Writing $3\,\dfrac 1 8$ as an improper fraction, we have

\[ \dfrac{(3\times 8)+1}{8} = \dfrac{25}{8} . \]

Writing $1 \, \dfrac{3}{4}$ as an improper fraction, we have

\[ \dfrac{(1\times 4) + 3}{4} = \dfrac {7} 4 . \]

Therefore, our problem now is to work out the following:

\[ \dfrac{25}{8} - \dfrac {7} 4 \]

To subtract two fractions with unlike denominators, we need to express each fraction as an equivalent fraction with a common denominator.

In this question, we can make a common denominator of $8.$

To put $\dfrac 7 4$ over a denominator of $8,$ we multiply the numerator and denominator by $2$:

\[ \dfrac{7}{4} = \dfrac{7\times 2}{4\times 2} = \dfrac{14}{8} \]

We can now subtract the fractions.

\begin{align*} \dfrac{25}{8} - \dfrac{14}{8} = \dfrac{\bbox[2px,lightgray]{11}}{8} \end{align*}

So the missing number is $11.$

Writing $9\,\dfrac{3}{4} $ as an improper fraction, we have

\[ 9\,\dfrac{3}{4} = \dfrac{(9 \times 4) + 3}{4} = \dfrac{39}{4} . \]

Writing $7 \, \dfrac{2}{5}$ as an improper fraction, we have

\[ 7 \, \dfrac{2}{5} = \dfrac{(7 \times 5) + 2}{5} = \dfrac {37}{5} . \]

Therefore, our problem now is to work out the following:

\[ \dfrac{39}{4} - \dfrac {37}{5} \]

To subtract two fractions with unlike denominators, we need to express each fraction as an equivalent fraction with a common denominator.

In this question, we can make a common denominator of $20.$

To put $\dfrac {39}{4}$ over a denominator of $20,$ we multiply the numerator and denominator by $5$:

\[ \dfrac {39}{4} = \dfrac{39 \times 5}{4 \times 5} = \dfrac{195}{20} \]

To put $\dfrac {37}{5}$ over a denominator of $20,$ we multiply the numerator and denominator by $4$:

\[ \dfrac {37}{5} = \dfrac{37 \times 4}{5 \times 4} = \dfrac{148}{20} \]

We can now subtract the fractions. \[ \dfrac{195}{20} - \dfrac{148}{20}= \dfrac{\bbox[2px,lightgray]{47}}{20} \]

So the missing number is $47.$

Writing $3\,\dfrac 1 {15}$ as an improper fraction, we have

\[ \dfrac{(3\times 15)+1}{15} = \dfrac{46}{15} . \]

Writing $2 \, \dfrac{1}{5}$ as an improper fraction, we have

\[ \dfrac{(2\times 5) + 1}{5} = \dfrac {11}{5} . \]

Therefore, our problem now is to work out the following:

\[ \dfrac{46}{15} - \dfrac {11}{5} \]

To subtract two fractions with unlike denominators, we need to express each fraction as an equivalent fraction with a common denominator.

In this question, we can make a common denominator of $15.$

To put $\dfrac {11} 5$ over a denominator of $15,$ we multiply the numerator and denominator by $3$:

\[ \dfrac{11}{5} = \dfrac{11\times 3}{5\times 3} = \dfrac{33}{15} \]

We can now subtract the fractions. \begin{align*} \dfrac{46}{15} - \dfrac{33}{15} = \dfrac{13}{15} \end{align*}

Writing $4 \, \dfrac{3}{10}$ as an improper fraction, we have \[ 4 \, \dfrac{3}{10} = \dfrac{(4 \times 10) + 3}{10} = \dfrac{43}{10} . \]

Writing $2 \, \dfrac{1}{2}$ as an improper fraction, we have \[ 2 \, \dfrac{1}{2} = \dfrac{(2 \times 2) + 1}{2} = \dfrac{5}{2} . \]

Therefore, our problem now is to work out the following: \[ \dfrac{43}{10} - \dfrac{5}{2} \]

To subtract two fractions with unlike denominators, we need to express each fraction as an equivalent fraction with a common denominator.

