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?
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.
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.