To add two mixed numbers where the fractional parts have unlike denominators, we apply the following steps:

Step 1: First, we add whole numbers to whole numbers.

Step 2: Then, we add fractions to fractions, converting them to equivalent fractions with a common denominator where necessary.

Step 3: Combine the results from steps 1 and 2 to form a mixed number.

So, let's find the value of 1\,\dfrac{1}{2} + 2\,\dfrac 1 4 .

Step 1: Adding the whole numbers, we get 1+2 = {\color{red}{3}}.

Step 2: We now add the fractions:

\dfrac{1}{2} + \dfrac 1 4

Here, we can make a common denominator of 4. To put \dfrac 1 2 over a denominator of 4, we multiply the numerator and denominator by 2 :

\dfrac{1}{2} = \dfrac{1\times 2}{2\times 2} = \dfrac{2}{4}

We can now add the fractions. We keep the denominator the same, and we add the numerators.

\begin{align*} \dfrac{2}{4} + \dfrac{1}{4} = \color{blue} \dfrac{3}{4} \end{align*}

Step 3: We now combine the results to get our final answer:

{\color{red}{3}} + {\color{blue}{\dfrac 3 4}} = 3\,\dfrac 3 4

FLAG

What is the value of 5\,\dfrac{1}{2} + 2\,\dfrac 1 6 ?

EXPLANATION

To add two mixed numbers, we add whole numbers to whole numbers and fractions to fractions.

Step 1: Adding the whole numbers, we get 5+2 = {\color{red}{7}}.

Step 2: Now we add the fractions:

\dfrac{1}{2} + \dfrac 1 6

To add two fractions with unlike denominators, we must express each fraction as an equivalent fraction with a common denominator.

Here, we can make a common denominator of 6.

To put \dfrac 1 2 over a denominator of 6, we multiply the numerator and denominator by 3\mathbin{:}

\dfrac{1}{2} = \dfrac{1\times 3}{2\times 3} = \dfrac{3}{6}

We can now add the fractions. We keep the denominator the same, and we add the numerators.

\begin{align*} \dfrac{3}{6} + \dfrac{1}{6} = \dfrac{4}{6} \end{align*}

Notice that we can simplify the above fraction by dividing the numerator and denominator by 2\mathbin{:}

\dfrac{4}{6} = \dfrac{4\div 2}{6\div 2} = \color{blue} \dfrac{2}{3}

Step 3: Combining our results gives:

{\color{red}{7}} + {\color{blue}{\dfrac 2 3}} = 7\,\dfrac 2 3

FLAG

$2\,\dfrac{3}{8} + 2\,\dfrac 1 4 = $

a
$3\,\dfrac 1 4$
b
$3\,\dfrac 3 8$
c
$4\,\dfrac 3 4$
d
$4\,\dfrac 5 8$
e
$4\,\dfrac 3 8$

$1\,\dfrac{3}{10} + 2\,\dfrac 1 5 = $

a
$3\,\dfrac 3 5$
b
$4\,\dfrac 2 5$
c
$4\,\dfrac 1 2$
d
$3\,\dfrac 1 2$
e
$3\,\dfrac 2 5$

Calculate the value of 2\,\dfrac{1}{6} + 5\,\dfrac 1 8 .

EXPLANATION

To add two mixed numbers, we add whole numbers to whole numbers and fractions to fractions.

Step 1: Adding the whole numbers, we get 2+5 = {\color{red}{7}}.

Step 2: Now we add the fractions:

\dfrac{1}{6} + \dfrac 1 8

To add the fractions, we need to find the least common multiple of the denominators ( 6 and 8 ).

  • Multiples of 6{:} \quad 6,12,18,{\color{purple}{24}},\ldots

  • Multiples of 8{:} \quad 8,16,{\color{purple}{24}},\ldots

Therefore, the least common denominator of 6 and 8 is {\color{purple}{24}}.

