Free 8th Grade Math Worksheets—Printable w/ Answers — Mashup Math - Free Printable

Let’s go through each problem step by step.

The key idea here is the Associative Property of Multiplication. It means that when you multiply three or more numbers, it doesn’t matter how you group them — the answer stays the same.

For example:
(2 × 3) × 4 = 2 × (3 × 4) → both equal 24.

In this worksheet, we’re using that rule to fill in missing numbers so both sides are equal — even though the grouping changes.

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Problem 1:
(__ × 7) × 5 = __ × (3 × 5)

We want both sides to be equal. On the right side, we have “3 × 5” grouped together. So on the left, we need to match that structure.

Left side: (__ × 7) × 5
Right side: __ × (3 × 5)

To make them equal, the first blank on the left should be 3, and the blank on the right should be 7.

Check:
(3 × 7) × 5 = 21 × 5 = 105
7 × (3 × 5) = 7 × 15 = 105 ✔

Answer: 3, 7

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Problem 2:
(6 × 2) × __ = (__ × 4) × 6

Left: (6 × 2) × ?
Right: (? × 4) × 6

We can rearrange to match groups. Let’s look at what’s already there.

On the left, 6 and 2 are multiplied first. On the right, ? and 4 are multiplied first, then times 6.

So if we set the second blank to 2, then right becomes (2 × 4) × 6 = 8 × 6 = 48
Then left must also be 48 → (6 × 2) × ? = 12 × ? = 48 → ? = 4

Wait — let’s try matching positions.

Actually, think of it as swapping which number is grouped with which.

Original: (6 × 2) × ___ = ___ × (4 × 6)? No — wait, right side is (__ × 4) × 6

Better approach: Use associative property directly.

We know: (a × b) × c = a × (b × c)

But here, the order of numbers might change too? Actually, no — multiplication is commutative too, but the worksheet is focusing on associative.

Looking again:

Left: (6 × 2) × X
Right: Y × (4 × 6)

Wait — actually, looking at the original image text:

It says:
2.) (6 x 2) x __ = (__ x 4) x 6

Ah! Right side ends with “x 6”, not “x (4 x 6)” — I misread earlier.

So:
Left: (6 × 2) × A
Right: B × 4 × 6 → which is same as (B × 4) × 6

So: (6 × 2) × A = (B × 4) × 6

Compute left without A: 6×2=12 → 12 × A
Right: B×4×6 = 24 × B

Set equal: 12A = 24B → A = 2B

Now pick small integers. Try B=3 → A=6

Check:
Left: (6×2)×6 = 12×6=72
Right: (3×4)×6 = 12×6=72 ✔

Is there a simpler way? Maybe they expect matching numbers.

Notice: Left has 6,2,A; Right has B,4,6

If we set A=4 and B=2:

Left: (6×2)×4 = 12×4=48
Right: (2×4)×6 = 8×6=48 ✔

Yes! That works better — smaller numbers.

So blanks: 4, 2

Answer: 4, 2

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Problem 3:
(__ × 9) × 5 = (__ × 5) × 3

Left: (A × 9) × 5
Right: (B × 5) × 3

Left total: A × 9 × 5 = 45A
Right total: B × 5 × 3 = 15B

Set equal: 45A = 15B → 3A = B

Try A=1 → B=3

Check:
Left: (1×9)×5 = 9×5=45
Right: (3×5)×3 = 15×3=45 ✔

Perfect.

Answer: 1, 3

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Problem 4:
7 × (__ × 10) = __ × (7 × 15)

Left: 7 × (A × 10) = 7 × A × 10 = 70A
Right: B × (7 × 15) = B × 105 = 105B

Set equal: 70A = 105B → divide both by 35: 2A = 3B

Try B=2 → A=3

Check:
Left: 7 × (3 × 10) = 7 × 30 = 210
Right: 2 × (7 × 15) = 2 × 105 = 210 ✔

Answer: 3, 2

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Problem 5:
(__ × 17) × 21 = 21 × (__ × 12)

Left: (A × 17) × 21 = A × 17 × 21
Right: 21 × (B × 12) = 21 × B × 12

Since multiplication is commutative, we can ignore order for equality.

So: A × 17 × 21 = 21 × B × 12

Cancel 21 from both sides: A × 17 = B × 12

Find A and B such that 17A = 12B

Smallest whole numbers: LCM of 17 and 12 is 204

So A = 12, B = 17

Check:
Left: (12 × 17) × 21 = 204 × 21
Right: 21 × (17 × 12) = 21 × 204 → same ✔

Answer: 12, 17

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Problem 6:
__ × (26 × 14) = 14 × (__ × 8)

Left: A × (26 × 14) = A × 364
Right: 14 × (B × 8) = 14 × 8B = 112B

Set equal: 364A = 112B

Simplify: Divide both by 28 → 13A = 4B

Try B=13 → A=4

Check:
Left: 4 × (26 × 14) = 4 × 364 = 1456
Right: 14 × (13 × 8) = 14 × 104 = 1456 ✔

Answer: 4, 13

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Problem 7:
(29 × __) × 5 = 29 × (__ × 33)

Left: (29 × A) × 5 = 29 × A × 5
Right: 29 × (B × 33) = 29 × B × 33

Cancel 29 from both sides: A × 5 = B × 33

So 5A = 33B

Smallest integers: A=33, B=5

Check:
Left: (29 × 33) × 5 = 957 × 5 = 4785
Right: 29 × (5 × 33) = 29 × 165 = 4785 ✔

Answer: 33, 5

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Problem 8:
(__ × 6) × 5 = (__ × 4) × 5

Both sides end with ×5, so we can cancel 5.

Left: A × 6
Right: B × 4

So 6A = 4B → 3A = 2B

Try A=2, B=3

Check:
Left: (2×6)×5 = 12×5=60
Right: (3×4)×5 = 12×5=60 ✔

Answer: 2, 3

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Problem 9:
(__ × 45) × 78 = (__ × 27) × 78

Again, both sides ×78 → cancel 78.

Left: A × 45
Right: B × 27

So 45A = 27B → divide by 9: 5A = 3B

Try A=3, B=5

Check:
Left: (3×45)×78 = 135×78
Right: (5×27)×78 = 135×78 ✔

Answer: 3, 5

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Problem 10:
__ × (4 × 22) = (4 × 25) × __

Left: A × (4 × 22) = A × 88
Right: (4 × 25) × B = 100 × B

Set equal: 88A = 100B → simplify: divide by 4 → 22A = 25B

Smallest integers: A=25, B=22

Check:
Left: 25 × (4 × 22) = 25 × 88 = 2200
Right: (4 × 25) × 22 = 100 × 22 = 2200 ✔

Answer: 25, 22

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Final Answers:

1. 3, 7
2. 4, 2
3. 1, 3
4. 3, 2
5. 12, 17
6. 4, 13
7. 33, 5
8. 2, 3
9. 3, 5
10. 25, 22

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Final Answer:
1. 3, 7
2. 4, 2
3. 1, 3
4. 3, 2
5. 12, 17
6. 4, 13
7. 33, 5
8. 2, 3
9. 3, 5
10. 25, 22