Free math worksheets

- The commutative property changes order, not grouping. - Look for the same two factors written in the opposite order. - Do not choose an equation that splits one factor into a sum.

1. The commutative property changes the order of the factors without changing the product. 2. Choice a) changes \(4 \times 7\) to \(7 \times 4\).

a) \(4 \times 7 = 7 \times 4\)

- Which multiplication facts have each product? - Remember that switching the order of unequal factors creates the related commutative fact. - Is either product also made by a square fact?

1. For product \(16\), the factor pairs within \(1\) through \(10\) are \(2\) and \(8\), and \(4\) and \(4\). Including both orders gives \(2 \times 8\), \(8 \times 2\), and \(4 \times 4\). 2. For product \(36\), the factor pairs within \(1\) through \(10\) are \(4\) and \(9\), and \(6\) and \(6\). Including both orders gives \(4 \times 9\), \(9 \times 4\), and \(6 \times 6\).

Product \(16\): \(2 \times 8\), \(8 \times 2\), \(4 \times 4\) Product \(36\): \(4 \times 9\), \(9 \times 4\), \(6 \times 6\)

- Find one multiplication fact for each product, then switch the factors. - Which multiplication facts have the given product? - Which two factors make the given product?

1. For \(14\), use factors \(2\) and \(7\): \(2 \times 7\) and \(7 \times 2\). 2. For \(45\), use factors \(5\) and \(9\): \(5 \times 9\) and \(9 \times 5\). 3. For \(28\), use factors \(4\) and \(7\): \(4 \times 7\) and \(7 \times 4\).

a) \(2 \times 7\) and \(7 \times 2\) b) \(5 \times 9\) and \(9 \times 5\) c) \(4 \times 7\) and \(7 \times 4\)

- Switch the positions of the two factors. - Then evaluate the product using either order. - Which order is easier for you to calculate?

1. Switch the order of the factors, then evaluate each product. 2. The completed equations are \(10 \times 4 = 4 \times 10 = 40\), \(2 \times 6 = 6 \times 2 = 12\), \(5 \times 7 = 7 \times 5 = 35\), and \(3 \times 8 = 8 \times 3 = 24\).

a) \(10 \times 4 = 4 \times 10 = 40\) b) \(2 \times 6 = 6 \times 2 = 12\) c) \(5 \times 7 = 7 \times 5 = 35\) d) \(3 \times 8 = 8 \times 3 = 24\)

- Do you need to evaluate both expressions in part a? - Does switching the order of factors change the product? - Which factor pairs have a product of \(12\)?

1. In part a, the factors are switched, so the products are equal: \(7 \times 4 = 4 \times 7\). 2. In part b, switching the factors does not change the product, so \(3 \times 9 = 27\). 3. One possible pair for part c is \(3 \times 4 = 12\) and \(4 \times 3 = 12\).

a) \(=\) b) \(3 \times 9 = 27\) c) For example, \(3 \times 4 = 12\) and \(4 \times 3 = 12\).

- Look for two addends that make a multiple of \(100\). - The commutative property lets you change the order of addends. - The associative property lets you change how the addends are grouped.

1. Use the commutative property to place \(327\) and \(473\) together: \(327 + 473 + 158\). 2. Use the associative property to regroup: \((327 + 473) + 158\). 3. Calculate: \(800 + 158 = 958\).

\((327 + 473) + 158 = 800 + 158 = 958\)

- Look for two factors whose product is \(10\). - The commutative property lets you change the order of factors. - The associative property lets you change how the factors are grouped.

1. Use the commutative property to reorder the factors: \(2 \times 5 \times 7\). 2. Use the associative property to regroup: \((2 \times 5) \times 7\). 3. Calculate: \(10 \times 7 = 70\).

\((2 \times 5) \times 7 = 10 \times 7 = 70\); commutative and associative properties

- Look for pairs of addends that make \(100\). - You may reorder and regroup addends without changing the sum. - Add the two partial sums.

1. Reorder and regroup the addends: \((64 + 36) + (19 + 81)\). 2. Calculate the partial sums: \(100 + 100 = 200\).

\(200\)

- Look for addends that combine to make \(100\) or another multiple of \(100\). - Reorder and regroup the addends to place convenient pairs together. - Remember to include any addend that is not part of a pair.

1. Reorder and regroup the addends: \((235 + 165) + (88 + 12) + 50\). 2. Calculate the partial sums: \(400 + 100 + 50 = 550\).

\(550\)

- Look for pairs that make multiples of \(100\). - Reorder and regroup the addends to place those pairs together. - Add the partial sums.

