- 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) \(4 \times 7 = 7 \times 4\)

- What does one row contain? How many rows can you see? - Now turn your attention to the columns. Did any dot get added or removed?

1. The array has \(4\) rows of \(6\) dots: \(4\times6=24\). 2. It also has \(6\) columns of \(4\) dots: \(6\times4=24\). 3. Both ways count exactly the same dots, so the product does not change.

Rows: \(4\times6=24\). Columns: \(6\times4=24\). Both count the same \(24\) dots.

- 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 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

- 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.

- 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.

- 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.

- Which two factors are together inside the parentheses each time? - Compare the two totals after doing the first multiplication in each expression.

1. First multiply \(2\times5=10\), then \(10\times3=30\). 2. In the other expression, multiply \(5\times3=15\), then \(2\times15=30\). 3. The same three factors are multiplied. Regrouping them changes the first calculation but not the total.

\((2\times5)\times3=10\times3=30\). \(2\times(5\times3)=2\times15=30\). Both give \(30\) because the same factors are multiplied in a different grouping.

- What could you count as a group: pencils in one packet, or packets in all the boxes? - Can both methods count every pencil without counting any pencil twice?

1. Counting by boxes gives \(2\times(3\times4)=2\times12=24\). 2. Counting all packets first gives \((2\times3)\times4=6\times4=24\). 3. Both methods count the same pencils.

Box first: \(2\times(3\times4)=2\times12=24\) pencils, with \(12\) pencils per box. Packets first: \((2\times3)\times4=6\times4=24\) pencils, with \(6\) packets altogether. Both ways count the same pencils, so the total stays \(24\).

- Calculate inside each pair of parentheses first. - Compare the two subtraction results before stating the rule.

1. The multiplication expressions both equal \(24\). 2. \(20-(5-2)=20-3=17\). 3. \((20-5)-2=15-2=13\). 4. The subtraction results differ, so regrouping is not allowed in the same way.

No. \(20-(5-2)=17\), but \((20-5)-2=13\).

- Follow the branches from the numbers toward the final result. - Try writing each tree as an expression with parentheses. - Compare the products after working out both expressions.

1. Tree a) multiplies \(3\times4\) first: \(2\times(3\times4)=2\times12=24\). 2. Tree b) multiplies \(2\times3\) first: \((2\times3)\times4=6\times4=24\). 3. The factors stay in the same order; only the grouping changes. This is the associative property of multiplication.

a) \(3\) and \(4\) are multiplied first; the product is \(24\). b) \(2\) and \(3\) are multiplied first; the product is \(24\). c) Associative property of multiplication.

- 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) \((5 \times 2) \times 7 = 70\)

- Keep the factors in the order of levels, rows, and spaces per row. - In part b), change only which two factors are grouped first. - Check whether your two expressions count the same spaces.

1. Group the levels and the rows per level: \((4\times5)\times10=20\times10=200\). 2. Group the rows and spaces per row: \(4\times(5\times10)=4\times50=200\). 3. Both groupings count the same parking spaces. Regrouping factors without changing their order is the associative property of multiplication.

a) \((4\times5)\times10\) b) \(4\times(5\times10)\) c) Both products are \(200\). Regrouping the same factors does not change the product.

- 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.