We need a common denominator to add fractions.

The lowest common multiple of 3 and 6 is 6 itself, so that is the best choice of common denominator. Note that the second fraction in this example doesn't need to be changed, because it already has a denominator of 6.

$$ \begin{eqnarray} &\ & 3 \small\frac{2}{3}\normalsize+4\small\frac{5}{6}\normalsize\\[9pt] &=& 3 \small\frac{2^{\ \large\textcolor{red}{\textsf{×2}}\normalsize}}{3^{\ \large\textcolor{red}{\textsf{×2}}\normalsize}}+4\small\frac{5}{6}\normalsize \\[9pt] &=& 3 \small\frac{4}{6}\normalsize+4\small\frac{5}{6}\normalsize\\[9pt] &=& 7 \small\frac{9}{6}\normalsize\\[9pt] &=& 8 \small\frac{3}{6}\:\:\:\small\textsf{(because }\frac{6}{6}\small=\normalsize1\small\textsf{)}\normalsize\\[9pt] &=& 8 \small\frac{1}{2}\normalsize \end{eqnarray} $$

Note: We chose this example to show how much better it is not to use top-heavy fractions or (even worse!) the method known as 'smile and kiss' to add or subtract fractions. Let's take a quick look at just how bad these are...

Here is how this question would be done using improper (i.e. top-heavy) fractions:

$$ \begin{eqnarray} &\ & 3 \small\frac{2}{3}\normalsize+4\small\frac{5}{6}\normalsize\\[9pt] &=& \small\frac{11}{3}\normalsize+\small\frac{29}{6}\normalsize\:\:\:\small\textsf{(unnecessary extra step)}\normalsize\\[9pt] &=& \small\frac{11^{\ \textcolor{red}{\large\textsf{×2}}\normalsize}}{3^{\ \textcolor{red}{\large\textsf{×2}}\normalsize}}+\small\frac{29}{6}\normalsize \\[9pt] &=& \small\frac{22}{6}\normalsize+\small\frac{29}{6}\normalsize\\[9pt] &=& \small\frac{51}{6}\normalsize\:\:\:\small\textsf{(large number on top)}\normalsize\\[9pt] &=& \small\frac{51^{\ \textcolor{red}{\large\textsf{÷3}}\normalsize}}{6^{\ \textcolor{red}{\large\textsf{÷3}}\normalsize}}\normalsize\:\:\:\small\textsf{(awkward division)}\normalsize\\[9pt] &=& \small\frac{17}{2}\normalsize\\[9pt] &\small\textsf{or}\normalsize& 8 \small\frac{1}{2}\normalsize\normalsize \end{eqnarray} $$

That was unpleasant. The 'smile and kiss' method is even more horrible:

$$ \begin{eqnarray} &\ & 3 \small\frac{2}{3}\normalsize+4\small\frac{5}{6}\normalsize\\[9pt] &=& \small\frac{11}{3}\normalsize+\small\frac{29}{6}\normalsize\:\:\:\small\textsf{(unnecessary extra step)}\normalsize\\[9pt] &=& \small\frac{11\!\times\!6+29\!\times\!3}{3\!\times\!6}\normalsize\:\:\:\small\textsf{(unnecessary multiplication)}\normalsize\\[9pt] &=& \small\frac{66+87}{18}\normalsize\:\:\:\small\textsf{(not the easiest addition)}\normalsize\\[9pt] &=& \small\frac{153}{18}\normalsize\:\:\:\small\textsf{(large number on top)}\normalsize\\[9pt] &=& \small\frac{153^{\ \large\textcolor{red}{\textsf{÷9}}\normalsize}}{18^{\ \large\textcolor{red}{\textsf{÷9}}\normalsize}}\normalsize\:\:\:\small\textsf{(nasty simplification)}\normalsize\\[9pt] &=& \small\frac{17}{2}\normalsize\\[9pt] &\small\textsf{or}\normalsize& 8 \small\frac{1}{2}\normalsize\normalsize \end{eqnarray} $$

We hope we have made our point! For the rest of this page, we will stick with mixed numbers to add or subtract.

We need a common denominator to subtract fractions.

