## Fractions Part 9 – Fraction Problems

OK, let’s now use the skills in my previous posts to solve some example problems. I will show the problems first so you can try them on your own, then further down the post, you can see the solutions:

$\begin{array}{l} {{1}{.}\hspace{0.33em}\hspace{0.33em}{6}\frac{1}{4}\hspace{0.33em}\div\hspace{0.33em}{2}\frac{7}{8}}\\ {{2}{.}\hspace{0.33em}\hspace{0.33em}\frac{7}{9}\hspace{0.33em}{+}\hspace{0.33em}\frac{11}{12}}\\ {{3}{.}\hspace{0.33em}\hspace{0.33em}{6}\frac{1}{4}\hspace{0.33em}{+}\hspace{0.33em}{2}\frac{7}{8}}\\ {{4}{.}\hspace{0.33em}\hspace{0.33em}\frac{11}{12}\hspace{0.33em}{-}\hspace{0.33em}\frac{7}{9}}\\ {{5}{.}\hspace{0.33em}\hspace{0.33em}{6}\frac{1}{4}\hspace{0.33em}\times\hspace{0.33em}{2}\frac{7}{8}} \end{array}$

SOLUTIONS:

${1}{.}\hspace{0.33em}\hspace{0.33em}{6}\frac{1}{4}\hspace{0.33em}\div\hspace{0.33em}{2}\frac{7}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{25}{4}\hspace{0.33em}\div\hspace{0.33em}\frac{23}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{25}{4}\hspace{0.33em}\times\hspace{0.33em}\frac{8}{23}\hspace{0.33em}{=}\hspace{0.33em}\frac{{25}\hspace{0.33em}\times\hspace{0.33em}{8}}{{4}\hspace{0.33em}\times\hspace{0.33em}{23}}$

Now I will simplify this as much as possible now before I multiply to keep the numbers as small as possible:

$\frac{{25}\hspace{0.33em}\times\hspace{0.33em}{8}}{{4}\hspace{0.33em}\times\hspace{0.33em}{23}}\hspace{0.33em}{=}\hspace{0.33em}\frac{{25}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{4}\hspace{0.33em}\times\hspace{0.33em}{2}}{\rlap{/}{4}\hspace{0.33em}\times\hspace{0.33em}{23}}\hspace{0.33em}{=}\hspace{0.33em}\frac{{25}\hspace{0.33em}\times\hspace{0.33em}{2}}{23}\hspace{0.33em}{=}\hspace{0.33em}\frac{50}{23}\hspace{0.33em}{=}\hspace{0.33em}{2}\frac{4}{23}$

${2}{.}\hspace{0.33em}\hspace{0.33em}\frac{7}{9}\hspace{0.33em}{+}\hspace{0.33em}\frac{11}{12}$.  Since these fractions are being added and they have different denominators, we need to find a common denominator. To find the Least Common Denominator (LCD):

9 = 3 × 3,   12 = 2 × 2 × 3

LCD = 2 × 2 × 3 × 3 = 36

So now convert each of the fractions into equivalent ones with 36 as the denominator. So I will multiply the top and bottom of the first fraction by 4 and the other one by 3:

$\frac{7}{9}\hspace{0.33em}{+}\hspace{0.33em}\frac{11}{12}\hspace{0.33em}{=}\hspace{0.33em}\frac{{7}\hspace{0.33em}\times\hspace{0.33em}{4}}{{9}\hspace{0.33em}\times\hspace{0.33em}{4}}\hspace{0.33em}{+}\hspace{0.33em}\frac{{11}\hspace{0.33em}\times\hspace{0.33em}{3}}{{12}\hspace{0.33em}\times\hspace{0.33em}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{28}{36}\hspace{0.33em}{+}\hspace{0.33em}\frac{33}{36}\hspace{0.33em}{=}\hspace{0.33em}\frac{61}{36}\hspace{0.33em}{=}\hspace{0.33em}{1}\frac{25}{36}$

${3}{.}\hspace{0.33em}\hspace{0.33em}{6}\frac{1}{4}\hspace{0.33em}{+}\hspace{0.33em}{2}\frac{7}{8}$. First, the fractions need to be converted to improper ones. Then since 4 evenly divides into 8, we just need to convert the first fraction to an equivalent one with 8 as the denominator:

