4.8: Add and Subtract Fractions with Different Denominators (Part 1)
 Page ID
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left#1\right}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\) Find the least common denominator (LCD)
 Convert fractions to equivalent fractions with the LCD
 Add and subtract fractions with different denominators
 Identify and use fraction operations
 Use the order of operations to simplify complex fractions
 Evaluate variable expressions with fractions
Before you get started, take this readiness quiz.
 Find two fractions equivalent to \(\dfrac{5}{6}\). If you missed this problem, review Example 4.1.14.
 Simplify: \(\dfrac{1 + 5 \cdot 3}{2^{2} + 4}\). If you missed this problem, review Example 4.3.13.
Find the Least Common Denominator
In the previous section, we explained how to add and subtract fractions with a common denominator. But how can we add and subtract fractions with unlike denominators?
Let’s think about coins again. Can you add one quarter and one dime? You could say there are two coins, but that’s not very useful. To find the total value of one quarter plus one dime, you change them to the same kind of unit—cents. One quarter equals \(25\) cents and one dime equals \(10\) cents, so the sum is \(35\) cents. See Figure \(\PageIndex{1}\).
Figure \(\PageIndex{1}\): Together, a quarter and a dime are worth 35 cents, or \(\dfrac{35}{100}\) of a dollar.
Similarly, when we add fractions with different denominators we have to convert them to equivalent fractions with a common denominator. With the coins, when we convert to cents, the denominator is \(100\). Since there are \(100\) cents in one dollar, \(25\) cents is \(\dfrac{25}{100}\) and \(10\) cents is \(\dfrac{10}{100}\). So we add \(\dfrac{25}{100} + \dfrac{10}{100}\) to get \(\dfrac{35}{100}\), which is \(35\) cents.
You have practiced adding and subtracting fractions with common denominators. Now let’s see what you need to do with fractions that have different denominators.
First, we will use fraction tiles to model finding the common denominator of \(\dfrac{1}{2}\) and \(\dfrac{1}{3}\). We’ll start with one \(\dfrac{1}{2}\) tile and \(\dfrac{1}{3}\) tile. We want to find a common fraction tile that we can use to match both \(\dfrac{1}{2}\) and \(\dfrac{1}{3}\) exactly. If we try the \(\dfrac{1}{4}\) pieces, \(2\) of them exactly match the \(\dfrac{1}{2}\) piece, but they do not exactly match the \(\dfrac{1}{3}\) piece.
Figure \(\PageIndex{2}\)
If we try the \(\dfrac{1}{5}\) pieces, they do not exactly cover the \(\dfrac{1}{2}\) piece or the \(\dfrac{1}{3}\) piece.
Figure \(\PageIndex{3}\)
If we were to try the \(\dfrac{1}{12}\) pieces, they would also work.
Figure \(\PageIndex{4}\)
Even smaller tiles, such as \(\dfrac{1}{24}\) and \(\dfrac{1}{48}\), would also exactly cover the \(\dfrac{1}{2}\) piece and the \(\dfrac{1}{3}\) piece. The denominator of the largest piece that covers both fractions is the least common denominator (LCD) of the two fractions. So, the least common denominator of \(\dfrac{1}{2}\) and \(\dfrac{1}{3}\) is \(6\).
Notice that all of the tiles that cover \(\dfrac{1}{2}\) and \(\dfrac{1}{3}\) have something in common: Their denominators are common multiples of \(2\) and \(3\), the denominators of \(\dfrac{1}{2}\) and \(\dfrac{1}{3}\). The least common multiple (LCM) of the denominators is \(6\), and so we say that \(6\) is the least common denominator (LCD) of the fractions \(\dfrac{1}{2}\) and \(\dfrac{1}{3}\).
The least common denominator (LCD) of two fractions is the least common multiple (LCM) of their denominators.
To find the LCD of two fractions, we will find the LCM of their denominators. We follow the procedure we used earlier to find the LCM of two numbers. We only use the denominators of the fractions, not the numerators, when finding the LCD.
Find the LCD for the fractions \(\dfrac{7}{12}\) and \(\dfrac{5}{18}\).
Solution
Factor each denominator into its primes.  
List the primes of 12 and the primes of 18 lining them up in columns when possible.  
Bring down the columns.  
Multiply the factors. The product is the LCM.  LCM = 36 
The LCM of 12 and 18 is 36, so the LCD of \(\dfrac{7}{12}\) and \(\dfrac{5}{18}\) is 36.  LCD of \(\dfrac{7}{12}\) and \(\dfrac{5}{18}\) is 36. 
Find the least common denominator for the fractions: \(\dfrac{7}{12}\) and \(\dfrac{11}{15}\).
 Answer