The lowest common multiple of 4 and 3 is 12, so that is our common denominator.

$$ \begin{eqnarray} &\ & 5 \small\frac{1}{4}\normalsize-2\small\frac{1}{3}\normalsize\\[9pt] &=& 5 \small\frac{1^{\ \large\textcolor{red}{\textsf{×3}}\normalsize}}{4^{\ \large\textcolor{red}{\textsf{×3}}\normalsize}}\normalsize-2\small\frac{1^{\ \large\textcolor{red}{\textsf{×4}}\normalsize}}{3^{\ \large\textcolor{red}{\textsf{×4}}\normalsize}}\normalsize\\[9pt] &=& 5 \small\frac{3}{12}\normalsize-2\small\frac{4}{12}\normalsize\\[9pt] &=& 4 \small\frac{15}{12}\normalsize-2\small\frac{4}{12}\normalsize\:\:\:\small\textsf{(because }1=\frac{12}{12}\normalsize\textsf{)}\normalsize\\[9pt] &=& 2 \small\frac{11}{12} \end{eqnarray} $$

Note: If you don't like the method above of writing \(5\large\frac{3}{12}\) as \(4\large\frac{15}{12}\small,\) you could use the following alternative, in which we subtract the whole numbers and the twelfths separately:

$$ \begin{eqnarray} &...\ =& 5\small\frac{3}{12}\normalsize-2\small\frac{4}{12}\normalsize\\[9pt] &\phantom{...\ }=& 3 - \small\frac{1}{12}\normalsize\:\:\:\small\textsf{(subtract separately)}\normalsize\\[9pt] &\phantom{...\ }=& 2\small\frac{12}{12}\normalsize - \small\frac{1}{12}\:\:\:\small\textsf{(because }1=\small\frac{12}{12}\normalsize\textsf{)}\normalsize\\[9pt] &\phantom{...\ }=& 2 \small\frac{11}{12} \end{eqnarray} $$

Note: You could also answer this question by converting each mixed number into an improper fraction before giving them a common denominator, but we prefer the method above, as it keeps the numbers nice and small.

For multiplication or division, we do not use a common denominator.

When multiplying or dividing, any mixed numbers must first be converted into improper fractions (also known as 'top-heavy' fractions). In this example, both of the fractions need to be converted into improper fractions.

It is always a good idea to do any possible simplification before you multiply.

$$ \begin{eqnarray} &\ & 7 \small\frac{1}{2}\normalsize\times 1\small\frac{3}{5}\normalsize \\[9pt] &=& \small\frac{15}{2}\normalsize \times \small\frac{8}{5}\normalsize \\[9pt] &=& \small\frac{15^{\ \textcolor{red}{\large\textsf{÷5}}\normalsize}}{2^{\ \textcolor{blue}{\large\textsf{÷2}}\normalsize}}\normalsize \times \small\frac{8^{\ \textcolor{blue}{\large\textsf{÷2}}\normalsize}}{5^{\ \textcolor{red}{\large\textsf{÷5}}\normalsize}}\normalsize \\[9pt] &=& \small\frac{3}{1}\normalsize \times \small\frac{4}{1}\normalsize \\[9pt] &=& \small\frac{12}{1}\normalsize\:\:\:\small\left(\textsf{now multiply top and bottom}\right)\\[9pt] &=& 12\:\:\:\small\left(\textsf{any number over 1 is itself}\right) \end{eqnarray} $$

In this question, the first thing we need to do is to convert the mixed number \(3\frac{3}{4}\) into an improper fraction .

Then, to divide by \(\frac{7}{12}\small,\) we multiply by its reciprocal .

Remember to simplify by dividing top and bottom before you multiply.

$$ \begin{eqnarray} &\ & 3\small\frac{3}{4}\normalsize\div\small\frac{7}{12}\normalsize \\[9pt] &=& \small\frac{15}{4}\normalsize\div\small\frac{7}{12}\normalsize \\[9pt] &=& \small\frac{15}{4}\normalsize\times\small\frac{12}{7}\normalsize \\[9pt] &=& \small\frac{15}{\cancel{4}\,1}\normalsize\times\small\frac{\cancel{12}\,3}{7}\normalsize \\[9pt] &=& \small\frac{45}{7}\normalsize\\[9pt] &\small\textsf{or}& 6\small\frac{3}{7}\normalsize \end{eqnarray} $$

Note: You may give the final answer either as an improper fraction or a mixed number.

