(\dfrac{1}{3}\)? (
    Solution
    x
    Solution:
    \(\displaystyle \frac{1}{4}+\frac{1}{3}=\frac{3}{12}+\frac{4}{12}=\frac{7}{12}\)
    )
  4. Add: \(\displaystyle \frac{1}{6}+\frac{1}{9}\) (
    Solution
    x
    Solution: \(\displaystyle \frac{1}{6}+\frac{1}{9}=\frac{3}{18}+\frac{2}{18}=\frac{5}{18}\)

    Details:
    We are adding \(\dfrac{1}{6}\) to \(\dfrac{1}{9}\). We can represent that as \(\dfrac{1}{6}\) of a whole, added to \(\dfrac{1}{9}\) of a whole:
    This is a picture of 2 circles. The first has an area of one sixth shaded. The second has an area of one nineth shaded. Under the circles is the equation one sixth plus one ninth.

    Step 1: Find the least common denominator.
    There are two ways that we can find the common denominator: skip counting using multiples or by factoring each denominator. For this example, we will use multiples:

    Multiples of \(6\): \(6, 12, {\color{red}18}, 24, etc.\)

    Multiples of \(9\): \(9, {\color{red}18}, 27, etc.\)

    The smallest number that both \(6\) and \(9\) divide into evenly is \(\color{red}18\)

    Step 2: Write both fractions as an equivalent fraction with a denominator of \(18\).
    Let’s start with \(\dfrac{1}{6}\).
    \(6 \times {\color{red}3} = 18\)

    Then multiply the numerator and denominator of \(\dfrac{1}{6}\) by \(3\):
    \(\displaystyle \dfrac{1\times\color{RED}3}{6\times\color{RED}3}=\dfrac{3}{18}\)

    We do the same for \(\dfrac{1}{9}\).
    \(9 \times {\color{red}2} = 18\)

    Then multiply the numerator and the denominator by \(2\):
    \(\dfrac{1\times\color{RED}2}{9\times\color{RED}2}=\dfrac{2}{18}\)

    This is a picture of 2 circles with a plus sign between them. The first has an area of 3 eighteenths shaded, with a caption underneath with the fraction 1 times 3 over 6 times 3. The second has an area of 2 eighteenths shaded, with a caption underneath with the fraction 1 times 2 over 9 times 2.

    Which gives us \(\displaystyle \dfrac{3}{18}+\dfrac{2}{18}\):
    This is a picture of 2 circles. The first has an area of 3 eighteenths shaded. The second has an area of 2 eighteenths shaded. Under the circles is the equation three over 18 plus 2 over 18.

    Note that both circles now have 18 parts. \(\dfrac{1}{6}\) is the same amount as \(\dfrac{3}{18}\) and \(\dfrac{1}{9}\) is the same amount as \(\dfrac{2}{18}\).

    Step 3: Add fractions
    \(\displaystyle \dfrac{3}{18}+\dfrac{2}{18}=\frac{5}{18}\)
    This is a picture of 3 circles. The first has an area of 3 eighteenths shaded. The second has an area of 2 eighteenths shaded. The third has 5 eighteenths shaded. The equation 3 over 18 plus 2 over 18 equals 5 over 18 is written below the circles.
    )
  5. Subtract: \(\displaystyle \frac{1}{6}-\frac{1}{9}\) (
    Solution
    x
    Solution: \(\displaystyle \frac{1}{6}-\frac{1}{9}=\frac{3}{18}-\frac{2}{18}=\frac{1}{18}\)

    Details:
    We are subtracting \(\dfrac{1}{9}\) from \(\dfrac{1}{6}\). We can represent that as \(\dfrac{1}{6}\) of a whole, minus \(\dfrac{1}{9}\) of a whole:
     This is a picture of 2 circles. The first has an area of one sixth shaded. The second has an area of one ninth shaded. Under the circles is the equation one sixth minus one ninth.

    Step 1: Find the least common denominator.

    There are two ways that we can find the common denominator: skip counting using multiples or by factoring each denominator. For this example, we will use multiples.

    Multiples of \(6\): \(6, 12, {\color{red}18}, 24, etc.\)
    Multiples of \(9\): \(9, {\color{red}18}, 27, etc.\)

    The smallest number that both \(6\) and \(9\) divide into evenly is \(\color{red}18\)

    Step 2: Write both fractions as an equivalent fraction with a denominator of \(18\).

    Let’s start with \(\dfrac{1}{6}\), \(6 \times 3 = 18\) so we multiply the numerator and denominator of \(\dfrac{1}{6}\) by \(3\):
    \(\displaystyle \frac{1\times\color{RED}3}{6\times\color{RED}3}=\frac{3}{18}\)

    We do the same for \(\dfrac{1}{9}\), \(9 \times {\color{red} 2} = 18\), so we will multiply the numerator and the denominator by \(2\):
    \(\displaystyle \frac{1\times\color{RED}2}{9\times\color{RED}2}=\frac{2}{18}\)

    This is a picture of 2 circles with a fraction under each circle and minus sign between the fractions. The first has an area of 3 eighteenths shaded, with a caption underneath with the fraction 1 times 3 over 6 times 3. The second has an area of 2 eighteenths shaded, with a caption underneath with the fraction 1 times 2 over 9 times 2.

    Which gives us the following:
    This is a picture of 2 circles with a fraction under each circle and minus sign between the fractions. The first has an area of 3 eighteenths shaded, with a caption underneath with the fraction 1 times 3 over 6 times 3. The second has an area of 2 eighteenths shaded, with a caption underneath with the fraction 1 times 2 over 9 times 2.

    Note that both circles now have 18 parts. \(\dfrac{1}{6}\) is the same amount as \(\dfrac{3}{18}\) and \(\dfrac{1}{9}\) is the same amount as \(\dfrac{2}{18}\).

    Step 3: Subtract

    We are subtracting \(\dfrac{2}{18}\) from the \(\dfrac{3}{18}\):
    Image of a circle with 1 eighteenths highlighted blue and a wedge depicting 2 eighteenths pulled out of a section of 3 eighteenths. There are words underneath the circle reading, remove 2 eighteenths, with an arrow pointing to the 2 eighteenths wedge.

    Which gives us \(\displaystyle \frac{3}{18}-\frac{2}{18}=\frac{1}{18}\):

    This is a picture of 3 circles. The first has an area of 3 eighteenths shaded. The second has an area of 2 eighteenths shaded. The third has 1 eighteenths shaded. The equation 3 over eighteen minus 2 over eighteen equals 1 over eighteen is written below the circles.
    )
  6. Subtract: \(\displaystyle \frac{1}{4}-\frac{5}{8}\) (
    Video Solution
    x
    Solution: \(\displaystyle \frac{1}{4}-\frac{5}{8}=\frac{2}{8}-\frac{5}{8}=-\frac{3}{8}\)

    Details:

    (Video Source | Transcript)
    )

Need More Help?

  1. Study other Math Lessons in the Resource Center.
  2. Visit the Online Tutoring Resources in the Resource Center.
  3. Contact your Instructor.
  4. If you still need help, Schedule a Tutor.