\(60\)
Find the least common denominator for the fractions: \(\dfrac{13}{15}\) and \(\dfrac{17}{5}\).
 Answer

\(15\)
To find the LCD of two fractions, find the LCM of their denominators. Notice how the steps shown below are similar to the steps we took to find the LCM.
Step 1. Factor each denominator into its primes.
Step 2. List the primes, matching primes in columns when possible.
Step 3. Bring down the columns.
Step 4. Multiply the factors. The product is the LCM of the denominators.
Step 5. The LCM of the denominators is the LCD of the fractions.
Find the least common denominator for the fractions \(\dfrac{8}{15}\) and \(\dfrac{11}{24}\).
Solution
To find the LCD, we find the LCM of the denominators. Find the LCM of \(15\) and \(24\).
The LCM of \(15\) and \(24\) is \(120\). So, the LCD of \(\dfrac{8}{15}\) and \(\dfrac{11}{24}\) is \(120\).
Find the least common denominator for the fractions: \(\dfrac{13}{24}\) and \(\dfrac{17}{32}\).
 Answer

\(96\)
Find the least common denominator for the fractions: \(\dfrac{9}{28}\) and \(\dfrac{21}{32}\).
 Answer

\(224\)
Convert Fractions to Equivalent Fractions with the LCD
Earlier, we used fraction tiles to see that the LCD of \(\dfrac{1}{4}\) and \(\dfrac{1}{6}\) is \(12\). We saw that three \(\dfrac{1}{12}\) pieces exactly covered \(\dfrac{1}{4}\) and two \(\dfrac{1}{12}\) pieces exactly covered \(\dfrac{1}{6}\), so
\[\dfrac{1}{4} = \dfrac{3}{12} \quad and \quad \dfrac{1}{6} = \dfrac{2}{12} \ldotp \nonumber \]
We say that \(\dfrac{1}{4}\) and \(\dfrac{3}{12}\) are equivalent fractions and also that \(\dfrac{1}{6}\) and \(\dfrac{2}{12}\) are equivalent fractions.
We can use the Equivalent Fractions Property to algebraically change a fraction to an equivalent one. Remember, two fractions are equivalent if they have the same value. The Equivalent Fractions Property is repeated below for reference.
If \(a, b, c\) are whole numbers where \(b ≠ 0\), \(c ≠ 0\), then
\[\dfrac{a}{b} = \dfrac{a \cdot c}{b \cdot c} \quad and \quad \dfrac{a \cdot c}{b \cdot c} = \dfrac{a}{b}\]
To add or subtract fractions with different denominators, we will first have to convert each fraction to an equivalent fraction with the LCD. Let’s see how to change \(\dfrac{1}{4}\) and \(\dfrac{1}{6}\) to equivalent fractions with denominator \(12\) without using models.
Convert \(\dfrac{1}{4}\) and \(\dfrac{1}{6}\) to equivalent fractions with denominator \(12\), their LCD.
Solution
Find the LCD.  The LCD of \(\dfrac{1}{4}\) and \(\dfrac{1}{6}\) is 12. 
Find the number to multiply 4 to get 12.  \(4 \cdot \textcolor{red}{3} = 12\) 
Find the number to multiply 6 to get 12.  \(6 \cdot \textcolor{red}{2} = 12\) 
Use the Equivalent Fractions Property to convert each fraction to an equivalent fraction with the LCD, multiplying both the numerator and denominator of each fraction by the same number.  \(\begin{split} \dfrac{1}{4} \qquad & \dfrac{1}{6} \\ \dfrac{1 \cdot \textcolor{red}{3}}{4 \cdot \textcolor{red}{3}} \qquad & \dfrac{1 \cdot \textcolor{red}{2}}{6 \cdot \textcolor{red}{2}} \end{split}\) 
Simplify the numerators and denominators.  \(\dfrac{3}{12} \qquad \dfrac{2}{12}\) 
We do not reduce the resulting fractions. If we did, we would get back to our original fractions and lose the common denominator.
Change to equivalent fractions with the LCD: \(\dfrac{3}{4}\) and \(\dfrac{5}{6}\), \(LCD = 12\)
 Answer

\(\dfrac{9}{12}, \dfrac{10}{12}\)
Change to equivalent fractions with the LCD: \( \dfrac{7}{12}\) and \(\dfrac{11}{15}\), \(LCD = 60\)
 Answer

