Addition and Subtraction

Lesson

One way to get better at mathematics is to learn strategies that help you to work more efficiently, or strategies that make it easier. We are going to look at a few strategies to help you with addition and subtraction.

Commutative is a rather long word for a simple idea:

If we add together two numbers, for example, $5+28$5+28, this is the same as $28+5$28+5. That is, $5+28=28+5$5+28=28+5.

How does this make addition more efficient, or easier?

One strategy for adding two numbers together is to start with the first number, then count up from there. For $5+28$5+28, we would start with $5$5, then count up $28$28 numbers.

However, if we use the Commutative Law, we can instead think of it as $28+5$28+5, which means we start at $28$28, then count up $5$5.

Would you rather count up $28$28, or only count up $5$5?

Watch out!

The Commutative Law does not work for subtraction.

Use the commutative property of addition to fill in the missing number.

$19+15=15$19+15=15$+$+$\editable{}$

Whenever working with addition and subtraction (and multiplication and division), if we see brackets, we need to work on those first, before we solve those problems.

For example, $13-\left(2+3\right)$13−(2+3). We begin by working out $2+3$2+3, which is $5$5, and then we solve $13-5$13−5, which is $8$8. Writing it all out, we get:

$13-\left(2+3\right)$13−(2+3) | $=$= | $13-5$13−5 |

$=$= | $8$8 |

Simplify the expression by filling in the missing number.

$10-\left(2+1\right)$10−(2+1) is the same as $10$10$-$−$\editable{}$

Consider a question like this: $\left(23+3\right)+7$(23+3)+7

We could simply work out the $23+3$23+3, which gives us $26$26, and then add $7$7, which gives us $33$33.

But, did you notice something about $3+7$3+7? $3+7=10$3+7=10, and it is very easy to add $10$10 to a number.

Are you allowed to add $3+7$3+7 in this question? The $3$3 is inside the brackets, and the $7$7 is outside the brackets, so how can we add them?

There is a nice little rule, when there is no negative (minus) sign in front of the brackets. We can simply remove them. This means that $\left(23+3\right)+7$(23+3)+7 is the same as $23+3+7$23+3+7. That is: $\left(23+3\right)+7=23+3+7$(23+3)+7=23+3+7.

As we can remove the brackets, we can group the numbers however we like, and so yes, we can group $3+7$3+7 to give $10$10, and add $23+10=33$23+10=33.

Remember!

We can add numbers in any order.

Simplify and solve:

$\left(12+4\right)+6$(12+4)+6

Firstly, rewrite the expression without brackets:

$\left(12+4\right)+6=12$(12+4)+6=12$+$+$\editable{}$$+$+$6$6

Now solve.

Generalise the properties of addition and subtraction with whole numbers