${6}\frac{1}{4}\hspace{0.33em}{+}\hspace{0.33em}{2}\frac{7}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{25}{4}\hspace{0.33em}{+}\hspace{0.33em}\frac{23}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{{25}\hspace{0.33em}\times\hspace{0.33em}{2}}{{4}\hspace{0.33em}\times\hspace{0.33em}{2}}\hspace{0.33em}{+}\hspace{0.33em}\frac{23}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{50}{8}\hspace{0.33em}{+}\hspace{0.33em}\frac{23}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{73}{8}\hspace{0.33em}{=}\hspace{0.33em}{9}\frac{1}{8}$

${4}{.}\hspace{0.33em}\hspace{0.33em}\frac{11}{12}\hspace{0.33em}{-}\hspace{0.33em}\frac{7}{9}$. These are the same fractions as in the second problem, but now it’s a subtraction problem. We have already converted these two fractions into equivalent fractions with the same denominator, so we will use that result here:

$\frac{11}{12}\hspace{0.33em}{-}\hspace{0.33em}\frac{7}{9}\hspace{0.33em}{=}\hspace{0.33em}\frac{33}{36}\hspace{0.33em}{-}\hspace{0.33em}\frac{28}{36}\hspace{0.33em}{=}\hspace{0.33em}\frac{5}{36}$

${5}{.}\hspace{0.33em}\hspace{0.33em}{6}\frac{1}{4}\hspace{0.33em}\times\hspace{0.33em}{2}\frac{7}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{25}{4}\hspace{0.33em}\times\hspace{0.33em}\frac{23}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{575}{8}\hspace{0.33em}{=}\hspace{0.33em}{71}\frac{7}{8}$

How did you do?

## Fractions Part 8 – Fraction Division

So far I have avoided division of fractions. This is not hard, in fact, it’s almost as easy as multiplying but I would like to show why the method works.

First consider 6 ÷ 2. This problem is asking the question, “How many 2’s fit into 6?”. The answer is 3 sets of 2’s make up 6. If the problem was 6 ÷ 3, the question would be “How many 3’s fit into 6?”. The answer here is there are 2 sets of 3’s that make up 6. Of course, the answer is not always an integer, Sometimes there are leftover numbers. For example, how many 2’s can make up 7, in other words, 7 ÷ 2. The answer is there are 3 sets of 2’s that can fit into 7 but there will be 1 left over as sets of 2’s do not exactly make up 7. Now remember that fractions are indicating division as well so the above problems are equivalent to

$\frac{6}{2}{,}\hspace{0.33em}\hspace{0.33em}\frac{6}{3}{,}\hspace{0.33em}\hspace{0.33em}\frac{7}{2}$

So what would ${1}\hspace{0.33em}\div\hspace{0.33em}\frac{1}{2}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{\frac{1}{2}}$

mean? It means the same as before: how many one-halves fit into 1? Now the answer is 2 because there are two halves in a whole. What about ${2}\hspace{0.33em}\div\hspace{0.33em}\frac{1}{2}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{\frac{1}{2}}$? Can you see that the answer is 4? Because if there are two things split into halves, then there will be 4 halves.

Let’s now look at ${3}\hspace{0.33em}\div\hspace{0.33em}\frac{3}{4}\hspace{0.33em}{=}\hspace{0.33em}\frac{3}{\frac{3}{4}}$

Looks like there are 4 three-quarters that fit into 3, and that is correct. Let’s look at one more. What about $\frac{3}{4}\hspace{0.33em}\div\hspace{0.33em}\frac{1}{4}\hspace{0.33em}{=}\hspace{0.33em}\frac{\frac{3}{4}}{\frac{1}{4}}$. Can you see that there are 3 one-quarters that fit into three-quarters?

All of the above answers can be obtained by multiplying and remembering that integers like 6 can also be shown as a fraction: $\frac{6}{1}$ as 6 divided by 1 is still 6.

So now let’s look at the previous problems. You can convert a fraction division problem by inverting the fraction in the denominator then multiplying:

${1}\hspace{0.33em}\div\hspace{0.33em}\frac{1}{2}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{\frac{1}{2}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{1}\hspace{0.33em}\times\hspace{0.33em}\frac{2}{1}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{1}\hspace{0.33em}{=}\hspace{0.33em}{2}$

The inverted fraction is called a reciprocal. A simply way to remember how to do fraction division is the phrase: invert and multiply.