In this example, we will need to add first and then multiply \(\frac23\) by the result of the addition. So, in this case, it actually makes sense to convert the \(2\frac{1}{2}\) into an improper fraction .

$$ \begin{eqnarray} &\ & \small\frac{2}{3}\normalsize\left(\small\frac{3}{4}\normalsize+ 2\small\frac{1}{2}\normalsize\right) \\[9pt] &=& \small\frac{2}{3}\normalsize\left(\small\frac{3}{4}\normalsize+\small\frac{5^{\ \textcolor{red}{\large\textsf{×2}}\normalsize}}{2^{\ \textcolor{red}{\large\textsf{×2}}\normalsize}}\normalsize\right) \\[9pt] &=& \small\frac{2}{3}\normalsize\left(\small\frac{3}{4}\normalsize+\small\frac{10}{4}\normalsize\right) \\[9pt] &=& \small\frac{2}{3}\normalsize\left(\small\frac{13}{4}\normalsize\right) \\[9pt] &=& \small\frac{\cancel{2}\,1}{3}\normalsize\times\small\frac{13}{\cancel{4}\,2} \\[9pt] &=& \small\frac{13}{6}\normalsize\\[9pt] &\small\textsf{or}& 2\small\frac{1}{6}\normalsize \end{eqnarray} $$

Note: You may give your final answer either as \(\frac{13}{6}\) or \(2\frac16\small.\)

This question involves both a subtraction and an addition. These operations have the same priority, so we just work from left to right.

Although you could do the subtraction first and then add \(2\frac14\) to the answer, we prefer to keep everything together and use a common denominator that works for all three fractions.

The lowest common multiple of 2, 3 and 4 is 12, so that is our common denominator.

$$ \begin{eqnarray} &\ & 4\small\frac12\normalsize-1\small\frac23\normalsize+2\small\frac14\normalsize\\[9pt] &=& 4\small\frac{1^{\ \large\textcolor{red}{\textsf{×6}}\normalsize}}{2^{\ \large\textcolor{red}{\textsf{×6}}\normalsize}}\normalsize-1\small\frac{2^{\ \large\textcolor{red}{\textsf{×4}}\normalsize}}{3^{\ \large\textcolor{red}{\textsf{×4}}\normalsize}}\normalsize+2\small\frac{1^{\ \large\textcolor{red}{\textsf{×3}}\normalsize}}{4^{\ \large\textcolor{red}{\textsf{×3}}\normalsize}}\\[9pt] &=& 4\small\frac{6}{12}\normalsize-1\small\frac{8}{12}\normalsize+2\small\frac{3}{12}\normalsize\\[9pt] &=& 5\small\frac{1}{12}\normalsize\normalsize\\[9pt] \end{eqnarray} $$

In the final step, we subtracted and added the whole numbers and twelfths separately: \(4-1+2\) for the whole numbers and \(6-8+3\) for the twelfths. We were lucky in this question. We didn't have to deal with a negative number of twelfths and we didn't have to simplify at the end.

Note: Although a question like this hasn't yet appeared on an SQA paper, it would be perfectly fair.

Sometimes, we might be faced with a question that we would know how do if both numbers were fractions, but actually, one of them is a whole number.

When this happens, we can write the whole number as a fraction by putting it over 1.

$$ \begin{eqnarray} &\ & 2\small\frac{4}{7}\normalsize\div 12 \\[9pt] &=& 2\small\frac{4}{7}\normalsize\div\small\frac{12}{1}\normalsize \\[9pt] &=& \small\frac{18}{7}\normalsize\div\small\frac{12}{1}\normalsize \\[9pt] &=& \small\frac{18}{7}\normalsize\times\small\frac{1}{12}\normalsize \\[9pt] &=& \small\frac{\cancel{18}\,3}{7}\normalsize\times\small\frac{1}{\cancel{12}\,2}\normalsize \\[9pt] &=& \small\frac{3}{14}\normalsize\\[9pt] \end{eqnarray} $$

Note: Although a question like this hasn't yet appeared on an SQA paper, it would be perfectly fair.

Because we are multiplying, the second number must be converted into an improper fraction .

Make sure you simplify top and bottom before doing the multiplication.

$$ \begin{eqnarray} &\ & \small\frac{5}{12}\normalsize\times 2\small\frac{2}{9}\normalsize \\[9pt] &=& \small\frac{5}{12}\normalsize\times\small\frac{20}{9}\normalsize \\[9pt] &=& \small\frac{5}{\cancel{12}\,3}\normalsize\times\small\frac{\cancel{20}\,5}{9}\normalsize \\[9pt] &=& \small\frac{25}{27}\normalsize \end{eqnarray} $$

We need a common denominator for subtraction.