In this question, we can make a common denominator of $10.$

To put $\dfrac{5}{2}$ over a denominator of $10,$ we multiply the numerator and denominator by $5$: \[ \dfrac{5}{2} = \dfrac{5 \times 5}{2 \times 5} = \dfrac{25}{10} \]

We can now subtract the fractions. \[ \dfrac{43}{10} - \dfrac{25}{10} = \dfrac{18}{10} \]

Simplifying, we have \[ \dfrac{18}{10} = \dfrac{18 \div 2}{10 \div 2} = \dfrac{9}{5}. \]

Therefore, $4 \, \dfrac{3}{10} - 2 \,\dfrac{1}{2}=\bbox[3pt,Gainsboro]{\color{blue} \dfrac{9}{5}}.$

Writing $2 \, \dfrac{1}{5}$ as an improper fraction, we have \[ 2 \, \dfrac{1}{5} = \dfrac{(2 \times 5) + 1}{5} = \dfrac{11}{5} . \]

Writing $1 \, \dfrac{2}{3}$ as an improper fraction, we have \[ 1 \, \dfrac{2}{3} = \dfrac{(1 \times 3) + 2}{3} = \dfrac{5}{3} . \]

Therefore, our problem now is to work out the following: \[ \dfrac{11}{5} - \dfrac{5}{3} \]

To subtract two fractions with unlike denominators, we need to express each fraction as an equivalent fraction with a common denominator.

In this question, we can make a common denominator of $15.$

To put $\dfrac{11}{5}$ over a denominator of $15,$ we multiply the numerator and denominator by $3$: \[ \dfrac{11}{5} = \dfrac{11 \times 3}{5 \times 3} = \dfrac{33}{15} \]

To put $\dfrac{5}{3}$ over a denominator of $15,$ we multiply the numerator and denominator by $5$: \[ \dfrac{5}{3} = \dfrac{5 \times 5}{3 \times 5} = \dfrac{25}{15} \]

We can now subtract the fractions. \[ \dfrac{33}{15} - \dfrac{25}{15} = \dfrac{8}{15} \]

Therefore, $2 \, \dfrac{1}{5} - 1 \,\dfrac{2}{3}=\bbox[3pt,Gainsboro]{\color{blue} \dfrac{8}{15}}.$

Writing $4\,\dfrac 2 {3}$ as an improper fraction, we have \[ \dfrac{(4\times 3)+2}{3} = \dfrac{14}{3} . \]

Writing $2\, \dfrac{5}{6}$ as an improper fraction, we have \[ \dfrac{(2\times 6) + 5}{6} = \dfrac {17}{6} . \]

Therefore, our problem now is to work out the following: \[ \dfrac{14}{3} - \dfrac {17}{6} \]

To subtract two fractions with unlike denominators, we need to express each fraction as an equivalent fraction with a common denominator.

In this question, we can make a common denominator of $6.$

To put $\dfrac {14} 3$ over a denominator of $6,$ we multiply the numerator and denominator by $2$: \[ \dfrac{14}{3} = \dfrac{14\times 2}{3\times 2} = \dfrac{28}{6} \]

We can now subtract the fractions. \begin{align*} \dfrac{28}{6} - \dfrac{17}{6} = \dfrac{11}{6} \end{align*}

Finally, we convert back to a mixed number: \[ 11 \div 6 = 1\,\textrm{R}\,5 \]

So, our final answer is $\bbox[3pt,Gainsboro]{\color{blue}1\,\dfrac{5}{6}}.$

Writing $5\,\dfrac{1}{6}$ as an improper fraction, we have \[ \dfrac{(5 \times 6) + 1}{6} = \dfrac{31}{6}. \]

Writing $3\,\dfrac{2}{3}$ as an improper fraction, we have \[ \dfrac{(3 \times 3) + 2}{3} = \dfrac{11}{3}. \]

Therefore, our problem now is to work out the following: \[ \dfrac{31}{6} - \dfrac{11}{3} \]

To subtract two fractions with unlike denominators, we need to express each fraction as an equivalent fraction with a common denominator.

