### Add/Subtract fractions

Adding and subtracting fractions is one of the most difficult areas of arithmetic for most students to get their heads around.

The reason why it is so hard, is that the structure of the numbers are so completely different from all the other numbers you have met up until this point. But do not despair.

It is worth reminding ourselves, at this point, to think about the meaning of the words **Numerator** and **Denominator**.

Numerator is derived from the Latin word **numerus** meaning a number.

Denominator is derived from the Latin word **nomine** meaning a name.

### Worked example 1: Adding fractions whose denominators are the same.

- Notes
- $${1\over3} + {1\over3}$$
- $$ \begin{aligned}{{3 \over 7} + {2\over 7}} &= \\&= {{3 + 2} \over 7}\\&= {5 \over 7}\end{aligned}$$
When you think about the problem here, it can be helpful to view the numerator and denominator as follows

**The numerator is the number of things we have**and**The denominator is the name of the things we are talking about**.When considering the nature of fractions, try not to think of denominators as numbers at all; that way you won't be tempted to add the denominators together, which is a

**very**common error.So we have 3 of these

**seventh things**and another 2 of these**seventh things**. That will make up a total of**5**of these**seventh things**. - $${5\over7}$$

### Worked example 2: Adding fractions where one denominator is a factor of another.

- Notes
- $${2\over9} + {5\over9}$$
- $$ \begin{aligned}{{3 \over 8} + {3\over 4}} &= \\&= {{3 + 6} \over 8}\\&= {9 \over 8}\\&= 1{1\over 8}\end{aligned}$$
The two fractions have denominators of 8 and 4. Notice that 8 is a multiple of 4, so 8 will be the lowest common multiple (LCM) of 8 and 4.

It is a common stylistic error to use an LCM of 32 (8 × 4). While this will still give you the right answer, if your working is correct, it opens your work up to errors as the arithmetic becomes harder. It is worth remembering that in all mathematics, you should not simply look for a method that works, but rather, you should try to find the best, simplest method. There are no marks to be gained by making life difficult for yourself.

- $$7\over9$$
If you end up with a top-heavy fraction, it is fine to leave it in that form. Alternatively, you may change it to a mixed fraction.

### Worked example 3: Adding fractions where the denominators which share no factors in common.

- Notes
- $${3\over5} + {1\over5}$$
- $$ \begin{aligned}{{2 \over 5} + {1\over 3}} &= \\&= {{6 + 5} \over 15}\\&= {11 \over 15}\end{aligned}$$
The two fractions have denominators of 5 and 3. This means that they cannot be added together immediately as they are different things. For example 2 elephants and 1 litre of oxygen, cannot meaningfully be added together. It is hard to see what it is that you might have 3 of. Just as adding a time to a speed or a mass to an acceleration, the answer would be meaningless.

However, unlike with elephants and air, we are able to manipulate fractions using the principles of equivalent fractions ${2\over5}\times{3\over3}$ to become ${6\over15}$ by multiplying both parts (numerator and denominator) by 3, and ${1\over3}\times{5\over5}$ to become ${5\over15}$ by multiplying both parts by 5.

Another method which some find useful is as follows:

- Find the LCM.
- Divide the LCM by each denominator and multiply the result by the associated numerator
- Then add the new numerators together.

So in this example you would do for the ${2\over5}$ fraction: 5 into 15 goes 3 and 3 × 2 = 6 and for the ${1\over3}$ fraction: 3 into 15 goes 5 and 5 × 1 = 5.

- $$4\over5$$

### Worked example 4: Adding fractions where the denominators share a common factor.

- Notes
- $${5 \over 6} + {2 \over 15}$$
- $$ \begin{aligned}{{5 \over 6} + {2\over 15}} &= \\&= {{25 + 4} \over 30}\\&= {29 \over 30}\end{aligned}$$
The two fractions have denominators of 6 and 15. Notice that both 6 and 15 are multiples of 3, so the LCM will be 30 in this instance.

It is a common stylistic error to use an LCM of 32 (8 × 4). While this will still give you the right answer, if your working is correct, it opens your work up to errors as the arithmetic becomes harder. It is worth remembering that in all mathematics, you should not simply look for a method that works, but rather, you should try to find the best, simplest method. There are no marks to be gained by making life difficult for yourself.