Now let’s do the other ones:

${3}\hspace{0.33em}\div\hspace{0.33em}\frac{3}{4}\hspace{0.33em}{=}\hspace{0.33em}\frac{3}{\frac{3}{4}}\hspace{0.33em}{=}\hspace{0.33em}\frac{\rlap{/}{3}}{1}\hspace{0.33em}\times\hspace{0.33em}\frac{4}{\rlap{/}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{4}{1}\hspace{0.33em}{=}\hspace{0.33em}{4}$

And finally

$\frac{3}{4}\hspace{0.33em}\div\hspace{0.33em}\frac{1}{4}\hspace{0.33em}{=}\hspace{0.33em}\frac{\frac{3}{4}}{\frac{1}{4}}\hspace{0.33em}{=}\hspace{0.33em}\frac{3}{\rlap{/}{4}}\hspace{0.33em}\times\hspace{0.33em}\frac{\rlap{/}{4}}{1}\hspace{0.33em}{=}\hspace{0.33em}\frac{3}{1}\hspace{0.33em}{=}\hspace{0.33em}{3}$

I will do more examples in my next post including division with mixed numbers.

## Fractions Part 7 – Arithmetic on Mixed Fractions

So now let’s use the skills from the last two posts to work on mixed fractions. Consider:

${3}\frac{3}{8}\hspace{0.33em}{+}\hspace{0.33em}{6}\frac{7}{8}$

As the denominators are the same, one can do this problem by adding the whole parts first to get 9 then add the fractions together to get $\frac{10}{8}$ which after simplifying and converting to a mixed fraction is equal to $1\frac{1}{4}$. Then you add this to the 9 to get $10\frac{1}{4}$. But I want to show a more general method that is particularly useful when the denominators are different and/or the problem is a subtraction.

The first step is to convert the mixed fractions to improper ones as discussed in my last post:

${3}\frac{3}{8}\hspace{0.33em}{+}\hspace{0.33em}{6}\frac{7}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{{(}{8}\hspace{0.33em}\times\hspace{0.33em}{3}{)}\hspace{0.33em}{+}\hspace{0.33em}{3}}{8}\hspace{0.33em}{+}\hspace{0.33em}\frac{{(}{8}\hspace{0.33em}\times\hspace{0.33em}{6}{)}\hspace{0.33em}{+}\hspace{0.33em}{7}}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{27}{8}\hspace{0.33em}{+}\hspace{0.33em}\frac{55}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{82}{8}$

The addition in the last step was easy as the denominators are the same. Now all that is left is to simplify and convert to a mixed fraction. You can convert first then simplify which has the advantage of having smaller numbers to factor, so let’s do that:

$\frac{82}{8}\hspace{0.33em}{=}\hspace{0.33em}{82}\hspace{0.33em}\div\hspace{0.33em}{8}\hspace{0.33em}{=}\hspace{0.33em}{10}$ with remainder of 2 so

$\frac{82}{8}\hspace{0.33em}{=}\hspace{0.33em}{10}\frac{2}{8}\hspace{0.33em}$

So now all that is left is to simplify the fractional part:

$\frac{2}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{\rlap{/}{2}\hspace{0.33em}\times\hspace{0.33em}{1}}{\rlap{/}{2}\hspace{0.33em}\times\hspace{0.33em}{4}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{4}$

So ${3}\frac{3}{8}\hspace{0.33em}{+}\hspace{0.33em}{6}\frac{7}{8}\hspace{0.33em}{=}\hspace{0.33em}{10}\frac{1}{4}$ which is the same answer we got before. Isn’t maths consistent? (and fun!).

Now let’s do one with different denominators:

${6}\frac{3}{8}\hspace{0.33em}{-}\hspace{0.33em}{3}\frac{7}{12}$

We start by converting the problem into one with improper fractions:

${6}\frac{3}{8}\hspace{0.33em}{-}\hspace{0.33em}{3}\frac{7}{12}\hspace{0.33em}{=}\hspace{0.33em}\frac{{(}{8}\hspace{0.33em}\times\hspace{0.33em}{6}{)}\hspace{0.33em}{+}\hspace{0.33em}{3}}{8}\hspace{0.33em}{-}\hspace{0.33em}\frac{{(}{12}\hspace{0.33em}\times\hspace{0.33em}{3}{)}\hspace{0.33em}{+}\hspace{0.33em}{7}}{12}\hspace{0.33em}{=}\hspace{0.33em}\frac{51}{8}\hspace{0.33em}{-}\hspace{0.33em}\frac{43}{12}$

Now we need to find a common denominator between 8 and 12. To find the least common denominator (LCD), we first factor both numbers:

8 = 2 × 2 × 2,  12 = 2 × 2 × 3

As 2 and 3 are the only factors present, we now combine them, using them only the maximum number of times each appear in the above factorisation:

LCD = 2 × 2 × 2 × 3 = 24

So the common denominator we will use is 24. We want to convert each of the fractions in the problem into equivalent ones that have 24 as the denominator. For the first fraction, we will multiply top and bottom by 3 to get the 24 in the denominator. We will use 2 for the second fraction to get 24  in the denominator there as well:

$\begin{array}{l} {\frac{51}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{{51}\hspace{0.33em}\times\hspace{0.33em}{3}}{{8}\hspace{0.33em}\times\hspace{0.33em}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{153}{24}}\\ {\frac{43}{12}\hspace{0.33em}{=}\hspace{0.33em}\frac{{43}\hspace{0.33em}\times\hspace{0.33em}{2}}{{12}\hspace{0.33em}\times\hspace{0.33em}{2}}\hspace{0.33em}{=}\hspace{0.33em}\frac{86}{24}} \end{array}$

So now the problem becomes:

$\frac{51}{8}\hspace{0.33em}{-}\hspace{0.33em}\frac{43}{12}\hspace{0.33em}{=}\hspace{0.33em}\frac{153}{24}\hspace{0.33em}{-}\hspace{0.33em}\frac{86}{24}\hspace{0.33em}{=}\hspace{0.33em}\frac{67}{24}$

Now convert back to a mixed fraction:

$\frac{67}{24}\hspace{0.33em}{=}\hspace{0.33em}{67}\hspace{0.33em}\div\hspace{0.33em}{24}\hspace{0.33em}{=}\hspace{0.33em}{2}$ with a remainder of 19.

So the final answer is $2\frac{19}{24}$. The fractional part cannot be simplified any further.

So the steps to do multiplication would be just to convert the mixed fractions to improper ones, multiply these, and then convert back to a mixed fraction.

The general steps to do arithmetic on mixed fractions are:

1.  Convert all fractions to improper fractions.
2. If the problem is a multiplication one or if the denominators are the same, skip to step 5.
3. Find the LCD for the denominators.
4. Convert the improper fractions into equivalent ones with the LCD as the denominator.
5. Do the indicated arithmetic (multiplication, addition, or subtraction) on the improper fractions.
6. Convert the answer back to a mixed fraction if the numerator is greater than the denominator.
7. Simplify the fractional part if needed.

## Fractions Part 6 – Mixed Fractions

In my last post, I showed how to convert an improper fraction to a mixed one. Today, let’s do the opposite: convert a mixed fraction to an improper one. Why would one want to do this apart from the pure joy and ecstasy of doing maths? Because, this would be the first step in adding, subtracting, or multiplying mixed fractions together. Notice that I have been avoiding division of fractions. This will be the topic of another post – we are well and truly going to beat the number of Rocky movies!

Let’s first show how to convert a mixed fraction to an improper one then we will use this skill eventually in a problem. Consider

$2\frac{3}{4}$

Now please realise that this is maths shorthand for ${2}\hspace{0.33em}{+}\hspace{0.33em}\frac{3}{4}$. The denominator in the fraction part of this is indicating a whole of something divided into 4 pieces. This will be the denominator in our eventual improper fraction as well.

So to convert the 2 whole pieces into the number of total pieces if each whole was divided into 4 pieces is to multiply the 2 by the 4 to get 8 pieces. That is, if you divided 2 whole things into 4 pieces each, you get 8 total pieces.

Now we also have to remember that we have this fractional part of $\frac{3}{4}$.  Remember that this fraction means that we have 3 pieces of a whole that was divided into 4 pieces. So we need to add these 3 pieces to the 8 pieces we have so far to get 11. So we have 11 pieces of some items that were each divided into 4 pieces. In fractional form, this is  $\frac{11}{4}$. This is the answer. So to convert a mixed fraction to an improper one, you multiply the denominator part by the whole number part, then add the numerator part. Let’s do another one:

${5}\frac{3}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{{(}{8}\hspace{0.33em}\times\hspace{0.33em}{5}{)}\hspace{0.33em}{+}\hspace{0.33em}{3}}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{{40}\hspace{0.33em}{+}\hspace{0.33em}{3}}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{43}{8}$

In a picture, you

and then set the result over the denominator.

In my next post, we will use this to do some math on mixed fractions.

## Fractions Part 5 – Mixed, Improper, Proper Fractions

Now the fractions we have been working with are called proper fractions: these are fractions where the numerator is smaller than the denominator. These types of fractions are smaller than one, which is why they are called proper as they are a fractional part of one. This implies there are things called improper fractions, and there are. Not improper in the sense of your drunk uncle at a wedding, but improper because they are greater than or equal to one. These will be fractions with a numerator equal to or greater than the denominator.

