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.

## Least Common Multiple (or Denominator)

So let’s find the least common multiple (LCM) of 4 and 18. I am now calling the LCD a least common multiple because these numbers are not in a denominator (yet), but it’s the same thing. It’s called a “multiple” because what we are finding is the smallest number that the given numbers both divide into.

So the first step is to factor each of the numbers into its prime factors:

4 = 2 × 2

18 = 2 × 9 = 2 × 3 × 3

The next step to get the LCM (or LCD if the numbers are in a denominator), is to multiply all the prime numbers in the factorisation above, but only the maximum times each appears. So, there are only two factors in the above breakdown, 2 and 3. The “2” appears twice in the “4” but only once in the “18”. The “3” appears twice in the “18. So I will use the “2” only twice and the “3” twice:

LCM = 2 × 2 × 3 × 3 = 4 × 9 = 36

So 36 is the smallest number that 4 and 18 can divide into. Now 4 × 18 = 72 is also a number that they will divide into but 36 is the smallest.

I will use this result in my next post but let’s see some more examples.

What is the LCM of 9 and 24?

9 = 3 × 3

24 = 2 × 2 × 2 × 3

So again only “2” and “3” are present. I will use the “3” only twice in calculating the LCM:

LCM =  2 × 2 × 2 × 3 × 3 = 72

Let’s try finding the LCM of 5 and 15:

5 is already a prime number

15 = 3 × 5

So the LCM = 3 × 5 = 15

Notice that sometimes the LCM is one of the given numbers. Always check to see if one of the numbers divides into the other one. If it does, the bigger number is the LCM.

Let’s try another one. What’s the LCM for 16 and 24:

16 = 2 × 2 × 2 × 2

24 = 2 × 2 × 2 × 3

LCM = 2 × 2 × 2 × 2 × 3 = 48

12 = 2 × 2 × 3

15 = 3 × 5

LCM = 2 × 2 × 3 × 5 = 60

## 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.