\(\dfrac{35}{60}, \dfrac{44}{60}\)
Step 1. Find the LCD.
Step 2. For each fraction, determine the number needed to multiply the denominator to get the LCD.
Step 3. Use the Equivalent Fractions Property to multiply both the numerator and denominator by the number you found in Step 2.
Step 4. Simplify the numerator and denominator.
Convert \(\dfrac{8}{15}\) and \(\dfrac{11}{24}\) to equivalent fractions with denominator \(120\), their LCD.
Find the number that must multiply 15 to get 120.  \(15 \cdot \textcolor{red}{8} = 120\) 
Find the number that must multiply 24 to get 120.  \(24 \cdot \textcolor{red}{5} = 120\) 
Use the Equivalent Fractions Property.  \(\dfrac{8 \cdot \textcolor{red}{8}}{15 \cdot \textcolor{red}{8}} \qquad \dfrac{11 \cdot \textcolor{red}{5}}{24 \cdot \textcolor{red}{5}}\) 
Simplify the numerators and denominators.  \(\dfrac{64}{120} \qquad \dfrac{55}{120}\) 
Change to equivalent fractions with the LCD: \(\dfrac{13}{24}\) and \(\dfrac{17}{32}\), LCD \(96\)
 Answer

\(\dfrac{52}{96}, \dfrac{51}{96}\)
Change to equivalent fractions with the LCD: \(\dfrac{9}{28}\) and \(\dfrac{27}{32}\), LCD \(224\)
 Answer

\(\dfrac{72}{224}, \dfrac{189}{224}\)
Add and Subtract Fractions with Different Denominators
Once we have converted two fractions to equivalent forms with common denominators, we can add or subtract them by adding or subtracting the numerators.
Step 1. Find the LCD.
Step 2. Convert each fraction to an equivalent form with the LCD as the denominator.
Step 3. Add or subtract the fractions.
Step 4. Write the result in simplified form.
Add: \(\dfrac{1}{2} + \dfrac{1}{3}\).
Solution
Find the LCD of 2, 3.  
Change into equivalent fractions with the LCD 6.  \(\dfrac{1 \cdot \textcolor{red}{3}}{2 \cdot \textcolor{red}{3}} + \dfrac{1 \cdot \textcolor{red}{2}}{3 \cdot \textcolor{red}{2}}\) 
Simplify the numerators and denominators.  \(\dfrac{3}{6} + \dfrac{2}{6}\) 
Add.  \(\dfrac{5}{6}\) 
Remember, always check to see if the answer can be simplified. Since \(5\) and \(6\) have no common factors, the fraction \(\dfrac{5}{6}\) cannot be reduced.
Add: \(\dfrac{1}{4} + \dfrac{1}{3}\).
 Answer

\(\dfrac{7}{12}\)
Add: \(\dfrac{1}{2} + \dfrac{1}{5}\).
 Answer

\(\dfrac{7}{10}\)
Subtract: \(\dfrac{1}{2}  \left( \dfrac{1}{4}\right)\).
Solution
Find the LCD of 2 and 4.  
Rewrite as equivalent fractions using the LCD 4.  \(\dfrac{1 \cdot \textcolor{red}{2}}{2 \cdot \textcolor{red}{2}}  \left( \dfrac{1}{4}\right)\) 
Simplify the first fraction.  \(\dfrac{2}{4}  \left( \dfrac{1}{4}\right)\) 
Subtract.  \(\dfrac{2  (1)}{4}\) 
Simplify.  \(\dfrac{3}{4}\) 
One of the fractions already had the least common denominator, so we only had to convert the other fraction.
Subtract: \(\dfrac{1}{2}  \left( \dfrac{1}{8}\right)\).
 Answer

\(\dfrac{5}{8}\)
Subtract: \(\dfrac{1}{3}  \left( \dfrac{1}{6}\right)\).
 Answer

\(\dfrac{1}{2}\)
Add: \(\dfrac{7}{12} + \dfrac{5}{18}\).
Solution
Find the LCD of 12 and 18.  
Rewrite as equivalent fractions with the LCD.  \(\dfrac{7 \cdot \textcolor{red}{3}}{12 \cdot \textcolor{red}{3}} + \dfrac{5 \cdot \textcolor{red}{2}}{18 \cdot \textcolor{red}{2}}\) 
Simplify the numerators and denominators.  \(\dfrac{21}{36} + \dfrac{10}{36}\) 
Add.  \(\dfrac{31}{36}\) 
Because \(31\) is a prime number, it has no factors in common with \(36\). The answer is simplified.
Add: \(\dfrac{7}{12} + \dfrac{11}{15}\).
 Answer