In this question, we can make a common denominator of $6.$

To put $\dfrac{11}{3}$ over a denominator of $6,$ we multiply the numerator and denominator by $2{:}$ \[ \dfrac{11}{3} = \dfrac{11 \times 2}{3 \times 2} = \dfrac{22}{6} \]

We can now subtract the fractions. \begin{align*} \dfrac{31}{6} - \dfrac{22}{6} = \dfrac{9}{6} \end{align*}

Simplifying we have \[ \dfrac{9}{6} = \dfrac{9 \div 3}{6 \div 3} = \dfrac{3}{2}. \]

Finally, we convert back to a mixed number: \[ 3 \div 2 = 1\,\textrm{R}\,1 \]

So, our final answer is $\bbox[3pt,Gainsboro]{\color{blue}1\,\dfrac{1}{2}}.$

If we attempt to subtract the fractions by putting them over a common denominator, we get the following: \begin{align*} \dfrac{1}{15} - \dfrac{2}{5} &=\dfrac{1}{15} - \dfrac{3\times 2}{3\times 5} \\[5pt] &=\dfrac{1}{15} - \dfrac{6}{15} \end{align*}

However, we cannot carry out this subtraction since the first fraction is smaller than the second!

To get around this problem, we write each mixed number as an improper fraction.

• Writing $4\,\dfrac{1}{15}$ as an improper fraction, we have \[ 4\,\dfrac{1}{15} = \dfrac{(4 \times 15) + 1}{15} = \dfrac{61}{15}. \]

• Writing $2\,\dfrac{2}{5}$ as an improper fraction, we have \[ 2\,\dfrac{2}{5} = \dfrac{(2 \times 5) + 2}{5} = \dfrac{12}{5}. \]

Therefore, our problem now is to work out the following: \[ \dfrac{61}{15} - \dfrac{12}{5} \]

To subtract two fractions with unlike denominators, we need to express each fraction as an equivalent fraction with a common denominator.

In this question, we can make a common denominator of $15.$

To put $\dfrac{12}{5}$ over a denominator of $15,$ we multiply the numerator and denominator by $3{:}$ \[ \dfrac{12}{5} = \dfrac{12 \times 3}{5 \times 3} = \dfrac{36}{15} \]

We can now subtract the fractions. \[ \dfrac{61}{15} - \dfrac{36}{15} = \dfrac{25}{15} \]

Simplifying we have \[ \dfrac{25}{15} = \dfrac{25 \div 5}{15 \div 5} = \dfrac{5}{3}. \]

Finally, we convert back to a mixed number: \[ \dfrac{5}{3} = 1\,\textrm{R}\,2 = 1\,\dfrac{2}{3} \]

Therefore, $4\,\dfrac{1}{15} - 2\,\dfrac{2}{5} = \bbox[3pt,Gainsboro]{\color{blue}1\,\dfrac{2}{3}}.$

Writing $6 \, \dfrac{1}{8}$ as an improper fraction, we have

\[ 6 \, \dfrac{1}{8}= \dfrac{(6 \times 8) + 1}{8} = \dfrac{49}{8} . \]

Writing $2 \, \dfrac{1}{4}$ as an improper fraction, we have

\[ 2 \, \dfrac{1}{4} = \dfrac{(2 \times 4) + 1}{4} = \dfrac{9}{4} . \]

Therefore, our problem now is to work out the following:

\[ \dfrac{49}{8} - \dfrac{9}{4} \]

To subtract two fractions with unlike denominators, we need to express each fraction as an equivalent fraction with a common denominator.