1. Reorder and regroup the addends: \((125 + 75) + (430 + 170)\). 2. Calculate the partial sums: \(200 + 600 = 800\).

\(800\)

- Look for two addends that make a multiple of \(100\). - Think about which property lets you change how addends are grouped. - Regroup before calculating.

1. Use the associative property to regroup: \((357 + 143) + 89\). 2. Calculate: \(500 + 89 = 589\).

\(589\); associative property

- First identify whether the order of the addends changes. - Then identify whether the grouping changes. - Match each change to the name of a property.

1. The student changes \(15 + 38 + 85\) to \(15 + 85 + 38\) by using the commutative property to reorder addends. 2. The student then uses the associative property to group \(15\) and \(85\) so they are added first.

The commutative property changes the order of \(38\) and \(85\); then the associative property groups \(15\) and \(85\) to be added first.

- Count the rows and the dots in each row. - Then read the same array by columns. - Does the total number of dots change when the array is turned?

1. The array has \(3\) rows with \(7\) dots in each row, so \(3 \times 7 = 21\). 2. Reading the same array by columns gives \(7\) columns with \(3\) dots each, so \(7 \times 3 = 21\). 3. Switching rows and columns does not change the total number of dots, which demonstrates the commutative property.

a) \(3 \times 7 = 21\) b) \(7 \times 3 = 21\) c) Both equations describe the same array, so they have the same product.

- Count the rows in each half. - Express “two equal amounts” with both addition and multiplication.

1. The whole array has \(6\) rows of \(9\) dots, so \(6 \times 9 = 54\). 2. Each half covers \(3\) rows of \(9\) dots, so each half contains \(3 \times 9 = 27\) dots. 3. Two equations that show the relationship are \(6 \times 9 = 3 \times 9 + 3 \times 9\) and \(6 \times 9 = 2 \times (3 \times 9)\).

For example, \(6 \times 9 = 3 \times 9 + 3 \times 9 = 27 + 27 = 54\) and \(6 \times 9 = 2 \times 27 = 54\).

- Use the picture to count the rows and the dots in each row. - Then use the same picture to count columns and the dots in each column. - Compare the two products.

1. The array has \(4\) rows of \(7\) dots, so \(4 \times 7 = 28\). 2. It also has \(7\) columns of \(4\) dots, so \(7 \times 4 = 28\). 3. Turning the page changes only the viewing direction; no dots are added or removed. The two equations illustrate the commutative property.

\(4 \times 7 = 7 \times 4 = 28\). Turning the page changes only how the same array is viewed.

- Use the image to count the rows in the top half and the number of dots in each row. - Compare the number of rows in one half with the number in the whole array. - Say how the two products are related.

1. The top half has \(3\) rows of \(8\) dots, so one half contains \(3 \times 8 = 24\) dots. 2. The whole array has \(6\) rows of \(8\) dots, so \(6 \times 8 = 48\). 3. The whole consists of two equal halves, so \(48 = 2 \times 24\). Thus, \(6 \times 8\) is twice \(3 \times 8\).

One half: \(3 \times 8 = 24\) Whole: \(6 \times 8 = 48\) The whole product is twice the product for one half.

- Think about counting the same array by rows and by columns. - Check whether changing the order of the two factors changes the total number of unit squares.

1. The rectangle can be viewed as \(4\) rows of \(9\) squares or \(9\) columns of \(4\) squares. 2. Both products equal \(36\): \(4 \times 9 = 36\) and \(9 \times 4 = 36\).

Yes, both expressions are correct. The area is \(36\) square units.

- Notice that the factors stay in the same order; only the parentheses move. - Evaluate each grouping separately before comparing the products. - Recall which multiplication property is about changing grouping rather than changing order.

1. Jalen's grouping gives \(3 \times (2 \times 5) = 3 \times 10 = 30\). 2. Priya's grouping gives \((3 \times 2) \times 5 = 6 \times 5 = 30\). 3. Both groupings are correct because the associative property of multiplication allows the factors to be regrouped without changing their order or product.

Yes. Both groupings give \(30\). The associative property of multiplication explains why the grouping can change without changing the product.

- Multiply the number of rows by the sum of each row. - Think about whether regrouping the same addends can change their total. - Each number appears once when you add by rows and once when you add by columns.

1. There are three row sums of \(15\), so the total is \(3 \times 15 = 45\). 2. Regrouping the same nine addends by columns does not change their sum. Therefore, the three column sums also have a total of \(45\).

a) \(45\) b) \(45\). The same nine numbers are being added, only grouped differently.

- Look for two addends that combine to make a multiple of \(100\). - The commutative property lets you change the order of addends. - The associative property lets you change how addends are grouped.