The lowest common multiple of 5 and 3 is 15, so that's what we use.

$$ \begin{eqnarray} &\ & 6\small\frac{1}{5}\normalsize-2\small\frac{1}{3}\normalsize\\[9pt] &=& 6\small\frac{1^{\ \textcolor{red}{\large\textsf{×3}}\normalsize}}{5^{\ \textcolor{red}{\large\textsf{×3}}\normalsize}}-2\small\frac{1^{\ \textcolor{red}{\large\textsf{×5}}\normalsize}}{3^{\ \textcolor{red}{\large\textsf{×5}}\normalsize}}\normalsize\\[9pt] &=& 6 \small\frac{3}{15}\normalsize-2\small\frac{5}{15}\normalsize\\[9pt] &=& 5 \small\frac{18}{15}\normalsize-2\small\frac{5}{15}\normalsize\:\:\:\small\textsf{(because }1=\frac{15}{15}\textsf{)}\normalsize\\[9pt] &=& 3 \small\frac{13}{15}\normalsize \end{eqnarray} $$

Note: If you don't like the method above of writing \(6\large\frac{3}{15}\) as \(5\large\frac{18}{15}\small,\) you could use the following alternative, in which we subtract the whole numbers and the fifteenths separately:

$$ \begin{eqnarray} &...\ =& 6\small\frac{3}{15}\normalsize-2\small\frac{5}{15}\normalsize\\[9pt] &\phantom{...\ }=& 4 - \small\frac{2}{15}\normalsize\:\:\:\small\textsf{(subtract separately)}\normalsize\\[9pt] &\phantom{...\ }=& 3\small\frac{15}{15}\normalsize - \small\frac{2}{15}\:\:\:\small\textsf{(because }1=\small\frac{15}{15}\normalsize\textsf{)}\normalsize\\[9pt] &\phantom{...\ }=& 3 \small\frac{13}{15} \end{eqnarray} $$

Although you could expand the bracket in the same way as you would do for an algebra expression, that would be more complicated than just doing the addition in the brackets, and then multiplying the answer by \(\frac34\small.\)

$$ \begin{eqnarray} &\ & \small\frac{3}{4}\normalsize\left(\small\frac{1}{3}\normalsize+\small\frac{2}{7}\normalsize\right) \\[9pt] &=& \small\frac{3}{4}\normalsize\left(\small\frac{1^{\ \textcolor{red}{\large\textsf{×7}}\normalsize}}{3^{\ \textcolor{red}{\large\textsf{×7}}\normalsize}} + \frac{2^{\ \textcolor{red}{\large\textsf{×3}}\normalsize}}{7^{\ \textcolor{red}{\large\textsf{×3}}\normalsize}}\normalsize \right) \\[9pt] &=& \small\frac{3}{4}\normalsize\left(\small\frac{7}{21} + \small\frac{6}{21} \right)\normalsize\\[9pt] &=& \small\frac{\cancel{3}\,1}{4}\normalsize\times\small\frac{13}{\cancel{21}\,7}\normalsize \\[9pt] &=& \small\frac{13}{28} \end{eqnarray} $$

We mustn't use mixed numbers for division. So our first step is to convert \(2\frac38\) to an improper fraction .

The second step is to multiply by the reciprocal of \(\frac{5}{16}\small.\) Remember to simplify before multiplying.

$$ \begin{eqnarray} &\ & 2\small\frac{3}{8}\normalsize\div\small\frac{5}{16}\normalsize \\[9pt] &=& \small\frac{19}{8}\normalsize\div\small\frac{5}{16}\normalsize \\[9pt] &=& \small\frac{19}{8}\normalsize\times\small\frac{16}{5}\normalsize \\[9pt] &=& \small\frac{19}{\cancel{8}\,1}\normalsize\times\small\frac{\cancel{16}\,2}{5}\normalsize \\[9pt] &=& \small\frac{38}{5}\normalsize\\[9pt] &\small\textsf{or}\normalsize& 7\small\frac{3}{5}\normalsize \end{eqnarray} $$

Note: You may give your final answer either as \(\frac{38}{5}\) or \(7\frac{3}{5}\small.\)

We mustn't use mixed numbers for division. So our first step is to convert \(1\frac56\) to an improper fraction .