- $${29 \over 30}$$

### Worked example 5: Subtracting fractions whose denominators are the same

- Notes
- $$ {3 \over 7} - {2\over 7} = $$
- $$ \begin{aligned}{{3 \over 7} - {2\over 7}} &= \\&= {{3 - 2} \over 7}\\&= {1 \over 7}\end{aligned}$$
When you think about the problem here, it can be helpful to view the numerator and denominator as follows

**The numerator is the number of things we have**and**The denominator is the name of the things we are talking about**.When considering the nature of fractions, try not to think of denominators as numbers at all; that way you won't be tempted to add the denominators together, which is a

**very**common error.So we have 3 of these

**seventh things**and we take off 2 of these**seventh things**. That will leave us with**1**of these**seventh things**. - $${1 \over 7}$$

### Worked example 6:

- Notes

### Worked example 7: Decimal eg

- Notes
- undefined
- $$ \begin{aligned}{{3 \over 4} - {3\over 8}} &= \\&= {{6 - 3} \over 8}\\&= {3 \over 15}\end{aligned}$$
The two fractions have denominators of 8 and 4. Notice that 8 is a multiple of 4, so 8 will be the lowest common multiple (LCM) of 8 and 4.

It is a common stylistic error to use an LCM of 32 (8 × 4). While this can still give you the right answer, if your working is correct, it opens your work up to errors as the arithmetic becomes harder. It is worth remembering that in all mathematics, you should not simply look for a method that works, but rather, you should try to find the best and simplest method. There are no marks to be gained by making life difficult for yourself.

- $${3 \over 15}$$

### Worked example 8: Subtracting fractions where the denominators share a common factor.

- Notes
- $$ {5 \over 6} - {2\over 15} = $$
- $$ \begin{aligned}{{5 \over 6} - {2\over 15}} &= \\&= {{25 - 4} \over 30}\\&= {21 \over 30}\end{aligned}$$
The two fractions have denominators of 6 and 15. Notice that both 6 and 15 are multiples of 3, so the LCM will be 30 in this instance.

It is a common stylistic error to use an LCM of 32 (8 × 4). While this will still give you the right answer, if your working is correct, it opens your work up to errors as the arithmetic becomes harder. It is worth remembering that in all mathematics, you should not simply look for a method that works, but rather, you should try to find the best, simplest method. There are no marks to be gained by making life difficult for yourself.

- $${21 \over 30}$$

### Exercise 1: Adding: denominators the same

**Instructions:**Add together the following fractions, giving your answers as simply as possible.

- Q1$${1\over3} + {1\over3}$$$$2\over3$$
- Q2$${3\over5} + {1\over5}$$$$4\over5$$
- Q3$${2\over9} + {5\over9}$$$$7\over9$$
- Q4$${2\over5} + {2\over5}$$$$4\over5$$
- Q5$${1\over9} + {4\over9}$$$$5\over9$$

### Exercise 2: Adding: denominators have no common factors

**Instructions:**Add together the following fractions, giving your answers as simply as possible.

- Q1$${3 \over 8} + {2 \over 9}$$$${43 \over 72}$$
- Q2$${1\over3} + {1\over 2}$$$$5\over6$$
- Q3$${2\over5} + {2\over3}$$$${16 \over 15} = 1{1 \over 15}$$
- Q4$${1\over5} + {1\over2}$$$${7 \over 10}$$
- Q5$${2\over5} + {1\over3}$$$${11 \over 15}$$

### Exercise 3: Adding: one denominator is a factor of the other

**Instructions:**Add together the following fractions, giving your answers as simply as possible.

- Q1$${1 \over 6} + {1 \over 4}$$$${5 \over 12}$$
- Q2$${1 \over 2} + {1 \over 4}$$$${3 \over 4}$$
- Q3$${2 \over 3} + {1 \over 6}$$$${5 \over 6}$$
- Q4$${3 \over 10} + {3 \over 20}$$$${9 \over 20}$$
- Q5$${3 \over 4} + {1 \over 8}$$$${7 \over 8}$$