For example $\frac{5}{3}$ is an improper fraction. Now they can be added, subtracted, multiplied, and simplified like any other fraction. But when the result of an operation with fractions results in an improper fraction, you are expected to convert this to a mixed fraction – a whole number plus a proper fraction.  So $\frac{5}{3}$ is equal to $1\frac{2}{3}$

So how do you convert an improper fraction to a mixed one? Just take out the whole parts and leave the resulting proper fraction. You do this by dividing: $\frac{5}{3}\hspace{0.33em}{=}\hspace{0.33em}{5}\hspace{0.33em}\div\hspace{0.33em}{3}\hspace{0.33em}{=}\hspace{0.33em}{1}$ plus a reminder of 2. So in this case, the improper fraction is one whole plus two thirds left over or $1\frac{2}{3}$.

Now try $\frac{24}{11}$:

$\frac{24}{11}\hspace{0.33em}{=}\hspace{0.33em}{24}\hspace{0.33em}\div\hspace{0.33em}{11}\hspace{0.33em}{=}\hspace{0.33em}{2}$ with a remainder of 2. So $\frac{24}{11}\hspace{0.33em}{=}\hspace{0.33em}{2}\frac{2}{11}$.

One more example:

$\frac{66}{33}\hspace{0.33em}{=}\hspace{0.33em}{66}\hspace{0.33em}\div\hspace{0.33em}{33}\hspace{0.33em}{=}\hspace{0.33em}{2}$.

Sometime there is no remainder and you just have a whole number. By the way, this result could also have been obtained by factoring:

$\frac{66}{33}\hspace{0.33em}{=}\hspace{0.33em}\frac{\rlap{/}{3}\rlap{/}{3}\hspace{0.33em}\times\hspace{0.33em}{2}}{\rlap{/}{3}\rlap{/}{3}\hspace{0.33em}\times\hspace{0.33em}{1}}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{1}\hspace{0.33em}{=}\hspace{0.33em}{2}$

I’m starting to feel like the Rocky movie series – will I get to part 7?

Here are more examples of adding fractions with different denominators:

1. $\frac{4}{15}\hspace{0.33em}{+}\hspace{0.33em}\frac{3}{5}$

So two different denominators. We have to find equivalent fractions that have the same denominator before we can add these. As mentioned in my last post, first check to see if one of the denominators evenly divides into the other. In this case, 5 divides into 15 three times so 15 can be our common denominator. We just have to convert the second fraction. We do this by multiplying top and bottom by 3 because multiplying the 5 by 3 will give us the 15 we want:

$\frac{3}{5}\hspace{0.33em}{=}\hspace{0.33em}\frac{{3}\hspace{0.33em}\times\hspace{0.33em}{3}}{{5}\hspace{0.33em}\times\hspace{0.33em}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{9}{15}$

So now,

$\frac{4}{15}\hspace{0.33em}{+}\hspace{0.33em}\frac{3}{5}\hspace{0.33em}{=}\hspace{0.33em}\frac{4}{15}\hspace{0.33em}{+}\hspace{0.33em}\frac{9}{15}\hspace{0.33em}{=}\hspace{0.33em}\frac{13}{15}$

Now I’ve been concentrating on adding fraction but subtracting them is just as easy:

2. $\frac{7}{10}\hspace{0.33em}{-}\hspace{0.33em}\frac{3}{14}$

Now 10 does not divide into 14 so we will have to find the LCD of these two numbers:

10 = 2 × 5

14 = 2 × 7

LCD = 2 × 5 × 7 = 70

So now let’s convert the two fractions to equivalent fractions with 70 as the denominator. For the first fraction, I want to multiply the top and bottom by 7 and the second fraction I want to use 5:

$\begin{array}{l} {\frac{7}{10}\hspace{0.33em}{=}\hspace{0.33em}\frac{{7}\hspace{0.33em}\times\hspace{0.33em}{7}}{{10}\hspace{0.33em}\times\hspace{0.33em}{7}}\hspace{0.33em}{=}\hspace{0.33em}\frac{49}{70}}\\ {\frac{3}{14}\hspace{0.33em}{=}\hspace{0.33em}\frac{{3}\hspace{0.33em}\times\hspace{0.33em}{5}}{{14}\hspace{0.33em}\times\hspace{0.33em}{5}}\hspace{0.33em}{=}\hspace{0.33em}\frac{15}{70}} \end{array}$