In this question, we can make a common denominator of $8.$

To put $\dfrac{9}{4}$ over a denominator of $8,$ we multiply the numerator and denominator by $2$:

\[ \dfrac{9}{4} = \dfrac{9 \times 2}{4 \times 2} = \dfrac{18}{8} \]

We can now subtract the fractions.

$$ \dfrac{49}{8} - \dfrac{18}{8}= \dfrac{31}{8} $$

Finally, we convert back to a mixed number:

\[ \dfrac{31}{8} =3 \, \textrm{R} \, 7 = 3 \, \dfrac{7}{8} \]

Writing $4\,\dfrac{1}{12}$ as an improper fraction, we have \[ 4\,\dfrac{1}{12} = \dfrac{(4 \times 12) + 1}{12} = \dfrac{49}{12}. \]

Writing $1\,\dfrac{3}{4}$ as an improper fraction, we have \[ 1\,\dfrac{3}{4} = \dfrac{(1 \times 4) + 3}{4} = \dfrac{7}{4}. \]

Therefore, our problem now is to work out the following: \[ \dfrac{49}{12} - \dfrac{7}{4} \]

To subtract two fractions with unlike denominators, we need to express each fraction as an equivalent fraction with a common denominator.

In this question, we can make a common denominator of $12.$

To put $\dfrac{7}{4}$ over a denominator of $12,$ we multiply the numerator and denominator by $3{:}$ \[ \dfrac{7}{4} = \dfrac{7 \times 3}{4 \times 3} = \dfrac{21}{12} \]

We can now subtract the fractions. \[ \dfrac{49}{12} - \dfrac{21}{12} = \dfrac{28}{12} \]

We can simplify this fraction by dividing the numerator and denominator by $4{:}$ \[ \dfrac{28}{12} = \dfrac{28 \div 4}{12 \div 4} = \dfrac{7}{3} \]

Finally, we convert back to a mixed number: \[ \dfrac{7}{3} = 2\,\textrm{R}\,1 = 2\,\dfrac{1}{3} \]

Therefore, $4\,\dfrac{1}{12} - 1\,\dfrac{3}{4} = \bbox[3pt,Gainsboro]{\color{blue}2\,\dfrac{1}{3}}.$

If we attempt to subtract the fractions by putting them over a common denominator, we get the following: \begin{align*} \dfrac{1}{6} - \dfrac{2}{3} & = \dfrac{1}{6} - \dfrac{2 \times 2}{3 \times 2}\\[5pt] & = \dfrac{1}{6} - \dfrac{4}{6} \end{align*}

However, we cannot carry out this subtraction since the first fraction is smaller than the second!

To get around this problem, we write each mixed number as an improper fraction.

• Writing $7\,\dfrac{1}{6}$ as an improper fraction, we have \[ 7\,\dfrac{1}{6} = \dfrac{(7 \times 6) + 1}{6} = \dfrac{43}{6}. \]

• Writing $3\,\dfrac{2}{3}$ as an improper fraction, we have \[ 3\,\dfrac{2}{3} = \dfrac{(3 \times 3) + 2}{3} = \dfrac{11}{3}. \]

Therefore, our problem now is to work out the following: \[ \dfrac{43}{6} - \dfrac{11}{3} \]

To subtract two fractions with unlike denominators, we need to express each fraction as an equivalent fraction with a common denominator.

In this question, we can make a common denominator of $6.$

To put $\dfrac{11}{3}$ over a denominator of $6,$ we multiply the numerator and denominator by $2{:}$ \[ \dfrac{11}{3} = \dfrac{11 \times 2}{3 \times 2} = \dfrac{22}{6} \]

We can now subtract the fractions. \[ \dfrac{43}{6} - \dfrac{22}{6} = \dfrac{21}{6} \]

Simplifying we have \[ \dfrac{21}{6} = \dfrac{21 \div 3}{6 \div 3} = \dfrac{7}{2}. \]

Finally, we convert back to a mixed number: \[ \dfrac{7}{2} = 3\,\textrm{R}\,1 = 3\,\dfrac{1}{2} \]

Therefore, $7\,\dfrac{1}{6} - 3\,\dfrac{2}{3} = \bbox[3pt,Gainsboro]{\color{blue}3\,\dfrac{1}{2}}.$