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Year 7 Maths Algebra and Equations

Law of Arithmetic

20 practice questions 2 video lessons Theory + worked examples

Law of Arithmetic looks at two rules that let you rearrange a sum or a product without changing the answer. In NSW Year 7 maths (NESA outcome MA4‑ALG‑C‑01) these are the commutative law, where the order does not matter, and the associative law, where the grouping does not matter.

You will swap and regroup numbers to spot friendly pairs such as \(25 \times 4 = 100\) or \(38 + 2 = 40\), split \(50\) into \(5 \times 10\) to multiply quickly, and see that the laws hold for negative numbers too. It is the trick behind adding up a shopping bill in your head. This subtopic is part of Year 7 Algebra and Equations.

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Practice questions

Every question with a fully worked solution.

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  • Law of Arithmetic - Video - Commutative law of addition Watch
  • Law of Arithmetic - Video - Associative law of addition Watch
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Theory

You can change the order (commutative law) and the grouping (associative law) of the numbers in a sum or a product, and the answer stays the same. This Year 7 maths page shows both laws, how to use friendly pairs and multiples of ten to work faster, and why the laws do not work for subtraction and division.

The commutative law says the order does not matter when you add or multiply: \(3 + 5 = 5 + 3\), so \(a + b = b + a\); and \(3 \times 4 = 4 \times 3\), so \(a \times b = b \times a\) (or \(ab = ba\)).

The associative law says the grouping does not matter when you add or multiply three or more numbers. Brackets show which pair is worked out first: \((3 + 5) + 2 = 3 + (5 + 2)\) and \((3 \times 4) \times 2 = 3 \times (4 \times 2)\). So \((a + b) + c = a + (b + c)\) and \((a \times b) \times c = a \times (b \times c)\).

The pronumerals \(a\), \(b\) and \(c\) (letters that stand for numbers) can be any numbers, so the laws work for integers too: \(-8 + 15 = 15 + (-8) = 15 - 8 = 7\).

Both laws work for adding and multiplying only.

Array of 3 rows of 4 Twelve dots in 3 rows of 4.
\(3\) rows of \(4\): \(3 \times 4 = 12\).
Array of 4 rows of 3 Twelve dots in 4 rows of 3: the same array turned.
Turn the array: \(4\) rows of \(3\) is still \(12\) dots, so \(3 \times 4 = 4 \times 3\).

Friendly pairs make a round number:

Friendly sumsFriendly products
\(38 + 2 = 40\)\(2 \times 5 = 10\)
\(56 + 4 = 60\)\(4 \times 25 = 100\)
\(62 + 8 = 70\)\(5 \times 8 = 40\)
\(73 + 27 = 100\)\(2 \times 50 = 100\)
Multiples of ten. Split into factors, then regroup: \(9 \times 50 = 9 \times 5 \times 10 = 45 \times 10 = 450\).
A sign moves with its number. In \(-9 + 25\) the \(9\) is negative, so swap to \(25 + (-9) = 25 - 9 = 16\).

How to use the laws to work faster

  1. Look for friendly pairs that make \(10\), \(50\), \(100\), ...
  2. Re-order the numbers so the pairs sit together (commutative law).
  3. Group each pair with brackets (associative law).
  4. Work out the brackets, then the rest.
Example 1 — Friendly pairs in a sum
Evaluate using the laws.
(i) \(43 + 8 + 7\)
(ii) \(-15 + 40\)
Solution
\(43 + 8 + 7\)\(=\)\((43 + 7) + 8\)
\(\)\(=\)\(50 + 8 = 58\)
\(-15 + 40\)\(=\)\(40 + (-15)\)
\(\)\(=\)\(40 - 15 = 25\)
43+8+7=(43+7)+8=58,−15+40=40−15=25

(i) \(58\)
(ii) \(25\)

Example 2 — Multiples of ten
Use the laws to evaluate
(i) \(8 \times 40\)
(ii) \(30 \times 70\)
Solution
\(8 \times 40\)\(=\)\((8 \times 4) \times 10\)
\(\)\(=\)\(32 \times 10 = 320\)
\(30 \times 70\)\(=\)\((3 \times 7) \times (10 \times 10)\)
\(\)\(=\)\(21 \times 100 = 2100\)
8×40=32×10=320,30×70=21×100=2100

(i) \(320\)
(ii) \(2100\)

Example 3 — Adding four numbers
Evaluate \(22 + 5 + 78 + 16\).
Solution

Re-order so the friendly pair sits together: \(22 + 5 + 78 + 16 = 22 + 78 + 5 + 16\). Then group the pairs.

\((22 + 78) + (5 + 16)\)\(=\)\(100 + 21\)
\(\)\(=\)\(121\)
22+5+78+16=(22+78)+(5+16)=100+21=121

The friendly pair is \(22 + 78 = 100\). The answer is \(121\).

Example 4 — A friendly product
Evaluate \(25 \times 17 \times 4\).
Solution
\(25 \times 17 \times 4\)\(=\)\(25 \times 4 \times 17\)
\(\)\(=\)\((25 \times 4) \times 17\)
\(\)\(=\)\(100 \times 17\)
\(\)\(=\)\(1700\)
25×17×4=(25×4)×17=100×17=1700

\(1700\)

Example 5 — Two friendly pairs
(i) A canteen has \(25\) trays of muffins. Each tray has \(4\) rows of \(9\). How many muffins is that?
(ii) Evaluate \(2 \times 13 \times 50 \times 3\).
Solution

(i) The friendly pair is \(25 \times 4 = 100\).
(ii) The friendly pair is \(2 \times 50 = 100\).

\((25 \times 4) \times 9\)\(=\)\(900\)
\((2 \times 50) \times (13 \times 3)\)\(=\)\(3900\)
25×4×9=900,(2×50)×(13×3)=100×39=3900

(i) \(900\) muffins
(ii) \(100 \times 39 = 3900\)

Example 6 — Not for subtraction
Show that the laws do not work for subtraction and division.
Solution
\(10 - 4\)\(=\)\(6\)
\(4 - 10\)\(=\)\(-6\)
\((20 - 8) - 2\)\(=\)\(10\)
\(20 - (8 - 2)\)\(=\)\(14\)
\(12 \div 2\)\(=\)\(6\)
10−4=6 but 4−10=−6,(20−8)−2=10 but 20−(8−2)=14

Changing the order or the grouping changes the answer. Also \(12 \div 2 = 6\) but \(2 \div 12 = \dfrac{1}{6}\).

Common pitfalls

Swapping in a subtraction. \(10 - 4 = 6\) but \(4 - 10 = -6\). Order matters for subtraction and division.
Regrouping a subtraction. \((20 - 8) - 2 = 10\), but \(20 - (8 - 2) = 14\).
Losing a negative sign. \(-9 + 25 = 25 + (-9) = 16\), not \(25 + 9\).
Forgetting a number. After re-ordering, check that every number from the question is still there.

Frequently asked questions

What is the commutative law?

The order does not matter when adding or multiplying: \(a + b = b + a\) and \(ab = ba\).

What is the associative law?

The grouping does not matter when adding or multiplying: \((a + b) + c = a + (b + c)\).

Does the commutative law work for subtraction?

No. \(10 - 4 = 6\) but \(4 - 10 = -6\).

What are friendly pairs?

Numbers that add or multiply to a round number, such as \(73 + 27 = 100\) or \(4 \times 25 = 100\).

How do you multiply by a multiple of ten in your head?

Split it into factors: \(9 \times 50 = 9 \times 5 \times 10 = 450\).