So now,

$\frac{7}{10}\hspace{0.33em}{-}\hspace{0.33em}\frac{3}{14}\hspace{0.33em}{=}\hspace{0.33em}\frac{49}{70}\hspace{0.33em}{-}\hspace{0.33em}\frac{15}{70}\hspace{0.33em}{=}\hspace{0.33em}\frac{{49}\hspace{0.33em}{-}\hspace{0.33em}{15}}{70}\hspace{0.33em}{=}\hspace{0.33em}\frac{34}{70}$

So subtracting is just as easy since you just need to subtract the numerators. But we’re not quite done here. The numerator and denominator are both even numbers so at least a 2 is a common factor:

$\frac{34}{70}\hspace{0.33em}{=}\hspace{0.33em}\frac{{17}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{2}}{{35}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{2}}\hspace{0.33em}{=}\hspace{0.33em}\frac{17}{35}$

And this is as simple as it can get. One more example:

3. $\frac{10}{21}\hspace{0.33em}{+}\hspace{0.33em}\frac{9}{24}$

Again, we will need the LCD to get the required denominator as 21 does not divide into 24:

21 = 3 × 7

24 = 2 × 2 × 2 × 3

LCD = 2 × 2 × 2 × 3 × 7 = 168

So now let’s convert the two fractions to equivalent fractions with 168 as the denominator. For the first fraction, I want to multiply the top and bottom by 8 and the second fraction I want to use 7:

$\begin{array}{l} {\frac{10}{21}\hspace{0.33em}{=}\hspace{0.33em}\frac{{10}\hspace{0.33em}\times\hspace{0.33em}{8}}{{21}\hspace{0.33em}\times\hspace{0.33em}{8}}\hspace{0.33em}{=}\hspace{0.33em}\frac{80}{168}}\\ {\frac{9}{24}\hspace{0.33em}{=}\hspace{0.33em}\frac{{9}\hspace{0.33em}\times\hspace{0.33em}{7}}{{24}\hspace{0.33em}\times\hspace{0.33em}{7}}\hspace{0.33em}{=}\hspace{0.33em}\frac{63}{168}} \end{array}$

So now,

$\frac{10}{21}\hspace{0.33em}{+}\hspace{0.33em}\frac{9}{24}\hspace{0.33em}{=}\hspace{0.33em}\frac{80}{168}\hspace{0.33em}{+}\hspace{0.33em}\frac{63}{168}\hspace{0.33em}{=}\hspace{0.33em}\frac{143}{168}$

So it’s not immediately obvious that there may be some simplification required. In this case, this fraction cannot be simplified but 143 is not a prime number. Can you find the factors of 143?

In my next post, we will continue with fractions (catching up to Rocky) and extend the knowledge we have so far.

So now that we know how to get the least common multiple (LCM) of two numbers, let’s apply this knowledge to add two fractions together that have different denominators.

So let’s look at:

$\frac{3}{4}\hspace{0.33em}{+}\hspace{0.33em}\frac{3}{18}$

In my last post, we found the LCM of the two numbers 4 and 18 to be 36. Now that these numbers are in the denominator, we can also call 36 the least common denominator (LCD) – two different terms to describe the same thing depending on the country, school district, or era you find yourself in.

From Part 2 of the posts on Fractions, I covered the method to add two fractions together that have different denominators. The trick is to find a common denominator but now we are more wise and can find a least common denominator. So we will use the LCM we found in my last post to solve the problem.

So I want to convert the two fractions into equivalent ones that have a denominator of 36. I do this my multiplying the top and bottom of the fraction by the same number needed to make the denominator 36:

$\begin{array}{c} {\frac{3}{4}\hspace{0.33em}\times\hspace{0.33em}\frac{9}{9}\hspace{0.33em}{=}\hspace{0.33em}\frac{{3}\hspace{0.33em}\times\hspace{0.33em}{9}}{{4}\hspace{0.33em}\times\hspace{0.33em}{9}}\hspace{0.33em}{=}\hspace{0.33em}\frac{27}{36}}\\ {\frac{3}{18}\hspace{0.33em}\times\hspace{0.33em}\frac{2}{2}\hspace{0.33em}{=}\hspace{0.33em}\frac{{3}\hspace{0.33em}\times\hspace{0.33em}{2}}{{18}\hspace{0.33em}\times\hspace{0.33em}{2}}\hspace{0.33em}{=}\hspace{0.33em}\frac{6}{36}} \end{array}$

Now that we have two equivalent fractions with the same denominator, the rest is easy as we just have to add the numerators together:

$\frac{3}{4}\hspace{0.33em}{+}\hspace{0.33em}\frac{3}{18}\hspace{0.33em}{=}\hspace{0.33em}\frac{27}{36}\hspace{0.33em}{+}\hspace{0.33em}\frac{6}{36}\hspace{0.33em}{=}\hspace{0.33em}\frac{33}{36}$

So are we done? That is the answer but textbooks and teachers will always tell you to “simplify” your answer. That means get the numbers as small as possible. The method of getting equivalent fractions above can be done in reverse to get simpler (but equivalent) fractions. If you can identify factors common to the numerator and denominator, these can be cancelled. Notice in our answer that there is a common factor of 3. That is:

$\frac{33}{36}\hspace{0.33em}{=}\hspace{0.33em}\frac{{11}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{3}}{{12}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{11}{12}$

So now we are done. In my next post I will do more examples.

## Factoring

Factors are things multiplied together. So numbers can have many different factors. For example, 12 can be seen as 6 × 2, or 3 × 4, or even 2 × 2 × 3. Notice that in the first two sets of factors, the 6 and the 4 can be further broken down to be 2 × 3 and 2 × 2, but the only factors of 2 are 2 × 1 and for 3, 3 × 1. (I am limiting myself to integer factors here because if other types of numbers are allowed, a number has an infinite number of factors).

So numbers like 2 and 3 that only have themselves and 1 as factors are called prime numbers. Other prime numbers are 5, 7, 11, 13, 17, 19, 23. Notice that the only even prime number is 2 as all other even numbers have 2 as a possible factor.

The process of finding the Lowest Common Denominator (LCD) of two or more fractions is done by factoring the denominators into its prime factors. This post is about factoring. I will then use this new skill to find the LCD in my next post.

Now factoring is easier if you know the times tables, commonly up to and including the 12 times tables. When a textbook says to factor a number, it usually means to get the number down to factors of just prime numbers like we did for 12 = 2 × 2 × 3.

Let’s try some:

9 = 3 × 3,  16 = 4 × 4 = 2 × 2 × 2 × 2,  20 = 4 × 5 = 2 × 2 × 5

So generally you just find any two factors of a number then break those factors down further until all that is left are prime numbers. When asked to do this by hand, the numbers are generally small, say less than 144 (12 × 12). Bigger numbers take more work and are usually done by computer programs. By the way, now that you know the factors of 12, you should be able to immediately write down the factors of 144:

144 = 12 × 12 = 2 × 2 × 3 × 2 × 2 × 3

Some more examples:

35 = 5 × 7,   40 = 4 × 10 = 2 × 2 × 2 × 5

100 = 10 × 10 = 2 × 5 × 2 × 5,  60 = 5 × 12 = 5 × 2 × 2 × 3

To find the LCD of two or more denominators, the first step will be to factor each denominator. I will do this in the next post.

$\frac{1}{3}\hspace{0.33em}{+}\hspace{0.33em}\frac{1}{4}$

You cannot add these directly because the sizes of the pieces of the whole are different sizes. The $\frac{1}{3}$ represents 1 piece (the numerator) of something divided into 3 pieces (the denominator) while the $\frac{1}{4}$ represents 1 piece of the same something divided into 4 pieces. It would be like adding 1 meter to 1 centimetre and getting 2 meters or centimetres? Either unit would be a wrong answer. You have to first convert one or both of the units to the same unit so that you can then add the numbers together. The same thing has to happen with fractions with different denominators: you got to find equivalent fractions where the size of the pieces (the denominators) are the same size.

In my last post, you saw that I can multiply top and bottom of a fraction by the same number to get an equivalent fraction. This is true because you are effectively multiplying by 1. It will be this skill that will help us add the two fractions.

Looking at the denominators 3 and 4, I want to find numbers to multiply them by to get the same number. If I multiply the 3 by 4 and the 4 by 3, I will get 12. This will be the new denominator that I want to use to convert each of the fractions.

So first, let’s convert the $\frac{1}{3}$ to an equivalent fraction with 12 in the denominator. To get the 12 in the denominator, I need to multiply the 3 by 4. But I cannot just do this in the bottom, I must also multiply the numerator by 4 as well to get an equivalent fraction:

$\frac{{1}\hspace{0.33em}\times\hspace{0.33em}{4}}{{3}\hspace{0.33em}\times\hspace{0.33em}{4}}\hspace{0.33em}{=}\hspace{0.33em}\frac{4}{12}$

So $\frac{4}{12}$ is the same fraction as $\frac{1}{3}$ only it now represents 4 pieces of something divided into 12 pieces. Let’s do the same with the $\frac{1}{4}$.