1. For a), use the commutative property to reorder the addends: \((245+355)+178=600+178=778\). 2. For b), use the associative property to regroup: \((64+136)+482=200+482=682\). 3. For c), use the commutative and associative properties: \((312+288)+399=600+399=999\).

a) \(778\); commutative property b) \(682\); associative property c) \(999\); commutative and associative properties

- Look for two factors that make a familiar product before multiplying all three. - You may change the grouping without changing the product. - When helpful, reorder factors so an easy pair is next to each other.

1. a) Group \(2\) and \(5\): \((2 \times 5) \times 7 = 10 \times 7 = 70\). 2. b) Group \(3\) and \(2\): \(4 \times (3 \times 2) = 4 \times 6 = 24\). 3. c) Reorder and group \(5\) and \(2\): \((5 \times 2) \times 7 = 10 \times 7 = 70\).

a) \((2 \times 5) \times 7 = 70\) b) \(4 \times (3 \times 2) = 24\) c) \((5 \times 2) \times 7 = 70\)

- Find pairs of addends that total \(100\). - Reorder the addends so each pair is together. - Add the three partial sums.

1. Pair addends that make \(100\): \(13 + 87 = 100\), \(26 + 74 = 100\), and \(39 + 61 = 100\). 2. Reorder and regroup the addends: \((13 + 87) + (26 + 74) + (39 + 61)\). 3. Add the partial sums: \(100 + 100 + 100 = 300\).

\(300\)

- Look for pairs that make multiples of \(100\). - Reorder the addends to place convenient pairs together. - Regroup the addends before calculating.

1. Reorder and regroup the addends: \((225+375)+(187+113)+100\). 2. Calculate the partial sums: \(600+300+100=1000\). 3. Reordering uses the commutative property, and regrouping uses the associative property.

\(1000\); commutative and associative properties

- Calculate the total using each grouping. - Think about whether addition changes when addends are reordered. - Think about whether addition changes when addends are regrouped.

1. Adding in the given order gives \(456 + 23 + 102 + 304 = 885\). 2. Lisa groups the two largest numbers and the two smallest numbers: \(456 + 304 = 760\) and \(23 + 102 = 125\). Then \(760 + 125 = 885\). 3. Both methods give the same result because the commutative and associative properties allow addends to be reordered and regrouped without changing the sum.

The sum is \(885\). Both methods give the same result because changing the order and grouping of addends does not change their sum.

- Keep the factors in the order \(4, 5, 10\). - Change only which pair is grouped first. - Compute both grouped expressions and compare the totals.

1. The first grouping is \((4 \times 5) \times 10 = 20 \times 10 = 200\). 2. Regrouping without changing factor order gives \(4 \times (5 \times 10) = 4 \times 50 = 200\). 3. Both groupings use the same factors in the same order and give \(200\). This shows the associative property.

a) \((4 \times 5) \times 10 = 200\) b) \(4 \times (5 \times 10) = 200\) c) Both groupings give \(200\) parking spaces.

- Keep the factors in the same left-to-right order. - Move the parentheses so the last two factors are multiplied first. - Your explanation should say what changes and what stays unchanged.

1. \((3 \times 2) \times 3 = 6 \times 3 = 18\). Regrouping gives \(3 \times (2 \times 3) = 3 \times 6 = 18\). 2. \((4 \times 2) \times 3 = 8 \times 3 = 24\). Regrouping gives \(4 \times (2 \times 3) = 4 \times 6 = 24\). 3. \((5 \times 2) \times 3 = 10 \times 3 = 30\). Regrouping gives \(5 \times (2 \times 3) = 5 \times 6 = 30\). 4. Only the grouping changes. The factors stay in the same order, so the associative property says the product stays the same.

a) \((3 \times 2) \times 3 = 3 \times (2 \times 3) = 18\) b) \((4 \times 2) \times 3 = 4 \times (2 \times 3) = 24\) c) \((5 \times 2) \times 3 = 5 \times (2 \times 3) = 30\) Only the grouping changes. The factors stay in the same order, and the product stays the same.

- Compare what changes: the order of factors or only the grouping symbols. - Recall which property is about order and which is about grouping. - For the final example, keep the grouping fixed and switch the order of two factors.

1. The equation is correct because regrouping the same factors does not change the product. 2. The property name is incorrect. Changing the grouping is the associative property, not the commutative property. 3. Using the same three factors, one equation that does show the commutative property is \((2 \times 3) \times 5 = (3 \times 2) \times 5\). The grouping stays the same while the order of \(2\) and \(3\) changes.

The equation is correct because only the grouping changes, so both sides have the same product. The property name is wrong: changing the grouping shows the associative property. One commutative example is \((2 \times 3) \times 5 = (3 \times 2) \times 5\), where the order of \(2\) and \(3\) changes.