\(\dfrac{79}{60}\)
Add: \(\dfrac{13}{15} + \dfrac{17}{20}\).
 Answer

\(\dfrac{103}{60}\)
When we use the Equivalent Fractions Property, there is a quick way to find the number you need to multiply by to get the LCD. Write the factors of the denominators and the LCD just as you did to find the LCD. The “missing” factors of each denominator are the numbers you need.
The LCD, \(36\), has \(2\) factors of \(2\) and \(2\) factors of \(3\). Twelve has two factors of \(2\), but only one of \(3\) —so it is ‘missing‘ one \(3\). We multiplied the numerator and denominator of \(\dfrac{7}{12}\) by \(3\) to get an equivalent fraction with denominator \(36\). Eighteen is missing one factor of \(2\) —so you multiply the numerator and denominator \(\dfrac{5}{18}\) by \(2\) to get an equivalent fraction with denominator \(36\). We will apply this method as we subtract the fractions in the next example.
Subtract: \(\dfrac{7}{15} − \dfrac{19}{24}\).
Solution
Find the LCD. 15 is 'missing' three factors of 2 24 is 'missing' a factor of 5 

Rewrite as equivalent fractions with the LCD.  \(\dfrac{7 \cdot \textcolor{red}{8}}{15 \cdot \textcolor{red}{8}}  \dfrac{19 \cdot \textcolor{red}{5}}{24 \cdot \textcolor{red}{5}}\) 
Simplify each numerator and denominator.  \(\dfrac{56}{120}  \dfrac{95}{120}\) 
Subtract.  \( \dfrac{39}{120}\) 
Rewrite showing the common factor of 3.  \( \dfrac{13 \cdot 3}{40 \cdot 3}\) 
Remove the common factor to simplify.  \( \dfrac{13}{40}\) 
Subtract: \(\dfrac{13}{24} − \dfrac{17}{32}\).
 Answer

\(\dfrac{1}{96}\)
Subtract: \(\dfrac{21}{32} − \dfrac{9}{28}\).
 Answer

\(\dfrac{75}{224}\)
Add: \( \dfrac{11}{30} + \dfrac{23}{42}\).
Solution
Find the LCD.  
Rewrite as equivalent fractions with the LCD.  \( \dfrac{11 \cdot \textcolor{red}{7}}{30 \cdot \textcolor{red}{7}} + \dfrac{23 \cdot \textcolor{red}{5}}{42 \cdot \textcolor{red}{5}}\) 
Simplify each numerator and denominator.  \( \dfrac{77}{210} + \dfrac{115}{210}\) 
Add.  \(\dfrac{38}{210}\) 
Rewrite showing the common factor of 2.  \(\dfrac{19 \cdot 2}{105 \cdot 2}\) 
Remove the common factor to simplify.  \(\dfrac{19}{105}\) 
Add: \( \dfrac{13}{42} + \dfrac{17}{35}\).
 Answer

\(\dfrac{37}{210}\)
Add: \( \dfrac{19}{24} + \dfrac{17}{32}\).
 Answer

\(\dfrac{25}{96}\)
In the next example, one of the fractions has a variable in its numerator. We follow the same steps as when both numerators are numbers.
Add: \(\dfrac{3}{5} + \dfrac{x}{8}\).
Solution
The fractions have different denominators.
Find the LCD.  
Rewrite as equivalent fractions with the LCD.  \(\dfrac{3 \cdot \textcolor{red}{8}}{5 \cdot \textcolor{red}{8}} + \dfrac{x \cdot \textcolor{red}{5}}{8 \cdot \textcolor{red}{5}}\) 
Simplify the numerators and denominators.  \(\dfrac{24}{40} + \dfrac{5x}{40}\) 
Add.  \(\dfrac{24 + 5x}{40}\) 
We cannot add \(24\) and \(5x\) since they are not like terms, so we cannot simplify the expression any further.
Add: \(\dfrac{y}{6} + \dfrac{7}{9}\).
 Answer

\(\dfrac{3y+14}{18}\)
Add: \(\dfrac{x}{6} + \dfrac{7}{15}\).
 Answer

\(\dfrac{5x+14}{30}\)
Contributors and Attributions
Lynn Marecek (Santa Ana College) and MaryAnne AnthonySmith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at http://cnx.org/contents/fd53eae1fa2...49835c3c@5.191."