If I multiply the 4 by 3, I will get 12, but again, this must also be done in the numerator:

$\frac{{1}\hspace{0.33em}\times\hspace{0.33em}{3}}{{4}\hspace{0.33em}\times\hspace{0.33em}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{3}{12}$

So now we have the original fractions converted to fractions with the same size pieces. We can now add them directly:

$\frac{1}{3}\hspace{0.33em}{+}\hspace{0.33em}\frac{1}{4}\hspace{0.33em}{=}\hspace{0.33em}\frac{4}{12}\hspace{0.33em}{+}\hspace{0.33em}\frac{3}{12}\hspace{0.33em}{=}\hspace{0.33em}\frac{7}{12}$

So that’s how fractions with different denominators are added together:

1. Find a common denominator
2. Convert each fraction by multiplying each denominator by the number needed to get the common denominator and multiply the corresponding numerator by the same number, then
3. Add the new equivalent fractions together by just adding the numerators.

Now for this problem, you found the common denominator by just multiplying the two denominators 3 and 4 together. This method will always work but you may end up with an equivalent denominator bigger than it has to be. In order to work with smaller numbers and reduce the work to simplify the resulting answer (that is get rid of common factors), we want to find what is called the least common denominator (LCD). This is also called the least common multiple (LCM). I will cover the way to find the LCD in my next post.

Before I begin with the main topic, let me review how fractions are multiplied together. If you remember, this is relatively easy as all you do is multiply the numerators together and the denominators together:

$\frac{3}{4}\hspace{0.33em}\times\hspace{0.33em}\frac{2}{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{{3}\hspace{0.33em}\times\hspace{0.33em}{2}}{{4}\hspace{0.33em}\times\hspace{0.33em}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{6}{12}$

Now let’s look at this problem a different way. Now multiplication is commutative which means the order of the things multiplied together does not matter. So we could develop the problem thus:

$\frac{3}{4}\hspace{0.33em}\times\hspace{0.33em}\frac{2}{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{{3}\hspace{0.33em}\times\hspace{0.33em}{2}}{{4}\hspace{0.33em}\times\hspace{0.33em}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{{2}\hspace{0.33em}\times\hspace{0.33em}{3}}{{4}\hspace{0.33em}\times\hspace{0.33em}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{4}\hspace{0.33em}\times\hspace{0.33em}\frac{3}{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{4}\hspace{0.33em}\times\hspace{0.33em}{1}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{4}$

Now that is a long line of work but this shows two things. One, you can create the same equivalent fraction by multiplying the numerator and the denominator by the same number:

$\frac{2}{4}\hspace{0.33em}\times\hspace{0.33em}{1}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{4}\hspace{0.33em}\times\hspace{0.33em}\frac{3}{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{6}{12}$

But you can also do the reverse – if there is a common factor in the numerator and the denominator, you can effectively cancel them, replacing them with an implied “1”:

$\frac{6}{12}\hspace{0.33em}{=}\hspace{0.33em}\frac{{2}\hspace{0.33em}\times\hspace{0.33em}{3}}{{4}\hspace{0.33em}\times\hspace{0.33em}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{4}\hspace{0.33em}\times\hspace{0.33em}\frac{3}{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{4}\hspace{0.33em}\times\hspace{0.33em}{1}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{4}$

So if you see this common factor, you can simply write:

$\frac{6}{12}\hspace{0.33em}{=}\hspace{0.33em}\frac{{2}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{3}}{{4}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{4}$

But notice there is another common factor:

$\frac{2}{4}\hspace{0.33em}{=}\hspace{0.33em}\frac{{1}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{2}}{{2}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{2}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{2}$

So $\frac{2}{4}{,}\hspace{0.33em}\frac{6}{12}{,}\hspace{0.33em}{\mathrm{and}}\hspace{0.33em}\frac{1}{2}$ are all equivalent fractions but $\frac{1}{2}$ is the one with the smallest numbers. These “smallest numbers” are found using things called prime numbers and I will talk about these later.

With this knowledge, let’s go back to the original problem and get to the final answer of $\frac{1}{2}$ in far fewer steps:

$\frac{3}{4}\hspace{0.33em}\times\hspace{0.33em}\frac{2}{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{\rlap{/}{3}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{2}}{\rlap{/}{2}\hspace{0.33em}\times\hspace{0.33em}{2}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{2}$

Remember that the cancelled factors are being replaced by “1”s. We can leave them out of the denominator since there is another number there but we must keep one of the “1”s in the numerator since there is no other number there.

We will use this skill of finding the factors of numbers in my next post. This will lead up to how to add fractions with different denominators together.