### Exercise 4: Adding fractions where the denominators share a common factor.

**Instructions:**Add together the following fractions, giving your answers as simply as possible.

- Q1$${1 \over 10} + {2 \over 15}$$$${7 \over 30}$$
- Q2$${2 \over 9} + {4 \over 15}$$$${22 \over 45}$$
- Q3$${3 \over 4} + {5 \over 6}$$$${19 \over 12} = 1{7 \over 12}$$
- Q4$${5 \over 6} + {7 \over 9}$$$${29 \over 18} = 1{11 \over 18}$$
- Q5$${5 \over 12} + {5 \over 8}$$$${25 \over 24} = 1{1 \over 24}$$

### Exercise 5: Subtracting: denominators the same

- Q1$${3 \over 5} - {1 \over 5}$$$${2 \over 5}$$
- Q2$${5 \over 9} - {2 \over 9}$$$${3 \over 9} = {1 \over 3}$$
- Q3$${4 \over 11} - {3 \over 11}$$$${1 \over 11}$$
- Q4$${4 \over 15} - {1 \over 15}$$$${3 \over 15} = {1 \over 3}$$
- Q5$${4 \over 15} - {2 \over 15}$$$${2 \over 15}$$

### Exercise 6: Subtracting: denominators have no common factors

**Instructions:**Subtract the following fractions, giving your answers as simply as possible.

- Q1$${1 \over 2} - {1 \over 3}$$$${1 \over 6}$$
- Q2$${2 \over 3} - {2 \over 5}$$$${1 \over 15}$$
- Q3$${3 \over 5} - {1 \over 6}$$$${13 \over 30}$$
- Q4$${2 \over 3} - {4 \over 11}$$$${10 \over 33}$$
- Q5$${3 \over 5} - {3 \over 8}$$$${9 \over 40}$$

### Exercise 7: Subtracting: one denominator is a factor of the other

**Instructions:**Subtract the following fractions, giving your answers as simply as possible.

- Q1$${1 \over 2} - {1 \over 4}$$$${1 \over 4}$$
- Q2$${1 \over 2} - {1 \over 6}$$$${1 \over 3}$$
- Q3$${1 \over 5} - {1 \over 10}$$$${1 \over 10}$$
- Q4$${2 \over 3} - {4 \over 9}$$$${2 \over 9}$$
- Q5$${5 \over 6} - {7 \over 12}$$$${1 \over 4}$$

### Exercise 8: Subtracting: denominators have a factor in common

**Instructions:**Subtract the following fractions, giving your answers as simply as possible.

- Q1$${1 \over 6} - {1 \over 8}$$$${1 \over 24}$$
- Q2$${2 \over 9} - {1 \over 12}$$$${5 \over 36}$$
- Q3$${5 \over 6} - {1 \over 10}$$$${11 \over 15}$$
- Q4$${3 \over 4} - {3 \over 10}$$$${9 \over 20}$$
- Q5$${5 \over 12} - {2 \over 9}$$$${7 \over 36}$$

### Exercise 9: Mixed exercise (examples from all of the above)

**Instructions:**Do the following calculations, giving your answers as simply as possible.

- Q1$${22 \over 47} + {5 \over 47}$$$${27 \over 47}$$
- Q2$${6 \over 11} - {2 \over 11}$$$${4 \over 11}$$
- Q3$${1 \over 29} + {22 \over 29}$$$${23 \over 29}$$
- Q4$${3 \over 7} - {2 \over 7}$$$${1 \over 7}$$
- Q5$${5 \over 6} + {1 \over 7}$$$${41 \over 42}$$
- Q6$${7 \over 10} - {4 \over 9}$$$${23 \over 90}$$
- Q7$${4 \over 5} + {1 \over 9}$$$${41 \over 45}$$
- Q8$${1 \over 2} - {1 \over 5}$$$${3 \over 10}$$
- Q9$${1 \over 6} + {5 \over 7}$$$${37 \over 42}$$
- Q10$${1 \over 5} + {1 \over 2}$$$${7 \over 10}$$