The second step is to multiply by the reciprocal of \(\frac34\small.\) Remember to simplify before multiplying.

$$ \begin{eqnarray} &\ & 1\small\frac{5}{6}\normalsize\div\small\frac{3}{4}\normalsize \\[9pt] &=& \small\frac{11}{6}\normalsize\div\small\frac{3}{4}\normalsize \\[9pt] &=& \small\frac{11}{6}\normalsize\times\small\frac{4}{3}\normalsize \\[9pt] &=& \small\frac{11}{\cancel{6}\,3}\normalsize\times\small\frac{\cancel{4}\,2}{3}\normalsize \\[9pt] &=& \small\frac{22}{9}\normalsize\\[9pt] &\small\textsf{or}\normalsize& 2\small\frac{4}{9}\normalsize \end{eqnarray} $$

Note: You may give your final answer either as \(\frac{22}{9}\) or \(2\frac{4}{9}\small.\)

This is a fairly simple addition. The only minor complication is when we obtain an improper fraction in the second-last step and need to use the fact that \(\frac{15}{15}\) equals the whole number \(1\) to simplify our answer.

$$ \begin{eqnarray} &\ & 2\small\frac{1}{3}\normalsize+\small\frac{4}{5}\normalsize\\[9pt] &=& 2\small\frac{1^{\ \textcolor{red}{\large\textsf{×5}}\normalsize}}{3^{\ \textcolor{red}{\large\textsf{×5}}\normalsize}}\normalsize+\small\frac{4^{\ \textcolor{red}{\large\textsf{×3}}\normalsize}}{5^{\ \textcolor{red}{\large\textsf{×3}}\normalsize}}\normalsize \\[9pt] &=& 2\small\frac{5}{15}\normalsize+\small\frac{12}{15}\normalsize\\[9pt] &=& 2\small\frac{17}{15}\normalsize\\[9pt] &=& 3\small\frac{2}{15}\normalsize \end{eqnarray} $$

We mustn't use mixed numbers for multiplication. So our first step is to convert \(1\frac57\) to an improper fraction .

Then, as always, we should simplify top and bottom before multiplying.

$$ \begin{eqnarray} &\ & \small\frac{3}{8}\normalsize\times 1\small\frac{5}{7} \\[9pt] &=& \small\frac{3}{8}\normalsize\times\small\frac{12}{7} \\[9pt] &=& \small\frac{3}{\cancel{8}\,2}\normalsize\times\small\frac{\cancel{12}\,3}{7}\normalsize \\[9pt] &=& \small\frac{9}{14}\normalsize \end{eqnarray} $$

This simple 2-mark question contains no complications of any kind.

$$ \begin{eqnarray} &\ & 5\small\frac{1}{2}\normalsize-1\small\frac{2}{7}\normalsize\\[9pt] &=& 5 \small\frac{1^{\ \textcolor{red}{\large\textsf{×7}}\normalsize}}{2^{\ \textcolor{red}{\large\textsf{×7}}\normalsize}}\normalsize -1\small\frac{2^{\ \textcolor{red}{\large\textsf{×2}}\normalsize}}{7^{\ \textcolor{red}{\large\textsf{×2}}\normalsize}}\normalsize \\[9pt] &=& 5 \small\frac{7}{14}\normalsize -1\small\frac{4}{14}\normalsize\\[9pt] &=& 4 \small\frac{3}{14}\normalsize \end{eqnarray} $$

Note: As explained in examples 1 and 2 above, we recommend that you do not use improper fractions to add or subtract fractions. However, if you insist on doing this question using improper fractions, you should get the final answer in the form \(\frac{59}{14}\small.\)

We mustn't use mixed numbers for division. So our first step is to convert \(2\frac16\) to an improper fraction .

The second step is to multiply by the reciprocal of \(\frac89\small.\) Remember to simplify before multiplying.

$$ \begin{eqnarray} &\ & 2\small\frac{1}{6}\normalsize\div\small\frac{8}{9}\normalsize \\[9pt] &=& \small\frac{13}{6}\normalsize\div\small\frac{8}{9}\normalsize \\[9pt] &=& \small\frac{13}{6}\normalsize\times\small\frac{9}{8}\normalsize \\[9pt] &=& \small\frac{13}{\cancel{6}\,2}\normalsize\times\small\frac{\cancel{9}\,3}{8}\normalsize \\[9pt] &=& \small\frac{39}{16}\normalsize \\[9pt] &\small\textsf{or}\normalsize& 2\small\frac{7}{16}\normalsize \end{eqnarray} $$

Note: You may give your final answer either as \(\frac{39}{16}\) or \(2\frac{7}{16}\small.\)