  • To put \dfrac 1 6 over a denominator of 24, we multiply the numerator and denominator by 4\mathbin{:} \dfrac{1}{6} = \dfrac{1\times 4}{6 \times 4 } = \dfrac{4}{24}

  • To put \dfrac 1 8 over a denominator of 24, we multiply the numerator and denominator by 3\mathbin{:} \dfrac{1}{8} = \dfrac{1\times 3}{8 \times 3} = \dfrac{3}{24}

We can now add the fractions. We keep the denominator the same, and we add the numerators:

\begin{align*} \dfrac{1}{6} + \dfrac 1 8 = \dfrac{4}{24} + \dfrac{3}{24} = \color{blue} \dfrac{7}{24} \end{align*}

Step 3: We combine the results to get our final answer:

{\color{red}{7}} + {\color{blue} \dfrac{7}{24}} = 7\,\dfrac {7}{24}

FLAG

What is the missing number in the following equality?

\[2\,\dfrac{1}{7} + 4\,\dfrac 2 3 = 6\,\dfrac{\,\fbox{$\phantom{0}$}}{21} \]

a
$16$
b
$4$
c
$17$
d
$5$
e
$10$

$ 2\,\dfrac{1}{8} + 4\,\dfrac 5 6 = $

a
$6\,\dfrac {23}{24}$
b
$6\,\dfrac {19}{24}$
c
$6\,\dfrac {7}{8}$
d
$7\,\dfrac {1}{24}$
e
$6\,\dfrac {11}{12}$

Sometimes, when we add two mixed numbers, the sum of the fractional parts is an improper fraction. In those cases, we must convert the improper fraction to a mixed number. Then, we can add the whole numbers and fractions as usual.

To illustrate, let's compute 1 \dfrac{2}{3} + 2 \dfrac{2}{3}.

Step 1: Adding the whole numbers, we get 1+2={\color{red}3}.

Step 2: Now, we add the fractions: \dfrac{2}{3}+\dfrac{2}{3}=\dfrac{4}{3}

Notice that this is an improper fraction. We can convert the fraction \dfrac{4}{3} to a mixed number in the usual way:

4\div 3 = 1\,\textrm{R}1 = {\color{blue}{1\,\dfrac 1{3}}}

Step 3: We combine the results to get our final answer:

{\color{red}{3}} + {\color{blue}{1\,\dfrac 1{3}}} = 4\,\dfrac 1{3}

Notice that we needed to add the whole numbers one final time.

FLAG

Find the value of 1\,\dfrac{1}{2} + 3\,\dfrac 3 4 .

EXPLANATION

To add two mixed numbers, we add whole numbers to whole numbers and fractions to fractions.

Step 1: Adding the whole numbers, we get 1+3 = {\color{red}{4}}.

Step 2: Now we add the fractions:

\dfrac{1}{2} + \dfrac 3 4

To add two fractions with unlike denominators, we must express each fraction as an equivalent fraction with a common denominator.

In this question, we can make a common denominator of 4.

To put \dfrac 1 2 over a denominator of 4, we multiply the numerator and denominator by 2 :

\dfrac{1}{2} = \dfrac{1\times 2}{2\times 2} = \dfrac{2}{4}

We can now add the fractions. We keep the denominator the same, and we add the numerators.

\begin{align*} \dfrac{2}{4} + \dfrac{3}{4} = \color{blue} \dfrac{5}{4} \end{align*}

Notice that this is an improper fraction. We can convert the fraction \dfrac{5}{4} to a mixed number in the usual way:

5\div 4 = 1\,\textrm{R}1 = {\color{blue}{1\,\dfrac 1{4}}}

Step 3: We combine the results to get our answer.

{\color{red}{4}} + {\color{blue}{1\,\dfrac 1{4}}} = 5\,\dfrac 1{4}

Notice that we needed to add the whole numbers one final time.

FLAG

What is the missing number in the following equality?

\[5\,\dfrac{3}{4} + 1\,\dfrac 5 6 = 7\,\dfrac{\,\fbox{$\phantom{0}$}}{12} \]

a
$5$
b
$1$
c
$11$
d
$7$
e
$2$

$3\,\dfrac{1}{2} + 4\,\dfrac 7 8 = $

a
$ 8\,\dfrac 3{8}$
b
$ 7\,\dfrac 5{8}$
c
$ 8\,\dfrac 1{8}$
d
$ 9\,\dfrac 3{8}$
e
$7\,\dfrac 1{8}$
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